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·EXERCISES IN
CHEMICAL CALCULATION
BY
H. F. COWARD, D.Sc.
CHIEF LECTURER IN CH&MISTRY, MUNICIPAL SCHOOL OF TECHNOLOGY, MANCHESTER
LEC1 URER IN THE VICTORIA UNIVERSITY OF MANCHESTER
AND
W. H. PERKINS, M.Sc.
ASSISTANT 1.ECTURER IN CHEMISTRY, UNIVERSITY OF LBEDS
SEVENTH IMPRESSION
*
Ver
14 26
POLAY
JAN
LONDON
EDWARD ARNOLD & CO.
41 & 43, MADDOX STREET, BOND STREET, W.
(All rights reserved}

·PRYTI.D IN GREAT BRITAIN BY WILLIAJI CLOWES AND SUNS, LIMITED, LONDUV AID PIG CLES.

·PREFACE In common with many other teachers of chemistry, the compilers of this voluine believe that the working of suitable numerical examples is the best aid to the understanding of the quantitative aspect of chemical phenomena. The ideal calculations are those made by the student from his own experimental data, but it is necessary for him to use the figures of other workers and the data of experiments which he himself cannot perform. For this reason the following set of numerical examples has been compiled. It is drawn from recent examination papers and from various original sources. A number of the examples available in examination papers have been rejected because they had no conceivable relation to possible experimental conditions, and were therefore misleading. In most of the chapters a concise explanatory statement precedes the questions for solution. It is not intended that this should take the place of wider reading and study, but rather that it should indicate the portions of the text-book relevant to the particular subject under discussion, and provide a brief summary of what should be found there. It is believed, however, that the matter is put in its correct order, that it is in accord with present-day kpowledge, and that certain of the beginner's difńculties are explained more clearly than is often the case. Dogmatic rules for obtaining an answer have, it is hoped been avoided.

·iv PREFACE Nowadays the teacher of chemistry finds it unnecessary, as a rule, to give preliminary lectures on elementary physics. At the same time, it is noteworthy that most students learn the application of Boyle's law and of Charles' law from their chemical training. Hence this book deals less with elementary physics than lias been the case in previous works of the same character, but contains some simple exercises on the two laws mentioned. It is assumed that the student has a working knowledge of the metric system, and it is desirable that he should understand the use of logarithms. It is also considered advisable to insist from the outset that careful attention be paid to the significance of the digits in experimental numbers and in results calculated from them, a subject which receives too little attention even in some original papers in the chemical and physical journals. Only in this way is it possible to avoid confusion in such a case as the choice between an approximate and an exact atomic weight for a particular purpose. In adhering to this principle, it has been necessary to make slight alterations in certain of the examples obtained from examination papers and other sources, e.g. the substitution of " 1.000 gram" or "1.0000 gram," as the case may seem to deinand, for the expression one gram.” Where any change in the statement of a copied question has been introduced, the word "adapted” is added. This is done in order to avoid omission of several slightly faulty questions which, with some alteration, become valuable. As regards the standard of difficulty, most of the chapters contain some questions suitable for beginners, and others for advanced students. The beginner should find little difficulty in choosing suitable ones for himself from the neighbourhood of those taken from Intermediate Science examination papers, ·PREFACE The thanks of the compilers are due to the authorities of the Universities of Manchester, Leeds, and London, of the Institute of Chemistry, and to the Controller of H.M. Stationery Office, for permission to reprint calculations from the examination papers under their control. Their thanks are also due to the Controller of H.M. Stationery Office, and to Messrs. Macmillan & Co., Ltd. for permission to reproduce the four-figure logarithm table at the end of the book; and, finally, to Mr. F. W. Atack, B.Sc. Tech., for checking many of the answers. H. F. C. W. H. P. May, 1912.

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·CUNTENTS PAOB OBATTER 1. INTRODUCTORY DATA. CALCULATIONS FROM EXPERIMENTAL 1 5 . II. THE LAWS OF CHEMICAL COMBINATION BY WEIGHT- CHEMICAL EQUIVALENTS III. THE RELATIONS BETWEEN PRESSURE, TEMPERATURE, AND VOLUME IN GASES-PARTIAL PRESSURES AND VAPOUR PRESSURES 15 . TIIN IV. DANSITY -- Avog.URO'S HYPOTHESIS AYD USE OF GASEOUS DENSITIES FOR MOLECULIR WEIGIIT DETERMINATIONS 23 32 . . . 43 V. ATOMIC WEquiTS. VL DEDUCTION OF PERCESTIGE COMPOSITION FROM FORMULA AND FORMULE FROM PERCENTAGE COMPOSITIONS VII. CALCULATION OF REACTING QUANTITIES FROM EQUA- TIONS. VIII. REACTING VOLUMES or GASES-GAS ANALYSIS IX. VOLUMETRIC ANALYSIS (SOLUTIONS) ORGANIC ANALYSIS 54 . 70 82 89 XI. THERSIOCHEMISTRY 99 . . XII. SOBE PRINCIPLES OF PUYSICAL CHEMISTRY 112 ANSWERS 131 SOME CoxVERSION NUMBERS 115 . 146 VAPOUR PRESSURE OF WATER DENBITY OF WATER 146 . INTERNATIONAL ATOMIO Welguts 117 . TABLE or LOJARITIMS 140 LNDEX 150

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·EXERCISES IN CHEMICAL CALCULATION CHAPTER I INTRODUCTORY-CALCULATIONS FROIL EXPERIMENTAL DATA ONE of the principal difficulties which confront the student in making numerical calculations from experi. mental data is that of deciding the number of figures he shall retain in his fipal result. If, for instance, the fraction is to be expressed in decimals, the ordinary arithmetical process may be carried out until the figure 1.571428 is obtained. As the quotient in dividing exactly 11 by exactly 7, this is a perfectly correct solution But in scientific measurements there is no such thing as exactly 11, because every measurement is subject to a certain error of observation. It is advantageous for such measurements to be expressed in figures which show not only the numerical result obtained, but also the accuracy with which the measurement has been made. When the weight of a crucible is said to be 11 grams it is usual to assume that the weight is known to the nearest gram, i.c. it is more than 10:5 and less than 11.5. Similarly, if the weight is given as 11:0 grams it is known to be more than 10 95 and less than 11:05, and so on. it is recognized that the number 38342 is subject to a possible error of + 0.00005 or about 1 part in 80,000. If this be borne in mind it will be quite clear that in calculations from such numbers all the possible errors Ir this way B ·2 EXERCISES IN CHEMICAL CALCULATION COLLAP. in the data will be transferred to the result of the calculation, Consequently the result will be subject to at least the same uncertainty as the least accurate of the data, and the final figure must be made to express that fact. It is therefore obvious that in adding together 10.7251 1.9 2:35 SLI 14.9751 the last three figures of the sum have no significance, for the number 1:9 has a possible error of + 0.05, and the may have at least this error. The result will therefore be correctly expressed by 15:0, though if the number is to be used in further calculation it is advisablo to retain one more figure and write 14.98. (In choosing the above figures a process of rounding off has been used in which it will be seen that the last figure retained has been increased by unity because the portion discarded was equal to more than half such a unit. If the sun had been 14:9215, the figures would have been 14.9 and 14.92 respectively.) The abovo arguments may now be illustrated by further examples, 760 12.7 = 747 0:031 • 17.624 = 0.0018 441.46 -0.921 479 0.012786 x 12.2 = 0.156 In the first case the result cannot be given as 7473 because the original 760 is presumably only correct to the nearest unit. In the other cases the calculation is only carried so far as to produce a figure in which the accuracy is of the same order as in the least accurate of the data. It will further be seen that recurring decimals have ·1.) CALCULATIONS FROM EXPERIMENTAL DATA 3 no practical significance since they represent quantities in which the figures after the decimal point may be continued indefinitely, i.e. in which the possible error is infinitely small. A difficulty which may sometimes arise is illustrated in multiplying 1728 by 1.2. The product, 20736 obviously contains three meaningless figures. It would not be much more satisfactory to express the result as 2100, i.e. 2100 + 0.5. A convenient method in such cases is to express the above result, for instance, as 2:1 x 103. It will be clear froin the results obtained above that it is not necessary to carry out the whole arithmetical process with all the figures available, since the same solution can be obtained by dispensing with unnecessary digits in the more accurate numbers, e.g. in 0.031 17.624, it is quite sufficient to use 0.031 – 17.6. Therefore, as a general rule, when a number of figures are to be subjected to multiplication or division it is advisable to (I.) Inspect the numbers given, and decide which one shows the smallest degree of accuracy. (II.) Round off all the other numbers so that they contain at most only one more significant figure than tle least accurate number. For example 0.0060 X 760.0 x 292-43 33.023 x 7633 x 273-0 x 0.0000899 In this caso the number 0 0060 is much less accurate than any of the others, and since this contains two significant figures the others may be rounded off to contain three, while the final result will only be correct to 'one part in 120. The expression is simplified to- 0.0060 X 760 X 292 33.0 X 763 x 273 x 0.0000899 ·4 EXERCISES IN CHEMICAL CALCULATION CONAP. I. the result being 2:15. This figure, however, is subject to an error greater than 0-005, which is, in fact, nearly 0.02. So the result might be given as 2:15 + 0.02. When the numbers for calculation have been suitably rounded off, the calculation itself should be carried out by abbreviated methods of multiplication and division, or better still by the use of logarithms. In the latter case it is necessary to use logarithm tables calculated to a suitable number of figures. Four-figure tables will suffice for most chemical calculations, although in special cases five-figure or even seven-figure tables may be necessary. The above arguments are all concerned with the accuracy of observations, e.g. the accuracy of a weighing, of the reading of a volume from the scale on a burette, or of a thermometer reading. This is quite distinct from error in the experimental processes connected with the observation. If, for instance, the amount of sodium chloride in a solution has been estimated by evaporating the water, the errors of observation (weighing) may have been only, say, 0.02 per cent.; the result is correctly calculated when carried out to this degree of accuracy (0-02 per cent.). The actual process, which is supposed to be the complete separation of water from sodium chloride, may have been imperfectly carried out, i.e. some water may have been retained in the residue, or some sodium chloride may have been lost. It would be mere coincidence that the errors due to these two causes shoulů balance each other, so that the process is subject to experimental errors which are quite distinct from the errors due to lack of sensitiveness of the measuring instrument (the balance). What concerns us in the following pages is that the. result, without containing any superfluous figures, shall be expressed as accurately as the data permit.

·7426 HAN WA CHAPTER II THE LAWS OF CHEMICAL COMBINATION BY WEIGUT -CHEMICAL EQUIVALENTS To understand the laws of chemical combination it is necessary to be able to deduce them from experimental data. Since it is on these laws that the whole quantitative study of chemistry depends, it is not surprising that their exactuess has been made the subject of searching investigation, Law of the Conservation of Mass.-1.000 gram of oxygen will unite with 0.126 gram of hydrogen to form water weighing 1.000 + 0.126 1.126 gram. Many such experiments have been performed with various substances, with the highest degree of accuracy possible to modern science. The result is the establishment of the Law of the Conservation of Mass, which states that in any chemical or physical change, the total mass of the system remains constant. (The mass of a body is measured by its weight.) Law of Fixed Ratios.—This law deals with the relation between the masses of substances taking part in a chemical action. Berzelius in considering the formation of lead sulphide froin its elements found that (I.) 25.000 grams of lead produced 28.855 grams of lead sulphide.

·6 EXERCISES IN CHEMICAL CALCULATION COHAP. (II.) 15.000 grams of lead produced 17.3125 grams of lead sulphide, i.e. one gram of lead produced (I.) = 1.1542 28.855 25.000 17.3125 15.000 gram of lead (11.) 1.1542 sulphide. Thus the combining quantities of lead and sulphur are in the constant ratio of 1:0 1542. The same conclusion is reached when other reactions are considered, and the more carefully the experiments have been performed, the more concordant have been the results. The constancy of ratio is not confined to the combination of two elements to form a compound, but is always found between the weights of any number of substances (elements or coinpounds) reacting in the same way. The general expression of these facts is the Law of Fixed Ratios, which has been stated as follows: Definite chemical compounds always contain their constituents in fixed and definite proportions, or- If several substances react together, then their masses as well as those of the new bodies formed always bear fixed proportions to each other. EXAMPLES.—(If logarithms are used for the examples in this section, seven-figure tables are necessary.) Show how the pumbers of questions 1 and 2 illustrato the exactuoss of the law of defuito proportions. 1. Berzelius synthesized silver sulphide. (Ostwald's Klas siker, 35, p. 89.) 2.605 grams silver gave 2.993 grams of sulphide 10.00 11.49 Stas did the same with greater precautions. (Brux, Acar, Sci. Bull., 1861, 255.)

·LAWS OF CHEMICAL COMBINATION 7 22 99 99 91.462 grams 2 . 31 " > 59-4225 grams of silver gavo 68.2482 grams of sulphido 104.139 119.6078 191.9094 220•4158 150.000 172.2765 249.076 286.061 2. Stas (luc. cit. 1861, pp. 237-248) prepared silver chloride from metallic silvor in a variety of difforent ways and obtained Weight of silver. Waight of silver chloride. I. 121.4993 grams II. 69.8674 92.8145 III. 101.519 134.861 IV. 108.549 144.207 V. 399.651 530 920 VI. 99.9925 132.8382 VII. 983140 130.602 3. Calculate tho percontage of oxygen in potassium chlorate from the following figures of Stas (loc. cit., pp. 320, 321):- Weight of chlorato. Weight of oxygen. I. 69.8730 grams 27.3636 grams II. 82:1260 32 1612 IIL 86-5010 33.8705 IV. 132-9230 52-0430 V. 127.2123 49.8102 VI. 59.727 23.3830 VII. 95.7975 37-5020 VIU 147.318 57.6840 [Note.--The weight of the chlorate is correct to the nearest half milligram; Stas' results, given in the answers, aro rounded off to correspond. The student may criticize Stas' last figaro in tho calculated percentago in one or two instances. 4. Berzelius found that 10.000 grams of silver produced 13 275 grams of silver chloride; 30.000 grams of lead produced 43.900 grams of lead sulphate; 10-000 grams of lend chloride on treatment with silver nitrate formed 10320 grams of silver chloride, which contained all the chlorine originally > 1 Ro ·8 EXERCISES IN CHEMICAL CALCULATION [CHAP. in the load chloride; and finally that 10.000 grams of lead chloride on treatment with sulphuric acid produced 10.920 grams of lead sulphate. What is tho composition of lead chloride deduced from these figures ? Law of Multiple Ratios. -Lead and oxygen unite to form several compounds, each being an example of the law of fixed ratios. Berzelius, on analyzing two oxides of lead, expressed their percentage compositions as follows: Load, Oxygen. A. 92.85 7.15 B. 86.51 13.19 The fact that the numbers are expressed in percentages obscures a simple relation which exists between the combining ratios. This becomes apparent when the figures are recast so as to have a constant quantity of one element in both analyses, e.g.-- 92.85 A contains = 13.0 of lead to 1:00 of oxygen by weight 7:15 86-51 B contains 6:41 13.49 The ratio 13:0:641 is 2:03:1, ie, the ratio of lead to oxygen is approximately twice as great in A aš in B. In all other cases, too, it is found that the quantities deduced in the above manner are connected by a simple numerical ratio, and the more rigorous the experimental conditions, the more conclusive are the results. Such results are all comprised in the Law of Multiple Ratios or Multiple Proportions, which may be stated as follows: If two substances, A and B, react in more than one ratio, the various masses of A which react with a fixed mass of B bear a simple ratio to each other. 5. Two oxides of nitrogen have the following compositions:-(i.) Nitrogen = 63.63 and oxygon = 36.36 per cont. (ü.) Nitrogon = 46.66 and oxygen = 53.33 por cent.

·11.) LAWS OF CBEMICAL COMBINATION 9 А ... Use these figures to illostrato the law of multiple proportions. (B. E. Stage I. Inorg., 1909) 6. It was found by analysis that the compounds of two elemente, A aud B, lad the following porcentage composition:- А B 18t 75 25 2nd 80 20 3rd 85.7 14:3 Show how these results illustrate the law of multiple proportions. (1st Year Hons. Manc.) 7. The following are analytical figures obtained by Hilpert and Dieckmann for two arsenidos of iron (Ber. 1911, 2378):- B Iron 27.3 per cent. 42-8 per cent. Arsenic ... 72.82 57.79 Show how theso figuros illustrate the law of multiple ratios. 8. The following pairs of compounds were investigated by Berzolius, who gave the percontago compositions:-- Sulphur, Oxygen, A. 40.58 59-42 Oxides of sulphur B. 50:55 49.45 Iron. Oxygen. A. 69.34 30.66 Oxides of iron B. 77.22 22.78 Carbon. Oxygen, A. 28.44 71.56 Oxides of carbon B. 44.28 55.72 Show how tho law of multiplo proportions holds in each caso. 9. Berzelius represonted the composition of copper sulphate and of basic coppor sulphate as follows:-- Copper sulphato. Basio copper sulphate, Copper oxido ... 50-9 por cent. 80.0 per cont. Sulphurio anhydrido 49-1 20.0 Show how these numbers illustrate the law of multiple ratios. 4. (2.97.22 Dr. ... { . 9) ·10 EXERCISES IN CHEMICAL CALCULATION COHAP. 10. Discuss the following experiments of Berzelius in the light of the law of multiple proportions : A. The artificial sulphide of iron was prepared by direct union of the elements. 2.00 grains of this body were taken and so treated that the sulphur was first converted into sulphuric acid and then separated as barium sulphate, of wbich 5-38 grams were obtained. Berzolius had previously shown that barium gulphato contained 34:00 per cent. of sulphuric anhydride, and that sulphuric anhydride contained 40.58 per cent. of sulphur. The iron contained in the artificial sulphido was precipitated in tho form of iron oxide (containing 69:34 per cont, of iron). From the 2.00 grams of sulphido were obtained 1.82 grams of oxide. B. The natural sulphide of iron (pyrites), which contained # littlo silica as impurity, was ronsted in a current of air to form the oxiile. 10:00 grams of pyrites gavo 6.67 grams of oxidized residue, which was found to contain 0:07 gram of silica, the remainder being iron oxido (containiug 69:34 per cent. of iron). 11. Berzelius found that (u) 10.00 grams of copper when complotely converted to black oxido gavo 12.45 grams of the latter, and (b) 10.00 granis of black copper oxido and 10:00 grams of copper when heated in a closed vessel with hydrochlorio acid loft an updissolved residue of 1.97 grams of copper. Assuming that tho copper oxide and copper combincil to form another oxide which dissolvod in the hydrochloric acid, use thcsc figures to illustrate the law of multiple proportions. Law of Equivalents.--Berzelius gave the percentage compositions of three substances thus- Sulphuric anbydride. Ferric oxide. Ferrous sulphide. ... I. Sulphur Oxygen 40:59 53.42 Iron Oxygen 69.34 30.66 Iron Sulphur 63 0 37.0 ... From the second column it is seen that the amount of oxygen which will combine with 63 parts of iron ·11.) LAWS OF CHEMICAL COMBINATION 11 63.0 = 30.66 x 69.34 27-8, i.e. 37.0 parts of sulphur (see third column or 27.8 parts of oxygen combine with the same amount (63 parts) of iron. These numbers, 37:0 and 27:8, are in the ratio 1:0.75, which is almost exactly twice the combining ratio of sulphur and oxygen in the 40-58 first compound, sulphuric anhydride, for =1:1.46. 59:42 In all such cases it is found that the ratio of the quantities of two substances which react with each other is the same as (or a simple multiple of) the ratio of the quantities which react with a third substance. Another example. -8 grams of oxygen combine with 1 gram of hydrogen to forin water, or with 23 grams of sodium to form an oxide of sodium. It has also been showu that 1 gram of hydrogen unites with 23 grams of sodium to form sodium hydride. All such cases are comprised in the law of equivalents (sometimes called the Law of Combining Proportions, or the Law of Reciprocal Proportions), thus-- The masses of different substances A and B which react with the same mass of another substance Care either identical with, or bear a simple ratio to, the masses which will react with one another, or to the masses which will combine with a fixed mass of another substance D. 12. Illustrate the law of equivalents from the following analyses : Putassium chloride. Potassium iodide. Chloride of iodine. Potassium ... Ohlorino 52.5 47.5 Potassium Iodine 23.6 76.4 Jodine Chlorine 78.2 21.8 13. Illustrate the laws of multiple and reciprocal proportions ·12 EXERCISES IN CHEMICAL CALCULATION CONTAP. . from the ascertained composition of the following compounds :- Marsh gas. Ethylene. Carbonio oxide, Carbonio acid. Water. 0 = 75 0 = 85.71 C = 42.86 C= 27.27 II = 11.11 H = 25 H = 14.29 0 = 57.14 0= 2.73 0 = 88.89 (General Class, Mano.) 14. Illustrate the law of equivalents and deduce the composition of sulphuric anhydride from the following experimental data of Berzelius :- (i.) 25.000 grams of lead gave 28.855 grams of lead sulphide. (ü.) 25.000 grams of lead gavo 26.925 grams of lend oxido. (iii.) 30.000 grams of lead gavo 43.90 grams of load sulphate. (iv.) 30.000 grams of leal oxido gave 40-77 grams of lead sulpbate. (v.) In experiments on the oxidation of lead sulphide it was shown that this substance may be converted into sulphato without any lead or sulphur being liberated or converted into any other substance, i.e. the process consists viorely in the taking up of oxygen by tho sulphido to form the sulpliate. Lead sulphate is formed by the direct union of lead oxide and sulphuric anhydride. The Meaning of " Simple Ratio.”—In the statement of the above laws the expression "simple ratio" has been used without any qualifying term to indicate the degree of simplicity which is implied. In general it will be found that the ratio will be capable of expression by two integers rarely greater than 5. In the case. of many carbon compounds and certain mineral substances, however, ratios of greater complexity may be obtained. Equivalents.-A slightly restricted but at the same time more useful meaning has been given to the term "equivalent." If a fixed weight of a particular element be taken as standard quantity, the equivalent of any other element is the weight which is capable of combining ·LAWS OF CHEMICAL COMBINATION 13 with or replacing the fixed weight of the standard element. The standard usually chosen is exactly 8 units of oxygen; the equivalent of hydrogen is then 1:008, for in the commonest compound of hydrogen and oxygen, namely water, the two elements are united in the ratio of 1.008 of hydrogen to 8 of oxygen. Similarly 32:7 is the equivalent of zinc, since 32-7 parts by weight of zinc will displace 1:008 part by weight of hydrogen from sulphuric acid. The law of multiple proportions, however, indicates the limitations of the use of equivalents. 1.008 part by weight of hydrogen is equivalent to 8 parts by weight of oxygen in the compound water ; but in hydrogen peroxide it is equivalent to 16 parts by weight of oxygen. An element or group of elements may thus have several equivalents, but they will always be in the ratio of simple whole numbers. Therefore in working the following examples it must be remembered that the equivalent weights refer to the elements as present in the particular compound under consideration, Equivalent weights of the elements, although thus restricted in use, are valuable because (i.) they express facts which are independent of any theory, and hence will not be abandoned on account of any new chemical theory that may be introduced in the future ; (ii.) they are at present by far the most exact numbers from which the relative weights of the atoms are deduced with the aid of the atomic and molecular theories. 16. 5.00 grams of motallic magnesium when dissolved in hydrochloric acid evolved 0-417 gram of hydrogen. What is the equivalent of magnesium ? (General Class, Manc.) 16. Calculato the equivalent of silver from the following data. 0.501 gram of silver dissolved in nitric acid gave ou Addition of hydrochloric acid 0-6655 gram of silver chloride, [Eqnivalent of chlorine is 35-46.] (London Pro. Sci. Vet., 1905, aduptod.) - ·14 EXERCISES IN CHEMICAL CALCULATION (ORAP. II. 17 In a determination of the equivalent of a metal it was found that 1000 parts by weight of it combined with 32.867 parts by woight of chlorino. Find the equivalent of the metal. (Intor. B.Sc., Manc.) 18. 1.560 grams of liquid chloride of phosphorus wore decomposed by excess of wator. The resulting liquid, after adding silver nitrate, filtering, washing and drying the precipitated silver chloride, gave 4.883 grams of silver chlorido. From theso data calculate tho equivalent of phosphorus, assuming the equivalents of silver and chlorine to bo 107-9 and 35.45 respectively, (London Inter. Intl., 1909.) 19. On the assumption that tho equivalent of mercury is 99.25, calculato to two placos of decimals tho cquivalents of oxygon, hydrogen, copper, sulphur, and chlorine from the following data. Mercuric oxido contains 92.59 per cent. of mercury; cupric oxido contains 79.9 per cent. of copper; capric chloride contains 47-3 per cent, of copper; sulphurettol hydrogen contuins 94.07 per cent. of sulphur; hydrogen chloride contains 97.23 per cent. of chlorino. (London Inter. Extl., 1907.) 20. 0·8648 gram of the black oxirle of copper was so treated that there resulted 0.7781 gram of the rod oxiile. By a further treatment the whole of this red oxide was converted into puro copper, which weighed 0:6919 gram. What is the equivalent of copper in each compound?

·www * 1426 WOWA CHAPTER III THE RELATIONS BETWEEN PRESSURE, TEMPERATURE, AND VOLUME IN GASES--PARTIAL PRESSURES AND VAPOUR PRESSURES CIIEMICAL calculations are primarily concerned with the mass or quantity of matter undergoing change. The mass of a solid or of a liquid is, as a rule, quickly obtained by direct weighing, but the weighing of a gas is a more troublesome operation, and is less often carried out in practice. Once the mass of a given volume, i.e. the density, of a gas has been determined it is simpler, instcad of weighing, to calculate the mass of another quantity of the gas from a measurement of its volume. The volume of the gas, however, varies so considerably with alterations in pressure and temperature, that a volume measurement gives no idea of quantity unless the temperature and pressure are defined. The variation in volume for a given change of temperature or pressure is nearly the same for all gases, so that the following general laws are almost independent of the nature of tho gas, within wide limits of temperaturo and pressure. Boyle's Law (or Marriotte's Law).--When a fixed quantity of gas at a constant temperature is subjected to an increased pressure, the volume decreases proportionately, and vice versa. If the pressure is doubled, the volume is halved, and if the pressure is made one-tenth of ·V a B P.V. 16 EXERCISES IN CHEMICAL CALCULATION (chap, its original amount, the volume becomes ten times as great. In general, the volume of a gas varies inversely as the pressure, provided the temperature remains unchanged. Expressed in symbols 1 1 or V = K. P where K is a constant; or PV = K, i.e. the product of the pressure into the volume is a constant quantity for the same mass of gas at a fixed temperature, In most chemical calculations the volume is measured in cubic centimetres (c.c.) and the pressure in millimetres (mm.) of mercury. For example, if 200 c.c. of air are measured at 760 mm., what will be the volume at 1000 mm., at constant temperature ? K = P V2 760 x 200 = 1000 x the new volume in c.c. 200 X 760 .. new volume = 152 c.c. 1000 The student will probably find it best to argue somewhat as follows: 200 c.c. of air at 760 mm), = 200 X 760 с.c. at 1 mm. 200 X 760 C.C. at 1000 mm. 1000 = 152 c.c. The law of Boyle is only approximately true, accurate experiments showing deviations which are too large to be attributed to crrors in measurement. For the calculations which appear in this and the following chapters, Boyle's law may be applied unless the contrary is expressly stated, since it is sufficiently accurate at ordinary pressures for all except the most refined measurements. 1.-A. gas 'occupies 100 c.c. at a pressure of 340 mm. of mercury. . What will be the volumo at pressures of 1000 mm ·m.] CHARLES' LAW 17 and 100 nm. respectively, assuming the temperature to be constant ? 2. 300 c.c. of gas measured at 740 mm. pressure are passed into a vessel of 500 c.c. capacity, which has previously beon evacnated. What will be the pressure in the new vessel ? 3. An oxygen cylinder contains gas at a pressure of 10:3 atmosphere. It is connected with another cylinder of onethird the capacity where tho pressure is 1:10 atmosphere. What will be the final pressure in the two vessels when the connecting tap is opened ? 4. Three separate simples of gas each occupy 80.00 c.c. at the following pressures :-216-4, 87.2, 4:3 mm. What would be the volume of each at 760 mm. pressure ? 5. A quantity of gas .is collected ovor morcury in a tube whose open lower cord stands in a reservoir of mercury. The volume of the gas is 87.0 c.c. The surface of the mercury in the tube stands 18.6 om, higher than the surface of the mercury in the reservoir. Tho barometer stands at 75.4 cm. What would be the volumo if the mercury were brought to the same level inside and outside the tube ? 6. A volume of gas together with a crystal of rock salt measuros 150.0 c.c. at 760 mm. pressure. On raising the pressure to 1000 mm. the combined volume becomes 116-4 c.c. What is the volume of the crystal ? Charles' Law (or Gay-Lussac's Law) deals with the change in volume of a fixed quantity of a gas with change in temperature. A gas will expand at(or 0:00366) of its volume at 0° Centigrade for every degree Centigrade through which it is heated, when the pressure is kept constant. It must be noted that the fraction at: is dependent on the standard reference temperature of 0° C. Thus if Charles' law is applied to the convenient quantity of 273 C.C. of gas measured at 0° C., it shows that the volume at 1° C. is 273 + 1 = 274 c.c., at -2° C. it will be 273 - % = 271 c.c., while at 10° C. it will be 273 + 10 = 283 6.0.

·18 EXERCISES IN CHEMICAL CALCULATION (ORAP. 213V 10° C. 311 c.c. $40.c. and so on, the volume at tº being 273 + 0 0.c., where t may be positive or negative. If the original volume at 0° C. was V C.C., its volume at 1° C. would be V + 37: V -2° C. V + 10, V For example, if a gas had a volume of 300 c.c. at 0° C. its volume at 10° C. would be 300 x 373 273 +6 In general, V c.c. at 0° C. become V C.C. at 273 ť°. If V, is the volume at 0° C. and Vé the volume at t° C., then 273 tot 273 It will now be seen that since V 273 + t Ve 273 + + and 273 V. 273 273 + 1 .. Vi 273 +6 EXAMPLE.-400 c.c, at 20' become at 15° C. V. To > 400 x 393 393 c.c. Note.--If t + 273 is written T and t + 273 = T', then it is clear that T and T represent temperatures to which the volume is proportional. These temperatures, obtained by adding algebraically 273 to the Centigrade values, are called absolute temperatures. Absolute zero or – 273° C. is the temperature at which, if Charles' Law were absolutely accurate, all gases would have zero volume. This being impossible Charles' Law evidently does not apply at very low temperatures, so that the methods of correction studied above must not be expected to apply in extreine cases. It is advisable for the studeut to distinguish between ·TIL] CHARLES' LAW 19 such approximate laws as those of Charles and Boyle, and laws which are accurate so far as the most delicate measurements test them, such as the law of Conservation of Mass, and the laws of Combination by Weight. can 7. 203 c.c. of hydrogen wore moasurel at a temperaturo of 20° C. What will bo tho volumes at 100°, 180°, 360°, and 500° C., tho pressure being constant ? 8. One samplo of gas occupied 75 C.C. at 25° C., and another sample 100 c.c. at 250° C. Which would havo the greater volumo at 100° C., the pressuro being constant ? 9. A litro of gas is measured at 20° C. At what temperature will its volume bo 1750 c.c.? 10. Which will show the greator expansion, 100 c.c. of gns at 0° C. bented to 20° C., or 300 c.c. at 600° C. heated to 620° C.? Simultaneous Corrections for Temperature and Pressure --The two corrections for temperature and pressure can be applied as in the following example :- 1000 c.c. of hydrogen are measured at 100° C. and 700 nim. pressure. What would be the volume at 0° C. and 760 mm. ? 1000 c.c. at 100% and 700 mm. = 1000 x 768 at 100° C. and 760 mm. = 1000 x 788 x 373 at 0° C. and 760 mm. = 674 c.c. The nature of the relations between pressure, temperature, and volume for a fixed quantity of gas can be expressed by PV = constant X T, or, as it is generally written, PV = KT. This expression comprises the laws of Boyle and Gny-Lussac, as may be seen by noting that (i.) If T is constant, then RT is constant and PV constant. प्रालि IL 167-b ·20 EXERCISES IN CHEMICAL CALCULATION [CHAP. (ü.) If P is constant, then V = constant x T, or V is proportional to the absolute temperature. It may further be deduced that (ii.) If V is constant, P = constant x T, or P is proportional to the absolute temperature. These corrections are so often to be applied later that only a few exercises in them are given in this chapter. Normal Temperature and Pressure. For the comparison of different volumes of gas, measured at various temperatures and pressures, it is convenient to adopt as standards, 0º C. and 760 mm. These are called normal temperature and pressure (N.T.P.). 11. A gas measured 302.4 0.C. at 17.6° C. and 731 mm. pressure. What is its volume at N.T.P. ? 12. For a given gas, P = 3.2 atmospheres, V = 22-4 litros, T = 291° absolute. What is P whon V = 100 litres and the temperature is 0° C.? 13. Of two samples of hydrogen, one has a volume of 251.3 C.C. at 12° C. and 171 mm., and the other 220 c.o. at 17° and 197 mm. Which contains the greater amount of matter? Partial Pressures and Vapour Pressure.--If a mixture of gases has a total pressure P, then P is the sum of a series of "partial pressures," Pi + Pa + P3 + ... etc. (Dalton's law). Each of these values represents the pressure which would be exerted by a single constituent gas occupying the same volume at the same temperature. For example, if a volume of air has a pressure of 7600 mm., made up of the following partial pressures :- Nitrogen 587.0 mm. Oxygen 157-6 Water vapour... 8.0 Argon... 7:1 Carbon dioxide 0:3 ... ... . . > ·PARTIAL PRESSURES 21 then if all the oxygen, for instance, were removed without change of volume the total pressure would be 760:0 - 1576 = 602.4 mm. If the water vapour were removed instead of the oxygen the pressure would be 752.0 mm. It is then evident from Boyle's Law that if the pressure had been kept at 7600 mm, while the water was removed the volume would have decreased to 76% of its original amount. The application of this principle is important when a gas has been measured over water because in such a case the volume is that of a mixture of the gas and water vapour. The partial pressure of water vapour in a gas in contact with water is independent of the total pressure of the gas and varies only with the temperature. If a little water is introduced into the vacuum at the top of a barometer tube at 15° C. the height of the mercury column will be reduced by 12.7 mm., unless all the water has previously been converted into vapour. The addition of a small but sufficient quantity of water to a dry gas measured at, say, 760:0 inm., will cause a similar rise in the pressure of the gas to 772:7 mm, if the volume remains constant. Consequently a given volume of gas at 760.0 mm. and 15° contains as much of the dry gas as would exist in the same volume at a pressure of 7473 mm. (For the values of the vapour pressure of water at various temperatures see p. 146.) EXAMPLE.--79-3 c.c. of air are measured over water at 11° and 753 mm. What would be the volume dry at N.T.P.? The vapour pressure of water at 11° is 10 mm. The pressure of dry air at 11° is 753 - 10 = 743 mm. Therefore at N.T.P. the volume is 79.3 x 748 3/4 = 74.5 c.c. If the water contains much dissolved matter its vapour pressure will usually differ considerably from that of pure water, and must be known exactly before this correction can be made.

·22 EXERCISES IN CHEMICAL CALCULATION (CHLAP. III, Note.—It is a common practice to say that 100 volumes of air (free from carbon dioxide and water vapour) contain 79 volumes of nitrogen and argon, and 21 volumnes of oxygen. This is a conventional way of saying that 100 volumes of air are formed by mixing these gases in the proportions given, although each gas is spread uniformly through the mixture, so that the volume of each gas is 100 C.C. The proportions of each are better expressed as percentage pressures, or as partial pressures, of the whole. 14. A mixture of equal volumes of dry oxygon and nitrogon is collected over water at 10° C. and 730.2 m. Calculate thio partial pressures in the mixture. 15. 170 c.c. of gas are measured over water at 15° C and 720 mm. What will be the volume dry at N.T.P.? 16. 27.3 c.c. of oxygen wore monsurod in & dry oudiometer over mercury at 0° C. and 367 m. pressure. The gas was then transferred to a vessel standing over water, and the water was brought to the same level inside and outside the vessel. What volume would the gas occupy at 27° C. and 700 mm. ? 17. Assuming dry air to be a mixture of 20:9 por cent. oxygon and 79.1 per cent. nitrogen by volume, calculato the percentage composition by volume of air saturated with water vapour at 18° C. and 760 mm. 18. The following quantities of grs are containod over norcury in separato measuring vessels :-- 70'0 c.c. of hydrogen at 15° C. and 720 mm. 10'0 c.c. of oxygen at 10° C. and 700 mm. 14'0 c.c. of nitrogen at 20° C. and 760 inm. All are mixed in anothor vessel at 18° C. and the volume is found to be 940 C.C. What is the total pressure in the Vossel and the partial pressure of cach constituent ?

·CHAPTER IV DENSITY.-AVOGADRO'S HYPOTIESIS AND THE USE OF GASEOUS DENSITIES FOR MOLECULAR WEIGHT DETERMINATIONS The Densities of Solids and Liquids.- The mass of unit volume of a substance is called its density. In the metric system of units this is the number of grams in 1 cubic centimetre, and since 1 c.c. of water at 4° C. weighs 1 gram, water las at this temperature a density of unity. The volumes and densities of all substances are affected to a greater or less extent by changes of pressure and temperature, so that any statement of a density must comprise a statement of the conditions under which it was determined. A large change in pressure is necessary to produce an appreciable change in the density of a solid or liquid, so that the effect of pressure changes is usually negligible. It is otherwise with the changes brought about by change of temperature, so that in most calculations dealing with density (e.g. determination of refractivities) the temperature of the determination must be stated. Relative Densities.--The ratio Weight of 1 c.c. of substance at ° Weight of 1 c.c. of water at t, is termed the relative density of the substance at tº come pared with the density of water at t, and is often found ·24 EXERCISES IN CHEMICAL CALCULATION COHAP. more convenient than the absolute density as defined above. It is sometimes called the specific gravity of the substance under the stated conditions of temperature, and is written, sp. gr. t°/1, , De, or d tº/t,. For mercury DTS 13.559, i.e. 1 c.c. of mercury at 15° weighs 13.559 times as much as 1 c.c. of water at 4°, Now, since 1 c.c. of water at 15° weighs 0-99913 gram (p. 146), then for mercury 13.559 1152 = 13-572. 0-99913 15° 1. A calibration mark is to bo made on the neck of a flask to indicate a content of 250 c.o. The flask having been counterpoised on a balance, what weight of water at 15-5° C. must be added so that the surface of the water shall show where the calibration mark is to be made ? 2. A vessel for measuring an exact volume of a gas is filled with mercury, the whole being at 15° 0. The mercury is then run out and its wcight found to be 1347.82 grams. The density of mercury at 15° being 13.5590, what is the capacity of the vessel at 15°, exprossed in c.c. ? 3. The following are determinations of the density of acetic acid made by various observers :- Oudeinans d 15°/4° = 1.0553 Perkin = 1.0568 d 15°/15° = 1.05704 Rainsay d 20°/4° 1.0491 de Visser d 16.6°/4° = 1.05315 Jones d 16.5°/4° = 1:0534 A rocont determination (Bousfield and Lowry, J.C.S., 1911, 1439) gives d 18°/4° = 1.05148 and d 20°/4° = 1.04922. From the latter figures calculate the coefficient of expansion of acetic acid, and apply to the older figuros, constructing a tablo bringing all to 18°/4° so as to indicate the degreo of concordance of the whole series. [Use table of densities of wator on. p. 146.] d 15°/15° ...

·1v.] DENSITIES 25 4. Given that tho spocifio gravity :f alcohol at 15° is 0.7947 referred to water at 15°, and that the specifio gravity of & mixture containing 50 por cent. by weight of alcohol is 0.9188 at the same temperature, find to the nearost c.c. the contraction in volume which takes place when a litre of alcohol is added to an equal weight of water at 15°. Application of Densities in Weighing. It is a wellknown physical principle that when a substance is Teighed in a fluid (liquid or gas) the apparent weight is less than the true weight by the weight of the fluid displaced. The true weight of a body can thereforo lo determined directly only by weighing in vacuo, against weights which have been standardized in vacuo. It is customary, however, when the true weight is required, to obtain it by applying a correction for the buoyancy of air to the ordinary weighings. 5. If two weights, each 50.0000 grams, ono of platinum (density 21:5) and the other of brass (density 8:5) were placed upon the opposite pans of a balance, what would be the difforence betwcen thoir apparent weights in air (a) at 0° C., (6) at 20° C.? (The weight of a litro of air at O' is 1.29 gram, at 20° 1.20 gram.) 6. 131.851 is the apparent weight in air of a quantity of silver chlorido, the weights used being made of platinum and standardized in vacuo. What is the true woight of the silver chloride ? (The density of silver chlorido is 5.55, and of platinum 21.5. The weight of a litre of air may be taken as 1.20 gram.) The Densities of Gases.--The weight of 1 c.c. of hydrogen at N.T.P. is 0.0000899 gram, of oxygen 0:00143, Other gases likewise give inconveniently small numbers, so for this and other reasons the densities of gases are expressed in other units than those used for solids and liquids. Since hydrogen is the lightest of all known ·26 EXERCISES IN CHEMICAL CALCULATION (CNAP. gases, its density has been taken as the standard, i.e. 1, so that Any (or Day or dog) = 15.90 indicates that oxygen (02) has a density 15.90 times as great as the density of hydrogen.* Obviously what has been said about the necessity for paying attention to the effect of changes of temperature and pressure on the densities of solids and liquids applies with much greater force to gases, since gases are subject to much greater changes in voluine than are solids and liquids, under similar changes of tenperature and pressure. The density of a gas is therefore expressed as the ratio The weight of a given volume at normal tenporature and pressuro The weight of the samo volume of hydrogen at normal ternp. and press. The most accurate determinations of the weight of a litre of hydrogen at N.T.P. give 0.0899 gram, but the number 0:0900 is usually sufficiently accurate, the error thus introduced being only 1 part in 899, or about 0:1 per cent. This number 0 0900 is usually written 0:09, but it is not forgotten that its degree of accuracy is much greater than would appear from this method of express. ing it. 7. Calculate the density of helium, compared with lydrogen as unity, from the following data given by Ramsay, Collio, and Travers (J.C.S., 1895, 67, 694): Volume of bulb 33-023 33.023 33.023 c.c. Temporaturo 19.43° 17.1° 18.33° C. Pressure 763.2 763.2 615.8 mm. Weight of gas 0.0061 0.0060 8. A dry fask of capacity about one litre filled with air weighs 54:428 grams. Part of the air is displaced by carbon dioxide, and the weight is then 51.814 grams. On examination * The density of air was at one time taken as unity. Since air is 14:4 times as heavy as hydrogen at the same temperature and pressure, any density oxpressed with air as unity can be expressed with hydrogen as unity by multiplying it by 14.4, ... I.. .. 0:00-19 gram ...

·19.] VATOUR DENSITIES 27 it is found that 610 c.c. of carbon dioxide are present. If tho measurements aro made at 15° and 750 mm., and the weight of a litre of air at N.T.P. is 1.29 gram, calculate the weight of a litre of carbon dioxido at N.T.P. '9. At what temperature will carbon dioxide bave the same density as air has at 0° C., the pressure being the same in each caso? Tho density at 0°C. of carbon dioxido= 21.8, of air=14.4. The Vapour Densities of Solid and Liquid Substances.-- The densities of vapours are expressed in the same way as those of gases, i.e. with hydrogen as unity. The expression Weight of a given volume of vapour at N.T.P. Weight of the same volume of hydrogen at N.T.P. is in this case not a statement of possible fact, because the vapour cannot exist at N.T.P. It is possible, however, to calculate what the numerator would be if the vapour diil not condense and if it followed Boyle's and Charles' laws when brought to N.T.P. The relative densities of a large number of vapours have been determined in this way because the values obtained have considerable theoretical significance, as will be shown in the next section. In the experimental inethods used (sce text-book), the volume of the vapour of a given weight of substance is measured, or an equivalent amount of air is displaced by the vapour, cooled, and measured at ordinary temperatures over water. EXAMI'LE. —In a Victor Meyer's apparatus, 0:1104 gram of chloroform displaced 23-2 c.c. of air measured over water at 18° and 772 mm. prossure. Vapour tension of water at 18° = 15.4 mm. Volume of air displaced, measured dry at N.T.P., 273 772 - 15.4 = 23.2 x X C.C. 291 760 ·28 EXERCISES IN CHEMICAL CALCULATION COMAP. Weight of this volume of hydrogen 0-09 273 772 - 15-4 x 23.2 x Х 1000 291 760 gram Whence vapour density of substanco 0 1104 0:09 = 56.6 273 772 - 15.4 x 23.2 x X 1000 291 760 ܕܽ 10. A porcelain vessel of 100 c.c. capacity was filled with a vapour at 1092° C. and atmospheric pressure, and the weight of the vapoar was found to be 0.2268 gram. What is tho vapour density of the substanco compared with hydrogen ? (1st Year Hons., Manc.) 11. A liquid weighing 0-2457 gram displaced, in V. Meger's apparatus, 30:0 c.c. of air measured over water at 17° C. when the barometer stood at 7745 mm. What is the vapour density of the substance referred to hydrogen? (1st Year Hons., Manc.) 12. In a Victor Moyer determination 0.0926 gram of a liquid gave 28.9 c.c. of gas, collected over water and measured at 16° C. and 753.6 mm. pressure. Calculate the vapour density of the liquid. (Inter. B.Sc., Manc.) 13. Iu a Victor Meyer determination of the vapour density of ether, 0-1036 gram of other displaced 32.7 c.c. of air, measured over water at 12º and 749 mm. pressure. What is the vapour density of ether? 14. In a Victor Moyer determination of the vapour density, 0.1301 gram of a substance displaced 30-1 c.c. of air, measured over water at 17° and 759 mm. pressure. What is the vapour density of the substance ? Avogadro's Hypothesis and the Use of Gaseous Densities for Molecular Weight Determinations. The densities of gases have a special significance for the chemist owing to the simple volume relationships of gases in chemical reactions. The law of gaseous volumes (Gay-Lussac) states ·18.) AVOGADRO'S HYPOTHESIS 29 that "When gases react they do so in simple proportions by volume, and the volume of any gaseous product is also simply related to the volumes of the reacting gases, if all the volumes are measured under the same conditions of temperature and pressure.' This law is not exact, for the most carefully determined combining volume ratios of gases are not precisely expressible by whole numbers but by numbers which differ slightly from whole numbers. The differences though small are greater than the possible error of these measurements, so that the law although of great importance is not rigorously accurate like the laws of chemical combination by weight. Gay-Lussac's law suggests, however, some hypothesis as to the structure of a gas, and the uniformity of the volume-temperature and volumepressure relationships adds force to the suggestion, which is in fact met by the hypothesis of Avogadro. This states that equal volumes of all gases measured under the same conditions of temperature and pressure contain equal numbers of molecules.* It has received support from the kinetic theory, which is based ou the assumptions that matter is molecular in structure and that the molecules while perfectly elastic, obey the ordinary laws of kinetics of such masses of substances as can be experimentally investigated. The student who wishes to make the applications of the molecular and atomic theories in the following chapters must master the arguments of those theories either in the classroom or from his text-book. They are restated here in an epitomized form at the beginning of each chapter in order to indicate those portions of the text-book which are to be consulted for the description of experimental proofs and for fuller theoretical treatment. . The molecule is the smallest part of a substance capable of Independent existence.

·30 EXERCISES IN CIIEMICAL CALCULATION CONAP. From Avogadro's hypothesis it follows that a gas or vapour density determination leads to the molecular weight, thusmass of a given volume of gas The density mass of the same volume of hydrogen 'mass of m molecules of the gas mass of n molecules of hydrogen By Avogadro's hypothesis m = n, so the ratio becomes Mass of 1 molecule of gas Mass of 1 molecule of hydrogen The denominator of the fraction is equal to 2 (or more exactly 2016) since the molecule of hydrogen contains in all probability two atoms, and the weight of cach atom is taken as unity (or more exactly 1.008). Thus the relative density of a gas or vapour is equal to half of its molecular weight, and the determination of vapour densities is valuable not only for substances which are gaseous at ordinary temperatures but for all substances which can be vapourized at temperatures where measurements are possible. (or 2:016 1 15. A quantity of gas woighing 0.062 gram occupies 25 c.o. at 100' and 760 mm. What is its molecular weight? 16. An oxide of nitrogen was found to boil at 22° C. Ono gram of tho liquid gave at 27' a light orange-coloured gas occupying 266 c.c.; on raising the temperature to 127° tho gas becamo dark red and occupied a volume of 709 c.c. ; on raising the temperature to 527' the gas became colourless and occupied & volume of 2127 c.c., the pressuro being that of the atmosphere throughout the experiment. What aro the vapour densities at the different temperatures, and what inferences would you draw as to the changes of the oxide ? (1st Year Hons., Manc.) 17. 120 cc. of the vapour of water weigh 0.056 gram at ·1v.] AVOGADRO'S HYPOTHESIS 31 180° C. and 740 mm. Show by means of these data that the molecular weight of wator is approximately 18. (London Hons. Int)., 1909.) An important relationship between the volume occupied by a gas and its molecular weight is established as follows: Since 1 litre of hydrogen at N.T.P. weighs 0:09 gram 2016 then = 22:4 litres of hydrogen at N.T.P. weigh 0.09 2:016 grams and 224 litres at N.T.P. of a gas of density A weigh 2.016 XA grams i.e. M grams, where M is the molecular weight of the substance. Hence the molecular weight in grams, or the gram-molecular weight of any gas, occupies 22:4 litres at 0° C. and 760 mm. This relationship may be used in all cases unless a high degree of accuracy is required. Note that it depends on Avogulro's hypothesis, which, as already stated, is not quite exact. Hence it is open to just the same small errors as that hypothesis. The extent of the variation is shown by the following values of the volume occupied by the gram-molecular weights of several gases (calculated from exact atomic weights, molecular weights and densities) at N.T.P. :- Hydrogen 22:43 litres Carbon dioxide 22:26 litres Oxygen 22:39 Chlorine 22:05 Carbon monoxido 22:39 Methane 22:44 9) 18. Calculate the cquivalent of a motal of which 0:18 gram when dissolved in dilute sulphurio acid ovolves 166.7 c.c. of hvarogen at N.T.P. (Bd. of E.I., Stage 1., Inorg., 1907.)

·CHAPTER V ATOMIC WEIGHTS Deduction of Approximate Atomic Weights from Molecalar Weights.—The atom being the sinallest amount of an element ever present in a molecule of any of its compounds, the atomic weight is the smallest weight of an element ever present in the molecular weight of any of its compounds. Thus the determination of the atomic weight of an element depends prinarily upon & knowledge of the molecular weights of as many of its compounds as possible. If the weight of an atom of hydrogen be taken as unity a series of approximate molecular weights, expressed in terms of this unit, is obtained by doubling the vapour densities (see Chapter IV) of the volatile compounds of the element under consideration. For example, in the case of nitrogen :- 1 2 3 5 Gus or vapuur Molecular density (wben weight (den. luydrogen =1). alty x 2). Per cent nitrogen in the substauce. Weight of nitrogen lu molecular welgbt of substance, . . ... . Nitrogen Ammonia Hydrazoio acid Nitrous oxide Nitric oxide Cyanogen Ethyl nitrate Nitrobenzene 14 8.5 21.5 22 15 26 45-5 61.5 28 17 43 44 80 52 91 128 100 82-4 97.7 63-6 46.7 53.8 15.4 11.4 28 14 42 28 14 28 14 14 ...

·ATOMIC WEIGHTS 33 The first column shows the compound, the second its vapour density; the third is the molecular weight deduced from the second by multiplication by 2; the fourth is the percentage of nitrogen in the compound as determined by gravimetric analysis independently of any theory; the fifth is the weight of nitrogen in the molecular weight of the compound, determined from the preceding two columns. It is evident that 14 is not only the least weight of nitrogen present in the molecular weight of any compound, but also that the compounds containing more than this have exactly 2 x 14 or 3 x 14, i.e. 2 or 3 atoms respectively present in the molecule. The atomic weight of nitrogen is thus 14. In a similar way the atomic weights of oxygen and carbon are founi: Cias of vapor density. Molecular weight. Weight nf Oxygen in molecular weight of compound, Welglt of carbon in molecular Meight of compound. ... 30 18 48 16 16. 16 48 ... ... Nitrous oxila Nitric oxide... Water Ozono Cyanngen Carbon monoxide Carbon dioxido Carbonyl chlorido 22 15 9 24 26 14 22 495 28 16 32 16 24 12 12 12 99 From these and from a very large number of similar determinations it is deduced that the atomic weight of oxygen is approximately 16, and of carbon 12. If a symbol is used to represent an atom of an element, e.g. H for hydrogen, 0 for oxygen, N for nitrogen, C for carbon, then formulæ are deduced for the compounds thus- Nitrous oxide contains 28 parts or 2 atoms of nitrogen ·34. EXERCISES IN CHEMICAL CALCULATION COHAP. and 16 parts or 1 atom of oxygen. Its formula is therefore written N20. Nitric oxide is NO, water is H20, ozone is 0g, cyanogen is C,N2, carbon monoxide is CO, carbon dioxide is CO2, ammonia is NH3, and hydrazoic acid is HN3 Against the above as a general method of obtaining atomic weights, the following objections may be adyanced : (1) Avogadro's hypothesis is not exactly true; hence the molecular weights obtained by its use are only approximate. (2) The determination of vapour densities is easy when the object is an approximate value, but extraordinarily difficult when an exact value is required. (3) Few volatile compounds of the metals are known. For these reasons, such atomic weights as can be obtained are approximate; it is only quite recently that D. Berthelot, Guye, and otliers have eliminated the two sources of error (1) and (2) so as to obtain accurate atomic weights for a few elements. Nevertheless, the knowledge of the approximate value has been of service for many years as an aid to the choice of an exact atomic weight. Exact Atomic Weights.--Since the time of Berzelius chemical analysis has indicated with considerable accuracy the equivalent weights of elements, or the ratios by weight in which they combine (Chapter II.). It is usually from these that the exact atomic weights are obtained The accurate combining ratio of hydrogen to oxygen in water is 2.016: 16.00, i.e. their atomic weights are in the ratio 1.008: 16.00, since the formula of water is H,O (see above). It has recently been decided by the International Committee on Atomic Weights to fix 0 = 16 exactly as the standard. This means that in calculations with any ·ATOMIC WEIGHTS 35 pretence to accuracy it is essential to use the value 1.008 for the atomic weight of hydrogen. If the approximate number 1 is used, as was the case in the early portion of this chapter, it must be remembered that an error of nearly 1 per cent. is thereby introduced. From the equivalent of any other element M its atomic weight may be deduced as follows. The element may combine with oxygen to form some of the oxides M20 MO M Og MO, MOS MO, M20MO, The atomic weight, A, of the element combines with the following weights of oxygen respectively: 8 8 x 2 8x3 8x4 8 x 5 8 x 6 8 X-7 8x8 Hence the equivalent of the element, i.e. the amount which combines with 8 parts by weight of oxygen, is- A А A A л A А A 2 3 5 6 7 8 If these oxides are assumed to be formed from water by the displacement of the hydrogen atoms by atoms of M, then in the eight compounds one atom of metal has replaced the following number of atoms of hydrogen (cf. formulæ), 1, 2, 3, 4, 5, 6, 7, 8. This number is the valency of the element. (Sce text-book for fuller discussion of this all-important subject.) It coincides with AA the denominator in the line A, 9.1. etc. atomic weight Hence Equivalent valency i.e. Atomic weight valency x equivalent. Since the valency is a simple whole number (from 1 to 8), and the equivalent is accurately known, the atomic weight is chosen by trial as that number which is nearest the approximate one as determined above, when in the formula is substituted for valency the numbers 1, 2, 3 ... 8 in turn, ·36 EXERCISES IN CHEMICAL CALCULATION Contap. EXAMPLE.-A metallic chloride was found to contain 25.60 per cent. of chlorine. What would be the atomic weight of the metal on the supposition that it was (a) a diad, (6) a triad ? (Cl = 35:46.) (Inter. B.Sc., Manc.) “Diad" and "triad” are terms sometimes used for divalent and trivalent elements respectively. 25.60 of chlorine unite with 100 - 25.60 of metal re. 74.40 of metal 35-46 .: 74.40 x of metal unite with 35-46 of chlorine 25.60 i.e. 103.06 of metal unite with 35:46 of chlorine, or replace 1-008 of hydrogen in hydrochloric acid. Hence since atomic weight = equivalent X valency then the atomic weight = (a) 103:06 x 2 = 2061 or () 103:06 x 3 = 309 2 Approximate Atomic Weights obtained by the Use of Dulong and Petit's Generalization. Only a few elements have a sufficient number of volatile compounds for the determination of their approximate atomic weights from vapour density observations. For most elements an independent rule or generalization enunciated by Dulong and Petit may be applied. This states that the product of the atomic weight and the specific heat (measured at ordinary temperatures) of a solid or liquid element is a constant, approximately 6:4. This generalization holds particularly for those elements which form the fewest volatile compounds, and whose atomic weights are thus least easy to determine by the previous method. The notable exceptions to Dulong and Petit's generalization are beryllium (glucinum), boron, carbon, and silicon, whose atomic weights have been determined by the previous method. It is especially important to note that the value of the atomic weight obtained from the equation- Atomic weight x specific heat = 6:4 C 1 ·ATOMIC WEIGHTS 37 X 8 is only very approximate, so that the exact number must be obtained from the equation- Atomic weight = valency x equivalent EXAMPLE,A metallic oxide was found to contain 52.94 per cent of the metal and 47:06 of oxygen. A rapid determination of the specific heat of the metal gavo 0.25. What is the exact atomic weight of the metal (0 = 16), and what is the formula of the oxide ? 52.94 of inctal combine with 47.06 of oxygen 52.94 8 47.06 52.91 i.e. equivalent of metal X 8 17.00 = 9:00 Also, froiu Dulong aud Petit's gencralization- Atomic weight x 0.25 = 6:4 approximately or, Atomic weight = 25:6 approximately. Now, atomic weight = valency x 9:00, and adopting values for the valency, in turn, of 1, 2, 3, and so on- Atomic weight either = 1 x 9.00 9.00 2 x 9:00 18:00 3 x 9:00 27.00 4 x 9:00 = 36.00; and so on. Since 27.00 is nearest the approximate number 23•6, the exact atomic weight is 27:00, the valency is 3, and the formula for the oxide is M,03. For atomic weiglits required in the following examples, see p. 147. 71. By tho action of hydrochloric acid on 0.1635 gram of a motal, 56 0.c. of hydrogon measured at N.T.P. were obtained. Calculate the atomic weight of the motal assuming the chloride to bave the formula MCly. (B.Sc., Manc.) 2. A metallic oxide is found to contain 21-33 per cent of or = or or = ·38 EXERCISES IN CHEMICAL CALOULATION (ORAP. oxygon, The specifio heat of the metal is found to be 0.056. What is the exact equivalent of the metal, and what is its probable atomic woight? (General Class, Mane.) 9. A metallic oxide was found to contain 48-0 per cent. of oxygen. What is the exact equivalent of the metal ? If the specific heat of the metal was found to be 0.123, what is tho probable atomic weight and valency of the motal, and what would be the probable nature of tho oxido? (Inter. B.Sc., Manc.) 4. An oxide of a metal contains 20 per cent. of oxygon, and forms salts with acids. What are the possiblo formulæ and molecular weights which could be assigned to the oxide ? If the specific heat of the metal was found to be 0.098, which of thogo molecular weights would you accopt? (Inter, B.Sc., Manc., 1906.) 6. The specific heat of a motul was found to be 0.237, and the chloride of this motal was found to contain 80.0 por cent. of chlorino. What is the atomio weight of the metal, and the formula of the chloride ? (Cl = 35.5.) (Goneral Class, Manc.) 1. 2:380 grams of tin gavo by treatment with concentrated nitric acid and subscquent ignition 3.020 grams of oxide. The specific heat of tin is 0.054. From these data calculate the equivalent and the atomic weight of this element. (Sch. of Technology, Manc., 1st Year.) V. On dissolving 0.444 gram of a metal in dilute sulphuric acid, 177 c.c. of hydrogen at 10° and 750 mm. were obtained. The specific beat of the metal was 0.107. Calculate its atomio weight. (London Iuter., Intl., 1907.) 8. One cubio centimetre of carbon disulphido (sp. gr. = 1.29) is transformed into vapour at the boiling point of water. What volume would it occupy in these circumstances, the pressuro being normal ? (Intor. B.Sc., Mano.) b. The following figures were obtained by Nilson and Pettersson in a determination of the vapour density of germapium chloride.

·ATOMIO WEIGHTS 39 Volume of gas corrected} 6-24 11.61 4.37 c.c. to N.T.P. Weight of gas ... 0.0601 0·1121 0.0421 grain Calculate the molocular weight of the chloride and compare it with that corresponding with the formula GeCl. 10. An element combines with fluorine to form a fluorido containing 83.8 per cent. of fluorine. The fluoride forms with hydrogen fluoride a monobasic acid. The specifio beat of the element at 50° C. was found to be 0.24, and at 250° the specific heat was found to be 0:37. What is the equivalent weight of the element and what is its probable atomic weight? (1st Yoar Hons., Manc.) In many cases the formula of a compound is established, and the exact atomic weight of an element in it is determined by observation of some chemical change which the compound undergoes. The atomic weights of the other elements present must be known. EXAMPLE.--Ammonium aluin (NH4)2Al2(SO4)4.24H20, on heating decomposes so that all the constituents except alumina are volatilized. Mallet (Phil. Trans., 1880) found that 8.2144 grams of the alum gave 0.9258 gram of alumina. The atomic weight of aluminium is deduced from these figures thus- (NH4)2Al2(SO4)4.241,0 produces Al2Og :: 852-75 + 2x of alum produce 48.00 + 2x of alumina, where x is the atomic weight of aluminiumn. 852-75 + 2x 8.2144 48.00 + 20 0.9258 whence X = 27.11. 11. On drying, 6.905 grams of crystallized borax left 3.652 grams of anhydrous borax; calculate from these data the atomic weight of boron assumiug the formula of crystallized borax to be Na B,0,,101,0, and that Na = 23.05, 0 = 16, I = 1.008. What will it be if H=1 be takon as the standard ? (London B.So. Pass, Extl., 1904.) - ·40 EXERCISES IN CHEMICAL CALCULATION CONAP. 12. When a quantity of potassium bromaurate was hoatod, 3.679 grams of gold and 2-219 grams of potassium bromide were obtained. Calculate the atomio weight of gold, the specific heat of the motal being 0.0324. (K = 39.10, Br = 79.92.) (B.Sc., Mano.) 43. Guye and Ter-Gazariau havo recently discovored that the most carefully rocrystallizod potassium chlorate contains at least 0·022 per cont. of potassium chloride. Taking account of this correction recalculato tho ntomic weight of silver from the following data of Stas. Percentage of oxygen in potassium chlorate = 39.154. Ratio of silver to potassium chlorido = 100 : 69.1143. (0= 16.) (B.Sc. lIons., Mauc.) 14. Harcourt and Baker (J.C.S., 1911, 99, 1312) determined tho atonic weight of tellurium by convorting it into the tetra-bromille (ToBr.). Calculato atomic woights from tlo following lata : Woight of Tellurium, 0.87822 0.59706 0:09189 0.02732 0-58307 Worl: vf Bromiuc combined 2-20103 1.49610 1.73472 1.37254 1.46162 - 16. According to Stas' doterminations - KCIO, contains 39-19 per cent. of oxygen, 100 grams Ag aro precipitated by 69·1 grams KCl, 100 grams Ag form 132.85 grams AgCl. From these data calculato tho atomic weight of chlorine. (1st Year LIons., Mauc.) 16. It has been found, when ono gram of hydrogen is burnt in excess of chlorino over water, that 35•288 grams of chlorine dissolvo in the wator as hydrochloric acid, and that 14 c.c. of oxygen aro liberated from tho aqueous vapour, What is the atomic weight of chlorino from this experiment? (1st Year Lons., Manc.) 17. According to Stas, when carbon monoxido is passed ·ATOMIC WEIGHTS 41 over hoated copper oxido, 2.750 grams of carbon dioxido aro formed for every gram of oxygen lost by the copper oxide. According to Leduc tho density of carbon monoxide is 13.92 compared with hydrogen. What values for the atomic weight of carbon would you calculato from these detorninations if H= 1.008? (1st Year Hons., Manc.) 18. The percentage composition of rubidium alum is A1,0.5, 9.81; SO3, 30-73; R1,0, 17.95; II,O, 41.51. The composition of another alum containing an cloment Mig M,03. 13.30; 80, 29-53; R6,0, 17.29; H,O, 39-88. Calculate tho atomic woight of tho cloment M, if Al = 27:1 and Rb = 85.5. (Lond. B.Sc., 1906.) 19. 1.0000 gram of calcito containing 1.00 per cent. of iron heatod in air yiolds 0:5620 of resicluc; dotermino the combining weight of calcium (Fe = 56, 0 = 12, 0 = 16.) [Assuno irou present as FOCO..] (B.Sc., Miquc.) 20. 1000 c.c. of mercury vapour at N.T.P. weigh 8.923 grains. [Assume that morcury vapour can exist at oº and 700 mm. without condonisation.] 118.394 grams of mercurio oxido yield when heated 109-63 grams of mercury. Assuming mercuric oxide is formed by the union of one atom of mercury with one atom of oxygen, what light do these data throw on the atomic and molecular weights of the metal? (Final, DInuc., 1911, alapted.) 21. If wo suppose that in Dumas' determination of the combiniug weights of oxygou and hydrogen, for every 18.016 frame of water forned, 0.001-1 gram of CO, was produced owing to tho prosence of marsh gas in the hydrogen ascd, what value would have been found for the atomic weight of hydrogen ? 22. In Dumas' exporimont on the combining weights of oxygon and hydrogen : if, for every 2.016 grams of hydrogen burnt by the copper oxide, 0.0010 gram of sulphur dioxido (carried ovor from the sulphuric acid drying tubos) had been docomposed in the heated bulb, what weight of water would

·42 EXERCISES IN CHEMICAL CALCULATION (cuap. v.
have beon collected, and if the solphur had all combined with
the copper what effect would the impurity have had on the
atomic weight determination ?
[The last two calculations have been adapted from
Mauchester examination questions, to suit the more modern
atomic weights.]

·CHAPTER VI PERCENTAGE COMPOSITIONS FROM FORMULA AND FORMULÆ FROM PERCENTAGE COMPOSITIONS Deduction of Percentage Compositions from Formulæ using Atomic Weights.-One of the consequences of the atomic theory is the representation of chemical compounds by means of formule composel of symbols of the elements, A knowledge of atomic weights enables the proportions (by weight) of the elements to be deduced from these formula. Thus H20, the formula for water, represents a compound containing 2.016 parts of hydrogen to 16.00 parts of oxygen by weight, or 88.8 per cent. of oxygen and 11.2 per cent. of hydrogen. For convenience the proportions are generally expressed iu percentages, which are calculated by simple proportion, The composition of potassium ferricyaniile, [KgFe(CN).), for instance, is deduced as follows: Weight of molecule = sum of weights of the atoms in it (3 x 39) +56+(6 x 12) + (6 x 14) = 329 These 329 parts of KgFe(CN), contain 3 x 39 parts of potassium, so that 3 x 39 Percentage of potassium - x 100 = 35.6 329 ·44 EXERCISES IN CHEMICAL CALCULATION [CHAP. Similarly- 56 Percentage of iron = X 100 17.0 329 6 x 12 Percentage of carbon = X 100 = 21.9 329 6 x 14 Percentage of nitrogen x 100 = 25.5 329 It has been shown in Chapter II., Example 9, that Berzelius represented copper sulphate as a compound of copper oxide and sulphuric anhydride. This method of resolving the formulæ of complex compounds into simpler molecules is often used to indicate the chemical character more clearly, e.g.-- Red lead, PW30s, is 2P0, PL02. Potassium hydrogen sulphate, KIISOs, is K,SO4.H_SO. [i.c. 2KHSO4]. Malachite, Cu,H,C'05, is Cu(Oll).CnC03, or 2010. CO2.H20. Mica, KA1,H.Sig012 (KA1311. (SiO4)3], is K20,3A1203. 211,0.6SiO2. Boracite, My7C1,B16030, is Mg(1,2 M43BO15, or MgCla. 6Mg0.8B208. In each of these cases there is a choice of the terms in which the percentage composition may be expressed. It is, of course, also possible to represent formulic, and to express percentage compositions in terms of the radicles present. Thus malachite (above) may be written Cu(OH)2CO3, and its composition may be expressed as percentages of Cu, OH, and COg. EXAMPLES. Find the percentage elementary compositions, of two or threo of the following, as accurately as possiblo, using International Atomic Weights (p. 147). 1. Potassium carbonato, K, 03.

·VI.] PERCENTAGE COMPOSITIONS 45 2. IIydrogen peroxide, FIO,. 3. Forric chloride, FeCl, or Fc,C1g. 4. Nitrosyl chloride, NOCI. 5. Cryolite, AlF3.3NaF or Nay AIF,. 6. Ammonium platinichlorido, (NLI,),PtClg. 7. Ethyl nitrito, C, H, NO. 8. Ethyl acetoacetate, CHZ.CO.CH,COOC, H., 9. Sodium benzone sulphonate, CH,SO, Na. 10. Express the percentage composition of malachite in torms of CO, CO2, and 11,0. 11. Express the percentage composition of kaulin, (H..11, Si.,09.11.,0), in terms of Al,O, SIO, HO (" constitutional"), 11,0 (“hydration"). 12. Express tho percentago composition of crystallized Bodium carbonato, Na CO2.10110, in terms of anhydrous sodium carbonato and water of crystallization. 13. Adio (J.C.S., 1900, 77, 1076) gave the composition of a potassium solium cobaltinitrito as follows: K = 172 Co = 13.1 Na 4.9 NO, = 61.2 IIO = 4.1 Comparo this with the numbers calculated from the formula K,NaCO(NO2)..11.0. 14. 0.0930 grams of a substance believed to have the formula C71120.N.,.HAUCI, was ignited, and loft a rosiiluo of 0·0272 grams of pure gold (Powor and Callan, J. C. S., 1911, 99, 2000). Show that these analytical figuros agree with the formnla assigned to the compound, 16. A substruco was found to contain 30:12 por cont. of barium. Which of the following two formuloo oxpresses the composition more nearly : (C 1.0,), BA or (C.11.0.),BA? (Smedley, J.C.S., 1911, 99, 1032.)

·46 EXERCISES IN CHEMICAL CALCULATION [CHAT. Deduction of Formulæ from Percentage Composition, using Atomic Weights.--The method of constructing a formula to correspond with a given percentage composition may be deduced by considering a definite case. D. Berthelot obtained the following figures on analyzing a compound containing mercury, nitrogen, and iodine Mercury = 73.77 Nitrogen = 2:53 Iodine = 23.56 These quantities may each be resolved into two factors, one being the atomic weight of the clement under consideration, thus- [The fractional numbers Mercury =7377=0:369 x 200 are found by dividing Nitrogen = 2:53 = 0.181 x 14 the percentage of each Iodine = 23.56= 0.185 x 127 element by its atomic weight.] The numbers 0.369, 0·181, and 0.185, are approximately in the ratio of the numbers of atoms present; they must be transformed into simple whole numbers, since fractions of atoms are never found in chemical compounds. They are clearly very nearly as 2:1:1, and the simplest, or empirical, formula of the compound is Hy2NI.* 100 po * A more precise proof runs as follows: If the elements A, B, O, form a compound A,B,Cr, and if the atomic weights of A, B, and o are a, b, and c, then the total weight of the molecule of the compound is pa + ab + rc. The percentage of A pa + ab + TC The percentage of B pa + ab + To 100 TC The percentage of C pa + 2b + rc The problem is, to determine p, , and , given the percentages of the clements, and a, b, and Co 100 qb ·VI.) PERCENTAGE COMPOSITIONS 47 It is not possible, of course, to decide without further information between this formula and some simple multiple of it, e.g. Hg.N,1, or HgNg1g. If the analysis is expressed in percentages of simpler compounds the process is similare.g. SiO2 = 47.0 = 0·78 X 60 (SiO2 = 28 + 32 = 60) Mg0= 31:3 = 0.78 x 40 (Mgo = 24 + 16 = 40) H20 = 21.2 = 1.18 x 18 (H2O = 2 + 16 = 18) 99.5 per cent. 0:78:0.78:1:18 = 2:2:3 very nearly, and the for mula is 2Mg0.2Si0,3H20, or 2MgSiO3.31,0. Since the usual analytical data for determining the composition of a compound are less accurate than the data from which the atomic weights are obtained, the proportions deduced are rarely quite exact. For this reason approximate atomic weights may be used in deducing formulac from analytical datil. The procedure to be followed in determining the formula of a body from its percentage composition is therefore (1) Divide each percentage by the corresponding atomic (or group) weight to obtain the ratio of the number of atoms (or groups) present. 100 pa rc Now per cent. of A: per cent. of B: per cent. of C 10) gb 100 TC pa + b + ro'pa + b + ro'pa + ab + ro pa qნ per cont. of A. per cent. of B por cent. of O Whence P:9:7 b Hence the proportion of the numbers of atoms presout (P:9:7) is obtained by dividing cach percentage by the corresponding atomio weight. Multiplication of the result by some constant number transforms these proportions into the simple whole numbers which show the empirical formula of the body, C ·48 EXERCISES IN CHEMICAL CALCULATION (OMAP. (2) Find the simplest whole numbers to express this ratio. If these are not obvious, divide each by the smallest and the ratio will then be apparent. > 16. An oxide of iron was found by analysis to have the following composition : Iron = 77.7, oxygen = 22.3. Find the simplest formula anil explain fully each step taken in the calculation. (Fe = 56.) (General Class, Manc.) 17. A solid hydrocarbon was found to contain 93.75 per cent. carbon and 6.25 per cent. hydrogen. Its vapour density was 6t. What formula would you resign to the substance? (Inter, B.Sc., Manc., 1910.) 18. A mineral contains-- Sodium 32.79 per cont. Aluminium 13:02 Fluorino 54:19 Calculato its simplest formul:, using the following Atomic weights: Na 23:05, Al = 27.1, F :: 19. Give the details of your calculations. (Bd. of Ed., Stago II., Iuorg., 1905.) 19. A doublo carbonate of nickel and potassium gave the following analytical data. (Reynolds, J.C.S., 1898, 204.) What is its formulu ? CO. 26.58 per cent. KO 28.82 per cout. Ni 22.64 11,0 21.86 20. Finil tho simplest fornuula of a substance with thc following porcentage composition : Copper 29:86 per cent. Nitrogen 13:27 per cent. Sulphur 15.17 Oxygen 37.91 Hydrogen 3.79 per cent. 21. A minoral is found to contain, in addition to oxygen, 5:1 per cent. of beryllium, 10:2 per cent. of aluminium, and 31•2 per cent. of silicon; dotermino its simplest formula,. and show how beryllium sulphate may be propared from it. (0 = 16, Bo = 9.1, Al = 27.1, Si = 28.4.) (2nd Year IIons., Manc., 1910.) P. ... ... 1) ·v1.) PERCENTAGE COMPOSITIONS 49 ... . 59 ... . 9) ... ... 22. The insolable Ca(NH)PO,H,O (stable at 100') on ignition gives cadmium pyrophosphate, Cd,P,Oz. A solution of a cadmium salt gave 0•4530 gram of the hyirated salt and 0.3712 gram of the pyro salt. How much cadmium was present ? (Miller and Pago, Chem. News, 1901, 313.) 23. An analysis of a mineral gave the following results :- Silicn 57:54 por cent. Limo 8:05 per cent. Alumina 16.28 Water 17.2:3 What is the simplest formula for a mineral of this composition ? (Si = 28, Al = 27, CA = 40, II = 1, 0 = 16.) (B.Sc., Manc., 111.) 24. The following is an analysis by Brauner (J.C.S., 1898, 958) of a doublo oxalato of thorinm and ammonium. What is its formula? Tho, 33.81 per cent. CO: (oxalic anhydride) 36-89 (NIT). 13:39 HO 10.11 25. Müller and Diefenthaler (Ber. 1910, 2322) give tho following analysis of a salt prepared by treating lead pitrato with potassium ferricyanide :- Pb = 52.86 Fe(CN)." = 27-07 NO, = 7.69 110 = 11.61 with the following confirmatory data :- Fc 7.27 C 9.23 What is tho formula of tho compound? Show that the confirmatory data ngreo with the formuln. 26. The following figuros aro taken from an analysis by Worner (J.O.S., 1888, 40+) of a basic doublo oxalate of potassium and chromium. Suggest a formula for tho substance. Potassium 21:00 per cent. Chromium 16.22 Oxalato radiclo (C,0.) 5+:39 27. Debus (J.C.S., 1888, 288) gave the following analysis > ·... ... ... ... 50 EXERCISES IN CHEMICAL CALCULATION [CHAP. of the hydrated potassiam salt of a thionic acid. What is the probable formula of the compound ? Potassium 21.64 per cont. Sulphur 43.97 Water pf bydration 7.81 28. Find the formula of tho hydrated mercuric potassium nitrite of which the following is a partial analysis. (Rây, J.C.S., 1907, 91, 2032.) Hg 35:19 por cent. K 20:24 per cent. N 12.00 per cent. 29. The percentage composition of chloride of silicon has been ascertained to be Si = 16:47 per cent. Cl = 83:53 per cent. Its vapour density is 85 (= 1). Assuming Cl = 35.5, calculate from these data the atomio poight of silicon. (Inter., Manc., 1910.) [N.B.-It is necessary to nssume that silicon chlorid, contains only one atom of silicon.] Analysis of Minerals.-In certain minerals one element may often replace another to any extent without altering the crystalline form of the mineral. These so-called 180- morphous relationships are expressed by writing the symbols of both elements and enclosing them in a square bracket, e.g. olivine is (Mg, Fe]z.SiO4, or 2[Mg, Fe]0.SiO2, indicating that the mineral is of the type Mg SiO with more or less of the Mg replaced by Fe. It is of course impossible to calculate a percentage composition for olivine unless the proportions of magnesium and iron are definitely stated. The deduction of formule for such minerals is carried out on the lines of the following determination for dolomite: Cao 31.37 = 0.560 X 56 MyO 21.23 0.525 x 40 CO, 47.67 = 1.083 X 44 Here the relations between the numbers are not simple, . ...

·VI.] PERCENTAGE COMPOSITIONS 51 ... ... but the Cao and the MgO are together equal to 1.085, which corresponds closely with the figure for CO2, so that the formula may be written, (CaO, MgO]C02, or (Ca, Mg]C0g. Another instance of isomorphous replacement is found in an analysis of soda felspar :- SiO 68:45 - 60 = 1.141 Al2O3 18.71 - 102 = 0.183 Fe2O3 0.27 - 160 = 0.0016 Cao 0-50 - 56 = 0.009 MgO 0:18 : 40 = 0:0045 K, 0:65 94 = 0·007 Nazo 11.24 62 = 0.181 In this case the relationships are not so exact and the analysis is not that of a pure substance. But arguing that the line and magnesia are combined with 0.0135 (i.e. 0009 + 0.0045) silica, tlien 0.188 moleculo of [NaK]30 and 0.185 of [Alle],03 are combined with 1.1275 of SiO2. These figures agree with the formula Nazo. Al2O3.6SiO2, or NaAlSig0s. 30. The following aro two analyses of spinel. What is its formula ? 60 79 A.403 isomorphous (07:25 5.26 Foo 21.74 MgO }isomorphous (27:01 12.84 31. Cabrcrite (Sachs, J.C.S., 1906, 90, 369) gave on analysis - A8,0 = 40-45 Nio = 26.97 Isomorphous Foo 1.10 MgO 6.16 Н,0 25-26 Total 99-94 Find a formula for this mineral. .. ..

·52 EXERCISES IN CHEMICAL CALCULATION (CHAT. 32. Natrolite (Harrington, J C.S., 1906, 90, 867), gave on analysis Sio, = = 47.09 A1,0,= 26-99 NA O = 16.16 K20 = 0.01 11,0 9.80 Total 100.35 Find a formula for this mineral. 33. Deduce the formula of tho mineral wolframite from the following analysis by Finlayson (J.C.S., 1910, 98,308). WO, ... 76.24 16:39 M40 FO Mno Cro 0:11 6-05 1:05 ... ... 34. Deduce the formula of the mineral datolito from the following analysis by Fromme (J.C.S., 1910, 98, 314) - Can 34.91 BO, 20.79 SiO, 37.6.7 AO, ... 1.18 ILO 5.90 It must be noted that only the salts of the so-called oxyacids can be resolved into two oxides for the purpose of expressing their composition. It is not possible to represent calcium chloride, for instance, as a compound containing calcium oxide together with an avhydride of hydrochloric acid. As the total amount of metal found in a mineral is usually represented in the percentage composition by the equivalent quantity of its oxide, and as the amount of a halogen is generally expressed as the percentage of the element, it is clear that the composition as put in this way will not be a true percentage composition since the sum will exceed 100 by the weight of oxygen wrongly attributed to the metal. The analysis of .

·v1.) PERCENTAGE COMPOSITIONS 53 ... apatite which has the formula Cas(PO4)3C1, or 9C90.- 3P,05.CaCl, might be expressed as- Cao 53.8 Corresponding to, though not P.O. 40.9 actually contained in, 100 Cl 6.8 parts of apatite. Total 101.5 If from this total is subtracted the amount of oxygen equivalent to the chlorine found (i.e. 1.5), which would be the amount wrongly associated with calcium, the total is exactly 100, and therefore analyses of minerals containing halogen are often corrected by subtracting from the sum of the percentages found, a quantity which is described as "O for F" or "O for Cl." 36. An analysis of sodalite was given as follows: silica = 37'1, alunina = 31.7, soda (N0,0) = 25-5, chlorine = 7-3 por cent. Deduce the formula of the mineral. (Si = 28, Al = 27, Na = 23, Cl = 35-5.) 36. A substance which contained zinc, phosphorus, oxygen, and chlorine, gave on analysis (Basset and Taylor, J.C.S., 1911, 1408) Zuo 21:2+ por conto 1,0 36.45 Ci 54.61 Total 112.30 Find its formula. 37. Doduco thu formula of the minoral cuspidine from the following analysis by Palache (J.C.S., 1910, 98, 219) :- SiO2 32.36 F 9.05 Mn() 0.71 Nazo 0.48 Cao 61.37 K,O 0.27 .. .00 .

·CHAPTER VII THE CALCULATION OF REACTING QUANTITIES FROM EQUATIONS The construction of chemical formula in accordance with the atomic and molecular hypotheses leads to the representation of chemical reactions by means of equations. For example- Zu + H2SO4 ZnSO4 + H, represents the reaction between zinc and sulphuric acid under certain conditions, The quantities of the four substances concerned in the reaction are calculated from the equation by the use of a table of atomic weights. It follows that 65-37 grams of zinc will react with 98.086 grams of sulphuric acid to produce 2.016 grams of hydrogen and 161:44 grams of zinc sulphate (anhydrous).* If it were necessary to determine the weight of zinc which could be dissolved in exactly 7 kilograms of 10.00 per cent. sulphuric acid the process would be- 7000 grams of 10 per cent. acid contain 700 grams hydrogen sulphate; 98.086 grams hydrogen sulphato require 65.37 grams zinc ; (froin the equation) 65.37 x 700 :: 700 grams hydrogen sulphate require 98086 = 466.5 grams zinc. The student should repeat this calculation, using • These figures should be chockad.

·OHAP. VII.] REACTING QUANTITIES 55 approximate atomic weights (H = 1, Zn = 65, S = 32) in order to see the nature of the error introduced by so doing. This type of calculation forms the basis of ordinary gravimetric analysis, as will be seen from the following example : A solution of sodium chloride gave with silver nitrate a precipitate of 0.2543 gram of silver chloride. How much sodium chloride was present in the solution ? NaCl + AgNO3 = NaNO3 + Agci 58.46 143.34 i.e. 143-34 grams of silver chloride result from 58:46 grams of sodium chloride; 58.46 x 0.2543 .: 0-2543 gram of silver chloride results froin 113:34 = 0.1037 gram of sodium chloride. For the following exercises, use the International Atomic Weights (p. 147) unless approximate values are given, or unless the data render the use of exact values of no gain to the accuracy of the calculation. 1. How much potassium chlorate must be heated to yield as much oxygen as would be obtainable from 500 grams of mercurio oxide ? (K = 39, CI = 35.5, O = 16, Hg = 200.) (Bd. of Ed., Stage I., Inorg., 1904.) 2. What products would be obtained, and what weight of ench, when 10 grams of (a) potassium chlorate, and of (6) white marble are separately heated to redness until no further change occurs ? (K = 39, C1 = 35.5, Ca = 40, C=12,0=16.) (Bd. of E., Stage I., Inorg., 1905.) 3. How mach anhydrous sodium carbonate should theo rotically be obtained from 100 lbs. of salt cake? 4. 100 0.c. of a solution of sulphuric acid was exactly noutralized by 32.0 grams of anhydrous sodium carbonate.

·56 EXERCISES IN CHEMICAL CALCULATION COHAP. 32, 0 Express the composition of the acid solution in torms of grams of sulphuric acid per litre. (Na = 23,0 = 12,0 = 16, S = 32, H = 1.) (Manchester Sch, of Tech., 1st Year, 1911.) 5. On passing puro carbon monoxide over honted copper oxide it was found that the loss in weight was 24:360 grams, and that the amount of carbon dioxide formod was 67.003 grams. From these data calculato the equivalont of carbon when 0 = 16. (Bd. of Ed., Stage II., Inorg., 1907.) 6. A solution of hydrobromic acid has a density of 1.055; 10 c.c. of this gave 1.878 gram of silver bromide; calculate the amount of II Br in 100 c.c., and in 100 grams of the solation. (II = 1, Br = 80, Ag = 108.) (London Intor. Pass, Extl., 1904.) 7. 0.200 gram of iron pyrites yields 0.700 gram of barium sulphate. What percentage of sulphur does the pyrites coutain? (Fe = 55.9, S = 10, Ba 137'4.) (Laboratory Class, Manc.) 8. A sample of conl contained 1.87 per cont. nitrogon. The ammonia evolvod on heating 100 tons of this coal gavo 2 tons 11 cwt. of ammonium sulphato. What percentage of tho nitrogen was ovolved from the coal as ammonia ? 9. 1.575 grain of a monobasic acid took 50 c.c. of a solution of NaOH containing 20 grams per litre for ncutralization. What is the molecular weight of the acid ? (N« = 23.) (Inter., Manc., 1911.) 10. A diad metal forms sevoral oxides; it was found that when 411 grams of an oxido (containing 9.34 per cent. of oxygen) were heated with nitric acid, that 397-2 grams of the nitrate had been forned whilst 143.4 grams of another oxido (containing 13.39 per cent. of oxygon) were left andissolved. Calculate the formula of the oxides and tho atomic weight of the metal, and writo out the equation representing the reaction which took place. (N = 14,0 = 16.) (London Pre. Sci., Veterinary, 1906.) 11. In a mixture of FeO and Fe,Oy, 0.1000 gram of KMnO, was roquirod for oxidation (in acid solution), and the F0,0,.

·VII.] REACTING QUANTITIES 57 (total iron by precipitation) weighed 0-8000 gram; what was the composition of the mixture ? (K = 39, Mn = 55, Fe=56). (Laboratory Class, Manc.) 12. Marignac found that 5.000 grams of strontium chloride containing 6 molecules of water of crystallization gavo 30442 grams of strontium sulphato. Calculate tho equivalent of strontium. (H = 1,0 = 16, S = 32, Cl = 35.5.) (London Inter. Pass, Extl., 1904.) 13. An organo-metallic compound containing only carbon, hydrogen, and the metal gave, on combustion, the following numbers : 0.267 gram of substance gavo 0.176 gram CO, and 0-108 gram 11.,0, and there remained behind 0.223 gram of the metallic oxide in tho combustion apparatus. The metal forms colourless salts. Find tho equivalent of the metal and deduce a probable atomic weight by help of the periodic arrangement of the elemcuts. (B.Sc. Ilons., Manc.) 14. Thallons chlorido (TICI) reacted with gold bromido (AuBry) so that (a) 0·232 gram of thallous chlorido gave 0.128 gram of gold. (b) 0.294 0.162 Construct the equation for the reaction. (Tl = 204, ci = 35.46, Au 197.2.) 15. D. Borar found (J.C.S., 1911, 99, 1415) that 1.000 gram of mercury reduced 0.2452 gram of potassium dichromate in the presence of hydrochloric acid. The chromium was convorted into chromic chlorido. Construct the equation for the roaction. 16. From the following data calculate what percentage of the whole hydrochloric acid oscapes from a chimney connected with salt cake plant in which 8 tons of 90 per cont, salt is docomposed in 24 hours :- (a) Hydrochlorio aoid in chimney gas is graiu por cubic foot. (6) Velocity of chimnoy gas corrected to standard temperature and pressure is 15 feet per second. (c) Diamotor of chimney at the point of obsurvation is 4 feet. (Llons. Mano., Tochl. Chem., 1894.)

·58 EXERCISES IN CHEMICAL CALCULATION (CBAP. 17. An analysis of a mixture of potassium chloride, bromide, and iodide is made by the following method :- One gram of the mixed salts is precipitated by silver nitrate, and the total weight of silver chloride, bromide, and iodide is determined. Another gram is similarly precipitated and the chloride is converted into bromide by digestion with potassium bromide. A third gram is precipitatod as before, and all the salts are converted into iodide by digestion with potassium iodide. The weights obtained are as follows :- AgCl + AgBr + AgI = 1.7713 grams AgBr + AgI = 2:1297 AgI = 2-6267 Calculate the percentages of the potassium chloride, bromide, and iodide in the original mixture. (K = 39, Cl = 35.5, Br = 80, I = 127, Ag 127, Ag = 108.) (Inst. Chem., Min., April, 1910.) 18. lIow would you soften & wator having the following composition : Grains per gallon. Calcium carbonate (CaCO,) 10 Calcium sulphate (CaSO.) 15 Magnesium sulphate (MgSO.) 5 Sodium sulphate (N&SO.) 5 Sodium chloride (NaCl) 5 (Iust. Clicm., Finnl, Mineral, 1911.) ... When two or more chemical reactions take place in succession, it is not usually necessary to make a separate calculation for each reaction. In the determination of manganese dioxide in a specimen of pyrolusite the mineral is treated with hydrochloric acid, the chlorine evolved being passed into potassium iodide where iodine is liberated, The quantity of free iodine is found by determination of the quantity of sodium thiosulphate required to combine with it. The following reactions take place :- ·VII.) REACTING QUANTITIES 59 Mn0, + 4HCI = MnCl2 + Cl2 + 2H,0 2KI + Cl2 = 2KCl + 21 2Na,,03 + 21 = Na $408 + 2NaI Inspection of these equations will show that 1 grammolecule of manganese dioxide (86.93 grams) will correspond with 2 gram-molecules of sodium thiosulphate (2 x 158-14 = 316:28). So if 0·5621 gram of pyrolusite corresponded with 1.207 gram of anhydrous sodium tbiosulphate, then the quantity of manganese dioxide present would be 1.207 X 86.93 = 0.3317 gram 316.28 0-3317 x 100 = 59.0 cent. 0-5621 рег or 19. What weight of chlorine would be produced by the action of hydrochloric acid on 10.00 grams of manganese dioxide? What weight of iodine would this chlorine displace from a solution of potassium iodide? (Mn = 55, Cl = 35.5, I = 127.) 20. How much pyrolusito (82 per cent, manganese dioxide), and how much hydrochloric acid (33 por cent. hydrogen chloride) will be required to furnish tho chlorine necessary to convert 40 grams of potassium hydroxide completely into chlorato and chloride ? 21. Platinio chlorido was added to & solution of ammonium chloride and the resulting precipitate, after ignition, left 1.700 grams of platinum. What weight of ammonium chlorido was presont in the solution ? (Pt = 195, N = 14 H = 1, C1 = 35'5.) (Bd. of Ed., Stage II., 1910.) 22. 0.8876 gram of a sample of pyrolusite yielded 0.8027 gram of carbon dioxide when it was warmod with oxalic acid and dilute sulphurio acid. What percentage of MnO, did it contain (Mn = 55.) (B.Sc., Mano.) 23. The gasos from the action of heat on 1.103 gram of ·60 EXERCISES IN CHEMICAL CALCULATION Cohap. lead nitrate (Pb(NO3)2] were passed into a solation of potassium iodido. What weight of iodine would be liberated ? (Pb = 207, N = 14,0 = 16, K = 39, I = 127.) (B.Sc., Manc., 1911.) [Note. The stadent will find it an instructive exorcise to test the result of his calculation experimontally.] 24. If 1.04 gram of silicon fluoride were entirely decomposed by water in a platinum dish, and the precipitate formed were collected and ignited, what would its weight be ? (B.Sc., Manc.) 25. Calculate the weight of quicklime required to noutra. lizo the hardness of 1,000,000 gallons of water containing 10 grains of calcium carbonate (dissolved as bicarbonato) per gallon, Certain Indirect Analyses.--The following example illustrates a common case of indirect analysis. 0.9031 gram of mixed sodium and potassium chlorides was treated with sulphuric acid, and the resulting mixture of sulphates weighed 1.0784 gram. What were the amounts of the two chlorides in the original mixture ? The equations involved are- 2NaCl + H2SO4 = Na2SO4 + 2HCI whence 116.92 grams of sodium chloride give 142.07 grams of sodium sulphate. 2KCl + H2SO4 = K2SO4 + 2HCI wbeuce 149.12 grams of potassium chloride give 174.27 grams of potassium sulphate. If x and 0.9031 - 3 represent the quantities of sodium and potassium chloride respectively-- 142:07 The x gram of sodium chloride give (0.9031 - x) gram of potassium chloride give 174.27 (0.9031 - x) 149.12 gram of sulphate 116.92 gram of sulphate.

·VIT.] REACTING QUANTITIES 61 and the weight of the mixed sulphates is 142.07 116.92 + 174.27 (0.9031 - 2) = 1.0784 149:12 Solving for a we get - 142.07 174.27 174.27 = 1.0784 - 0.9031 x 116.92 149.12 149.12 2(1.2151 1.1687) 1.0784 1.0553 X = 0.494 i.e. sodium chlorido = 0.494 gram, and potassium chloride = 0.409 The student is advised to apply some of the considerations in Chapter I to this problem. For instance, if the experimental error in the weight of the mixed sulphates may be 0.2 mgr., what error would that produce in the result calculated ? [6 mgr.] Example No. 30 (below) is a case where a small crror in weighing will produce a large error in the result calculated—so large that the method of experiment would in that case not be used by a good analyst. 26. A quantity of a mixture of potassium iodide and' sodium chlorido transformed into the normal sulphates is found to weigh the same as the original salts; determine tho composition of the mixture. (K = 39, Na = 23, CI I = 1265, S = 32, 0 = 16.) (Laboratory Class, Manc.) 3504, 27. A mixturo of potassium chloride and iodido is converted into potassium sulphate, and the voight of the sulphate is found to be the samo is that of the original mixture. In what proportions wore the salts present in the mixture ? = 39-1, C1 = 35-4, I = 126.9, S = 32, O = 16.) (Laboratory Paper, Manc.)

·62 EXERCISES IN CHEMICAL CALCULATION LOHAP. 28. A mixture of dry calcium sulphate and calcium carbonate, when strongly heated over the blowpipe, is reduced to 75-0 per cent. of its original weight. What aro the proportions of the two compounds in the mixture ? (Ca = 40, S = 32, C = 12.) (Inter, B.Sc., Leeds.) 29. 2.000 grams of a mixture of sodium and potassium chlorides gavo 4.300 grams of silver chloride. What was the percentage of each salt in the mixturo ? 30. A mixture of barium monoxide and barium peroxido contained 88-0 per cent. of barium. What was the percentage composition of the mixture? What would be the error in the calculated percontages if there were an error of (a) 1, (6) 0:1, in the percontage of barium? The Volume-Weight Relationships of Gases. So far, only the weights of the substances taking part in a chemical reaction have been calculated. If any reagent is gaseous, its volume is deduced from its weight, or vice versâ, by use of the principle laid down in Chapter IV. If, for instance, in the example referring to zinc and sulphuric acid (p. 54), it were required to find the volume of moist hydrogen produced at 15° C. and 750 mn., then Zn + H2SO4 = ZnSO4 + Hly 98.086 grams hydrogen sulphate give 2:016 grams or 22 4 litres of hydrogen at N.T.P. (see p. 31). 700 700 grams hydrogen sulphate givo 22:4 x 98-086 litres of hydrogen at N.T.P. 700 288 760 22-4 x Х 174 litres moist 98.086 X 273 7373 or at 150 and 750 mm.

·vnt.) REACTING QUANTITIES 63 The datum used for transforming weights into volumes, and vice versa, is either of the following, and practice in their use will soon show which is the more convenient for any given case. 1. The gram-molecular weight of any gas at N.T.P. occupies 224 litres (approximate, but used in all except the most acourate work. See p. 31). II. One litre of hydrogen at N.T.P. weighs almost exactly 0.09 gram. Hence one litre of any gas at N.T.P. weighs 0.09 x d grams, where d is the density of the particular gas referred to hydrogen as unity. molecular weight Now d is approximately 2:016 molecular weight or less exactly 2 molecular weight The approximation da is often 2 sufficiently accurate, but gives results differing by 0·8 per cent, from those obtained by use of datum I. (above). The answers given for the following exercises are calculated from the data : The gram-molecular weight of a gas occupies 22:4 litres at N.T.P., or molecular weight One litre of a gas weighs 0:09 x 2:016 gram at N.T.P. Since these data bear small inacouracies, exact atomic weights give as a rule results of only the same degree of accuracy as approximate atomic weights. These latter are therefore used, with the exception of H = 1·008, which differs by 0-8 per cent, from H = 1.000. The following approximate atomic weights are accurate to 1 in 1000 ·64 EXERCISES IN CHEMICAL CALCULATION COBAP. ... ... Barium Calcium *Carbon *Chlorine Copper *Hydrogen *Iron Lead Magnesium 137-4 40:1 12 35.5 63.6 1.008 55.8 207 24:3 Manganese... 55 Mercury 2006 *Nitrogen 14 *Oxygen 16 l'otassium ... 39.1 *Silver 108 *Sodium 23 *Sulphur 32-1 Zinc 654 31. What is the weight of 220 c.c. of each of the following gasos, measured at 0° C. and 760 mm. : (a) hydrogen, (b) carbon dioxide, (?) nitric oxido, (d) nitrous oxide ? 32. What is the volume occapied at 0° C. and 760 m prossure by 1.00 gram of each of the following gasos: (a) hydrogon, (b) carbon dioxide, (o) nitric oxide, (11) nitrons oxido ? 33. Calculate the volume (at normal temperaturo and prosenro) of oxygen obtainable by heating in a test tubo 1.00 gram of each of the following substancos: Potassium chlorate, red lead, potassium nitrate. 34. What weight of water will be produced by passing 10 litres of hydrogen moasured at 760 mm. and 0° C. over a 1.000 gram of heated copper oxido ? (Inter. B.Sc., Leede.) 35. What weight of oxygon could be obtained from 54 grams of mercuric oxido? What volume would this weight of oxygen occupy at - 91° C. and 7 atmospheros pressure ? (IIg = 200, 0 = 16.) (1st Dipl. Dental Surgery, Manc.) 36. If 15.0 grams of water are decomposed into the constituent gases, what volume will these gasos N.T.P.? (Bd. of Ed., Stage I., Inorg., 1909.) 37. What weight of zinc will be dissolved by 250 grams occupy at * Approximate atomic weights starred should be remembered by the student; the others are invariably (and the starred ones often) appended to an examination question when necessary.

·VII.) REACTING QUANTITIES 65 of dilute sulphuric acid containing 20 per cent. of H.So, and what volume of gas measured at N.T.P. will be evolved ? (Bd. of Ed., Stage I., Inorg., 1907.) 38. Give equations reprosenting the formation of hydrogen by the action of (a) steam, and (b), dilute sulphurio acid on iron. What weight of iron would be necessary in cach caso for the preparation of 300 litres of hydrogon measured at 27° C. and 740 mm. pressrre ? (General Class, Manc.) 39. What volume of oxygen, measured at 15° C. and 760 mm., can be obtain,d by the complete decomposition of (the bydrogen peroxide in] 20 c.c. of an aqueous solution containing 30 grams of hydrogen peroxide per litre ? (1st " Conjoint," March, 1910.) 40. An acidified solution of bydrogen peroxido, on treatment with excess of potassium permanganato, gavo 32.6 times its volume of oxygen. What volume of oxygen, measured at N.T.P., would it evolve (a) if heated alone, (b) if some finely divided manganese dioxido were introduced, if it wero neutralized by means of dilute alkali, and then finely divided manganose dioxide introduced ? 41. Calculate the volume of nitrous oxide, mcasured at N.T.P., obtainable from 20 grams of ammonium nitrate, and tho volumo of nitrogen it would contain. (Bd. of Ed., Stage I., Inorg., 1910.) 42. One gram of phosphorus was placed in a litre globe of air which was sealed up at N.T.P. The phosphorus wag ignited by gently warming the globe. What was the final prossuro in the glube when it had cooled to 13° C.? (Tho percentage of oxygen in air by volume = 20.9.) (General Class, Manc.) 43. A piece of soilium was complotely convertod into chloride by the absorption of 250 c.o. of chlorine (mensurod at 10° and 720 mm.). What was the weight of the sodium nsed and of the chloride formod ? (Bd. of Ed., Stage II., Inorg., 1907.) 44. What weight of common salt would be required to furnish sufficient hydrogen chloride to neutralize 100 grams ·66 EXERCISES IN CHEMICAL CALCULATION Conap. 210.) of a 30 per cent. solution of Caustio soda? What would be the volume of the gas at normal temperature aud pressure ? (Bd. of Ed., Stage I., Inorg., 1906.) 45. What weight of carbon dioxide could be made by burning 100 grams of carbon, and what volume would this gas occupy at 546° C. and 112 mm. pressure ? (General Olass, Mano.) 46. 10 grams of carbon are strongly heated in a current of steam. Assuming that the reaction takes place quantitatively with the formation of carbon monoxide and hydrogen, calculate the volume of cach gas produced, 'and express in litres at N.T.P. (Manc. Sch. of Tech., 1st Year Course, 1910.) 47. What volume of carbonic acid gas can be formed by the action of hydrochloric acid on 100 grams of the following: acid potassium carbonate, calcium carbonate, normal bismuth carbonate ? (K = 39, Ca = 40, Bi = (General Class, Manc.) 48. How much snlphuric acid will be required to decompobe 50 grams of anhydrous potassium carbonate, and what volume of carbon dioxide measured at 12º and 740 mm, will be obtained ? (Dauntsey Entrance Mod. Sch., Mano.) 49. If 11 grains of Baso, are obtained from the sulphur prosent in 5 cubic foet of coal gas, express the rosult in terms of number of grains of sulphur per 100 cubic feet of gas, and in milligrams of sulphur per litre, and as c.c. of H28 por litre. 50. What weight of sulphur would be sufficient to convert all the oxygen in the air of a room 4 metres high, 6 metres long, and 5 metros wide into sulphur dioxido? The air of the room is to be taken at 20° C. and 760 mm. pressure. [Assumo air to contain 21 per cont. of oxygen, by volume.] (1st "Conjoint," Mar., 1909, adapted.) 51. It was found that 224 c.c. of sulphur dioxide at N.T.P. just decolourised 100 c.o. of a solution of potassium permanganate in dilute sulphurio acid. Calculate the strength of the potassium permanganate solution in grams per litre. (Inter., Mano., 1911.)

·11.) REACTING QUANTITIES 67 741 mm. 52. In order to determine the strength of diluto sulphuric avid having the density 1.091, 2 grams of pure magnesite (MgCO) wero troated with 10 c.c. of the acid, After all action had coased it was found that 0.8743 gram of magnesite remained andissolved. Calculate the number of grams of 7,80, in (a) a litre, (b) a kilogram, of the dilute acid. What rolume of carbon dioxide at 13° C. and 740 mm. pressure would be produced in the above reaction ? (London Intor., Vetorinary, 1907.) 53. Exactly 3 grams of magnesito (MgCO3) were added to 5.000 grams of dilute sulphuric acid ; after all action had coased, it was found that 0.480 gram wis loft undissolved ; calculate tho percentage strength of the sulphuric acid. In this reaction what volume of carbon dioxide would be evolved, measured at (a) 0° C. and 760 mm., (6) 13° C. and ? (London Pro. Sc., Vet., 1905.) 54. 8.000 grams of sulphur aro burnt in oxygon : what is the volume of the rogulting gas under normal conditions, and what volume would the gas occupy at 91° C. and 380 mm. presguro? (S = 32.) (General Class, Manc.) 56. What weight of hydrogon sulphide and of sulphur dioxide could bo formed from 10.00 grams of sulphur, and wbat volume would those gases occupy ander standard conditions? (S = 32.) (Genoral Class, Muno.) 56. Calculato the volume of sulphur dioxide, measured at normal temperature and prossure, which the solution of 100 grams of copper in oxcoss of concentrated sulphuric acid should give according to the usual equntion. Why would the actual yield be less than this? (Manc. Sch. of Tech., 1st Year, 1911.) 57. What is the usual cquation representing the action of strong sulphurio acid on metallio copper? What volume of gas ought to be obtained by the solution of 76•2 grams of copper, the gas being measured at 21° C. and 791 mm.? (London Inter. Pass, Extl., 1905.) 68. 0-0239 gram of carbon was treated with hydrogen at ·68 EXERCISES IN CHEMICAL CALCULATION [CHAY. 1150°. (Bone and Coward, J.C.S., 1910, 97, 1224.) In the products of the reaction 42-43 c.c. of methane at 0° anil 760 mm. were found. What percentage yield is this, reckoned upon the carbon used ? 59. 100 c.c. i of gas vero completely decomposed by heated tin. The tin was converted into stannous sulphide with an increase in weight of 0.144 gram. The residual gas consisted of hydrogen, and when passed over hented copper oxide gave 0.081 gram of wator. Show how the formula of the gas may be doduced from these data, given that a litro of hydrogen at N.T.P. weighs 0.09 gram. (Inter., Manc., 1910.) 60. Calculate by the shortest method you know the volume of gas which would be evolved by dissolving in hydrochloric acid 100 grams of a mixture of 84 parts by woight of magnesium carbonate with 10.0 parts of calcium carbonato. Find also the weight of silver nitrate required to precipitate complotely the chlorido formed. (Ca = 40, Mg = 24, Ag = 108.) (Inter., B.Sc., Mano.) 61. 1.000 gram of a mixture of the carbonates of calcium and magnesium gavo 240 c.o. of carbon dioxide (N.T.P.) Calculato the composition of the mixture. (B.So., Mano) 62. What is tho ratio of tho volumes of hydrogon and carbon dioxide (both measured at 770 mm. and 15° C.) evolved respectively by the interaction of (a) 60 с.o. of 30 per cent. bydrochloric acid on an excess of zino, (b) 40 0.c. of 30 per cent. hydrochlorio acid on an excess of boiling sodium carbonate solution ? (Inter., B.Sc., Leeds.) 63. A liquid petroloum of specific gravity 0-7803 is found to contain 85.0 per cent. of carbon and 15.0 per cent. of hydrogen. What is the minimum volume of air necessary for the completo combustion of 10.0 litres of it, and what weight of water and volumo of oarbon dioxido will bo formed ? Air may be assumed to contain 21.0 por cent. of oxygen, and tho measurements to be made at N.T.P. (Inter., B.Sc., Leods.)

·VII.) REACTING QUANTITIES 69 64. A fuel used in a gas producer contains 12 per cent, of ash and 72 per cent. of carbon. The ashes made contain 5 por cent. of unburnt carbon; the gas produced contains 0.165 of an ounce of carbon por cubic foot of gas at 0° C. and 760 mm. pressure. What volume of gas, measured under these conditions, is boing produced per ton (2240 lbs.) of fuel used ? (Hons., Manc., Fuel, 1909.)

·CHAPTER VIII THE KEAQTING VOLUMES OF GASES-GAS ANALYSIS TUE equation 200 + 02 = 2CO, indicates that two molecules of carbon monoxide react with one molecule of oxygen to produce two molecules of carbon dioxide. Hence 2n molecules of carbon monoxide react with n molecules of oxygen to produce 2n molecules of carbon dioxide. These 2n molecules of carbon monoxide will occupy twice the volume of n molecules of oxygen and the same volumo as the 2n molecules of carbon dioxide, always assuming that the three gases arc measured under the same conditions of temperature and pressure (Avogadro's hypothesis). Thus a direct inspection of the equation enables the statement to be made that 2 volumes of carbon monoxide react with 1 volume of oxygen to produce 2 volumes of carbon dioxide. By simple proportion the reacting quantities can be calculated for any given amount of any one roagent. Thus- (A) 150 volumes of carbon monoxide will, on complete combustion, combine with 75 volumes of oxygen, and produce 150 volumes of carbon dioxide, the three measurements being all made at the same temperature and pressure. (B) A mixture of 50 volumes of carbon monoxide and 50 volumes of oxygen evidently contains an excess of oxygen beyond that required for complete combustion of the carbon monoxide. If, then, the mixture be inflamed, ·OHAP, VIII.] REAOTING VOLUMES OF GASES 71 the 50 volumes of carbon monoxide will unite with 25 volumes of oxygen, producing 50 volumes of carbon dioxide; with this will be mixed the 25 volumes of oxygen in excess of that required for complete combustion. The application of corrections for change in temperature and pressure when these are not constant may be made by the use of the principles explained in Chapter 111. 1. 100 c.c. of carbon monoxido are mixed with 40 c.o. of oxygen and oxploded. If the resulting mixture is shaken with caustic potash, what volume of gas will romain, and what gas will it be? (Bd. of Ed., Stage I., Inorg., 1907.) 2. If 10.0 litres of carbon monoxide at 0° and 760 mm. are burned, what volume of oxygon at the same temperature and pressuro will be consumed, and what will be the volumne of carbon dioxide formed if measured at 150 and 740 mm. ? (Bd. of Ed., Stage II., Inorg, 1905.) 3. Wbat volume of oxygen at N.T.P. is required for the combustion of 50 grams of carbon monoxiilo. (Iuter. B.Sc., Manc.) 4. How many litres of oxygon, measured at 20° and 750 mm., are necessary for the complote combustion of 150 grans of marsh gas? (Bi. of Ed., Stage II., 1909.) 5. What change in volume would take place in one litre of oxygon by- (a) burning sulphur in it, (b) converting 6 per cent. of it into ozone, c) combining it with twico its volume of carbon monoxide ? (London Inter., Intl., 1908.) 6. How many litres of hydrogen, measured at 20° and 750 mm., are necessary to combine with the oxygen erolvod by the complete decomposition of 500 grams of potassium chlorate ? (Dauntboy Entrance Medical Schol., Manc.). 7. How many grams of 10 per cent. hydrogen peroxido would be required to oxidize 12 grams of lead sulphide? If tho volume of oxygen obtained by heating this quantity of ·72 EXERCISES IN CHEMICAL CALCULATION Couar. hydrogen peroxide were combined with carbon monoxide, what would be the volume of the resulting gas measured at 15º and 750 mm. ? (London Inter. oterinary, 1907.) 8. A sample of gas had the following percentage composition : hydrogen, 45; marsh gas, 30; carbon monoxide, 20; Acetylene 5. 100 volumes of it were mixed with 160 volumeg of oxygen and the mixture oxploded. Calculate the volamo and the composition of the resulting mixture of gases (all boing supposcd dry). (London Inter. Ext)., 1906.) 9. Calculate the volume and composition of the gasos which would result from exploding a mixture of 150 volumes of oxygen and 100 volumes of gas having the percentage composition : hydrogen, 46; marsh gas, 28; carbon monoxide, 18; ethylene, 6; carbon dioxide, 2. (London Inter. Veterinary, 1907.) 10. Coal gas of the composition given below is burnt in air : HI ... Hydrogen (H.) Methanc (CII.) Ethane (C1) Ethyleno (C'II.) Benzene (CH) Carbon monoxide (CO) Carbon dioxide (CO) Carbon disulphide (CS) Nitrogen (N) Per cent. 48.0 26.0 2.0 3.0 1.0 15.0 20 0:1 2.9 ... ... .< 100.0 Name the products and give the quantities of each produced per litre at N.T.P. of gas burnt, stating your results in C.C. measured at 100° C. and 760 mm. (Inst. Chem. Inter., Jan., 1911.) 11. The density of liquid carbon dioxide was foun: to be 6-88 compared with water at 4° C. What volume of gas at N.T.P. would be formed by the evaporation of 100 c.c. of tho Liquid, and what volumo of carbou monoxide mcasured at ·VIII) FORMULA OF A GAS 73 27° and 720 mm. would bo formed by passing this gas over heated carbon ? (C = 12, 0 = 16.) (Genoral Class, Manc.) 12. If 100 c.c. of hydrogen and 100 c.c. oxygen, both at 27° O. and 216 mm., bo mixed and fired, at what temperatures would the products of their ignition occupy 300 c.c. at 216 mm. 13. Calculate the number of volumes of hydrogen (to four decimal places) which combine with one volume of oxygen when 6870-1 volumos of hydrogen, mixed with 3139.7 volumes of oxygen and exploded, left 9-2 volumes of oxygen uncombined. (London Inter. Pass, Extl., 1905.) 14. If the density of oxygen is 15.90 and the atomic weight of oxygon 15.86 [H = 1] what are the combining volumes of oxygen and hydrogen and wliat is the density of electrolytic gas? (Hons. B.Sc., Manc.) Determination of the Formula of a Pure Gas.-An example of the determination of the formula of a gas is found in the reaction between chlorine and excess of ammonia solution (see text-book for practical details). The experiment shows that 3 volumes of chlorine displace from an unknown volume of annmonia, 1 volume of nitrogen. Now, chlorine and hydrogen unite only in one ratio, namely, in cqual volumes. Hence the unknown volume of ammonia gas would yield on decomposition nitrogen and hydrogen in the ratio of - 1 volume nitrogen 3 volumes hydrogen 1 molecule nitrogen 3 molecules hydrogen (by Avogadro's hypothesis) 1 atom nitrogen 3 atoms hydrogen i.e. ammonia is NH3 or N,H, or some other simple multiple of NH3. The formula is finally proved $c be NH, by a determination of the density of the gas 2.e = ·74 EXERCISES IN CHEMICAL CALCULATION CORAP. ... ... ... 15. A mixture of chlorine with an oxide of chloriue gave on explosion an oxpansion from 100 c.o. to 130 c... The exploded gas was found to contain 70 c.c. of chlorino and 60 0.c. of oxygen. What was the oxide, and how mnch of it was there in the niixturo ? (1st Year Hons., Manc) 16. 12 c.c. of gaseous ammonia were completely burnt when fired with 18 co of an oxide of nitrogen, and 15 0.c. of nitrogen roinained in the eudiometor after the condensation of tho steam. What is the formula of the oxide of nitrogen ? (1st Year Hons., Manc.) 17. A gaseous compound of carbon and hydrogen gave the following results on combustion. Volume of gas 14.8 c.o. Volume of gas and excess of air 200-2 c.c. Volume after explosion 1780 c.c. Volume after removal of carbon dioxide 115 c.c. What was its ino lecular formulu ? 18. Find the molecular formula of a gaseous oxide of carbon from the following data reforring to the combustion of the substance with an excess of oxygen and consequent production of carbon dioxido. (Dicls and Wolf, Ber., 1906, 39, 689.) Volume of gas 9.86 c.o. Volume of gas and excoss of oxygen 101.5 C.C. Volume after explosion ... 104.1 C.C. Volume after treatment with potassium hydroxide 74.8 ... ... ... ... ... C.C. ... When dealing with a gaseous mixture whose constituents are known, the relative amounts of the latter may be deterrnined by gas-volumetric processes. EXAMPLE.When 100 c.c. of a mixture of nitrous and nitric oxides were mixed with 200 c.c. of hydrogen and fired in a eudiometer, 80 c.c. of nitrogen remained. What was the composition of the mixture ? (Inter. B.Sc., Manc.) The two equations are written for the combustion of the two oxides separately. Note that there is sufficient ·VI.] GAS ANALYSIS 75 hydrogen to decompose completely either oxide if present alone; therefore also to decompose any mixture of the two. Underneath each member of the equation is placed an expression showing the volume of it reacting with or resulting from * C.C. of nitrous oxide or y c.c. of nitric oxide as follows: N,O + Hg = N, + H2O X - 0 2NO + 2H2 = N2 + 211,0 Y Y y 0 2 C + y 100 Y 20+ 80 2 60 с.c. of nitrous oxide 40 c.c. of uitric oxide Then and wlience A similar calculation to the above is made in the analysis of nixtures of methane, hydrogen, and nitrogen by explosion with a measured excess of oxygen. After cooling to the original temperature the contraction is noted. The carbon dioxide is then absorbed by a solution of potassium hydroxide, and its amount thus found. Corrections for changes in temperature and pressure and for the presence of varying amounts of water vapour were formerly required in analyses of this kind, but at the present tine the best forms of apparatus are arranged to keep a constant temperature throughout the analysis, and the measurement is either that of volume at constant pressure or, what is almost the same, pressure at constant volume. Unless all the measurements are those of dry gases (an operation of some difficulty), it is usual to keep the walls of the measuring tube moist with water or dilute sulphuric acid. In this way, since the temperature does not vary, a constant pressure of aqneous vapour is ensured for all the ·76 EXERCISES IN CHEMICAL CALCULATION COHAP. measurements. If these are volumes at, say, 14° C. and 760 mm., then the same readings would represent quantities of dry gas at 14º and 748 mn., i.e. at a constant temperature and pressure. (The vapour pressure of water or dilute sulphuric acid at 14° is approximately 12 mm.) If the measurements are pressures at, say, 14° C. and at constant volume, the apparatus is so constructed that the readings express the difference in pressure between the moist gas and saturated aqueous vapour at the same temperature contained in the other limb of a U-tube manometer. This difference is obviously the partial pressure of the dry gas, so that the effect of the water vapour is automatically eliminated. To return to the calculation of the analysis, the example may be taken- Gas (hydrogen + methane + nitrogen)... 83.9 volumes 4845 Mixture after explosion 349.9 Mixture after treatnient with KOTI 322:1 Here the contraction = 484.5 3439 = 134.6 And the carbon dioxide = 349.9 - 322:1 = 27-8 Let there be x volumes of hydrogen and y volumes of methane in the original 83-9 volumes of gas. Now 211, + 0.2 = 211,0 Gas + oxygen C 2 0 2 The water produced has no effect on the volume measurement. Since z volumes of hydrogen and z volumes of oxygen disappear from the gas, the presence of volumes of hydrogen accounts for a contraction or explosion of 30 volumes. 2 ·vm.) GAS ANALYSIS 77 Further, CH4 + 20, = CO2 + 2H1,0 y 2y y 0 Since y volumes of methane and 2y volumes of oxygen disappear from the gas, and are replaced by y volumes of carbon dioxide, the presence of y volumes of methane accounts for a contraction of 2y volumes and for the formation of y volumes of carbon dioxide. The nitrogen is unaffected. 2y Then contraction due to combus 3x tion of hydrogen 2 And contraction due to combustion of methane 32 Total contraction = + 2y = 1316 2 Also, total carbon dioxide = y = 27.8 whence x, hydrogen 52:7 62.8 per cent. y, methane = 27.8 = 33.1 and 839 (x + y), nitrogen 4.1 3-4 = > Note that the nitrogen is determined by difference, 80 that such an analysis will of necessity give results adding up to 100 per cent. 19. If 100 volumes of air are mixed with 80 volumes of hydrogen and an electric spark is passed through the mixture, what will be the volumo and composition of the remaining gus measured at tho samo tomperaturo and pressure ? (BI. of Exl., Stage I., Inorg., 1910.) 20. To 100 c.c. of a mixture of oxygen and nitrogon, standing over water, 100 c.c. of hydrogen woro added, and the mixtuto was firod. After cooling the residual gases (nitrogen and hydrogen) occupied 32 c.o. What was the composition of the mixturo ? (General Class, Mano.) 21. 100 c.o. of a mixture of carbonio oxide and marsh gs were added to 200 0.0. of oxygen in a endiometer. After firing ·78 EXERCISES IN CHEMICAL CALCULATION COĦAP. ... ... and cooling the residual gas moasured 235 0.0. Calculate the composition of the mixture. (1st Year Hons. Class, Mano.) 22. 100 volumes of a mixture of oxygen, nitrous oxide and nitrogen were mixed with 150 of hydrogen, and exploded. On cooling, 90 volumes were left. To these 50 volumes of oxygen were added, and the mixture was exploded. The residue on cooling measured 95 volumes. What was the composition of the mixture ? (Dauntsey Entrance Med. Schol., Mado.) 23. A mixture of nitrous and nitrio oxides with nitrogen was analysed as follows. All readings were taken under the same conditions of temperature and pressuro ovor watcr. Volume of mixture 17.0 c.o. Volume of mixture and excess of hydrogen 36:1 Volume of mixture after reaction and cooling 17.0 The excess of hydrogen was then determined by explosion with oxygen. Volume after adding oxygen to residue 25.5 c.c. Volume after explodiny 19.2 What is the percentago composition of the mixture ? 24. 100 volumes of a mixture of carbonic oxide and acetylope were mixed with 250 volumes of oxygen in a eudiometer and fired. After cooling the residual gases occupied 260 volumes, and after trcatment with potash the residual oxygen occupied 120 volumes. What was the composition of the mixture ? (1st Year Hons., Manc.) 25. 100 c.c. of a mixture of carbon disulphide vapour and cyanogen were mixed with 300 c.c. of oxygen and fired. The contraction on cooling was 30 c.c., tho SO, formed was 60 C.C., the Co, formed was 170 c.c. What was the composition of tho mixture ? (1st Year Hons., Manc.) 26. To a mixture of acetylene, hydrogen, and carbon monoxide, occupying 20 c.o. (corrected), 50 c.c. of oxygen were addod and a spark passed. After cooling, the residual gases ... 17 + ·VIII.) GAS ANALYSIS 79 ... contained 16 0.c. of carbon dioxide and 36 c.o. of unburnt oxygen. What was the composition of the mixture ? (Hons., Manc., 1911.) 27. 40 c.c. of a mixture of hydrogen, mothane, and nitrogen were exploded with 10 c.c. of oxygon. After cooling, the residual gas measured 30.5 c.c. On treatment with caustic potash the volume diminished to 33.5 c.c., and on treatment with alkaline pyrogallol to 32 c.c. Calculate the percentage composition of the original mixturo. (B.Sc., Manc., 1910.) 28. Given a gaseous mixture of nitrogen, oxygen, carbon monoxide, and acetylene, how would you prove the presence of each of tho gaseous constituents ? If 0.027 gram of water was produced after explosion of 150 c.c. of the mixture with oxygen, what was the percentage of acetylene in the original mixture ? (B.Sc., Manc., 1911) 29. A mixture of carbon monoxide, bydrogen, ethylene and nitrogen was analysed with the following results :- Volume of gas 102.5 After treatment with bromine 89.55 After treatment with ammoniacal cuprous chloride 46.6 Aftor admixture with oxygen and air 411•1 After explosion 3440 Calculato the percentage composition. 30. 352.0 volumes of a coal mine gas, after treatment with caustio potash, becomo 351•0 volumes ; after explosion with a small amount of olectrolytic gas (2H, + 0,), 32-8 yolunos ; aftor treatment of this with caustio potash, 338-5 volumes. Calculato the percentages of carbon dioxide and of methane, assuming that tho only hydrocarbon present is methane. 31. A sample of gas known to contain carbon dioxide, carbon monoxide, ethylene, methane, hydrogen, and nitrogen, was analysed as follows: 86.0 volumes on treatment with caustio potash were reduced to 84:9 volumos; on further treatment with Nordhausen acid and caustic potash the volume was reduced to 82.6. This residual gas was thereapon exploded with 55•77 volumes of oxygen and 326.21 volumes of .. ...

·80 EXEROISES IN CHEMICAL CALCULATION [OnaP. 2 ... air (frce from carbon dioxide). The volume after explosion was 383.88; after treatment with caustio potash it was reduced - to 340-62. Assuming that the original gas contained 2-3 per cent. of nitrogen, calculate its composition from the foregoing analysis. (Fuel, Manc., 1902. 32. 163 volumes of a snmple of coal gas were reduced to 161.8 volumos over caustic potash, and afterwards to 155 volumes over fuming sulphuric acid. 95 volumes of tho residual gas wero exploded in a eudiomoter with 600 volumes of oxygen. After the gas had cooled it measured 539.5 volumes, and on treatment with caustic potash was further reduced to 492.5 volumes. Assuming that the gas contains no nitrogen, and that the only saturated hydrocarbon present is methano, calculate its composition from the foregoing figures. (Fuel Class, Manc., 1901.) 33. Calculate from the following data, the composition of & mixture of methano, othano, and nitrogen :-- Volume of gas taken 53.5 c.c. Volume of oxygen added 452.0 Volume after explosion 405.8 Volume after treatment with KONI 349.2 (London Ilong. Intl., 1908.) 34. A mixture of carbonic oxide, mothane, and ethane, measuring 10 c.c. was mixed with 40 c.c. of oxygen and fired in a moist oudiometer, After cooling, the residual gases were found to consist of 12 c... of carbonic acid and 23 0.c. of unconsumed oxygen. What was the composition of tho mixture ? (B.Sc. Hons., Manc.) 35. 60 с.c. of a mixture of carbon disulphide and carbon monoxide were mixed with 240 c.c. of oxygon in a eudiometer and fired; after cooling, the residual guscs, CO, SO, and (), occupied 238 C.C., of which 132 c.c. woro absorbod by potash. Wbut was the composition of the mixture, and what percentage of the Cs, vas burnt to 80,? (B.Sc. Hone., Manc.) 36. 1 litre of hydrogen containing small amounts of methane and of oitrogon was treated with palladium, which ...

·[1.] GAS ANALYSIS 81 ... ... ... ... ... romoved all but 38.8 c.c. of gas. 99.6 volumes of this was mixed with air and oxygen to a total volume of 566.5 and exploded. Its volume then was 413.2, and after treatmont' with potash, 402.1. Find the percentages of nitrogen and of methane in the original gas. 37. A samplo of gas containing carbon monoxide, carbon dioxide, nitrous oxide, and nitrogen was analysed with the following results :- Volume of gns used 148.9 Volume after treatment with moist potash 52.6 After addition of hydrogen 275.6 Aftor addition of clectrolytic gns (2H, + 0,) 310.4 After explosion * 245.1 After treatment with ammoniacal Cu,Cl, 235.7 Calculate tho porcentage composition. 38. A mixture of oxygen and carbon monoxide is exposed to the action of ultraviolet light. Ozono and carbon dioxide are the now gasos formed in the reaction. If 100.00 c.c. of gas 02 were taken in which the ratio was 1.015, how many c.c. CO of each of the gases will be found in the residue which CO2 measures 85.87 c.c. and has the ratio = 0.24192 CO (Chapman, Chadwick, and Ramsbottom, J.C.S., 1907, 944.) 39. It is proposed to analyze mixtures which contain (a) methano, carbon monoxido, hydrogon, and nitrogen ; (6) methane, ethane, hydrogen, and nitrogen, by combustion with excess of oxygen and measurement of (i.) volume of gascous mixture, (ii.) volume of oxygon added, (iii.) contraction after explosion, (iv.) volume of carbon dioxide produced, (v.) residual oxygen. Show in each case whether it is possible to calculate tho composition of the original mixture or not. Would it be possible to analyse these two mixtures by passing measured volumos of each over red-hot copper oxido and weighing soparately the water and carbon dioxide produced ? * A negligible amount of CO, was formed, ·CHAPTER LX VOLUMETRIC ANALYSIS (SOLUTIONS) ? The principles and methods of volumetric analysis can only be studied in the laboratory. The following sections, therefore, contain only an outline of the nomenclature, and illustrations of the methods which should be used in calculation. Normal Solutions of Acid and Alkali. A normal golution of an acid contains 1 gram-equivalent (i.e. 1.008 gram) of hydrogen replaceable by a metal, per litre. For example, normal hydrochloric acid contains (1.008 + 35:46) grams of HCl per litre. Normal sulphuric acid contains (2:016 + 32:07 + 64:00) grams of H2SO4 per litre, because the gram-molecular weight of sulphuric acid (2.016 + 32:07 + 04:00 giams) contains 2 gram-atons of hydrogen. A normal solution of an alkali is exactly equivalent to a normal solution of an acid. For example, since NaOH 33.00 + 16.00 + 1.008 + HOI NaCl + H2O 1.008 + 35:46 40-01 36:47 then normal caustic soda solution contains 40-01 grams of NaOH per litre, or 23.00 grams, i.e. the equivalent, of Na. Normal baryta solution contains 3(137.37 + 32:00 + 2.016) grams of Ba(OH), because 1 gram-molecule of Ba(OH), is equivalent to 2 gram-molecules of HCI- Ba(OH)2 + 2HCl = BaCl2 + 2H20 ·CHAP. IX.) VOLUMETRIC ANALYSIS 83 In certain reactions salts of weak acids (e.g. sodium carbonate, Na,C03, and borax, NagB401) behave as alkalies, and neutralize stronger acids, e.g.-- NazC0g + 2HCl = 2NaCl + H2O + CO2 The above definition of a normal solution of an alkali still holds in such cases, e.g. normal sodium carbonate contains }(46.00 + 12:00 + 48:00) = 53.00 grams of the anhydrous salt per litre. Normal Solutions of Oxidizing Agents. -A normal solution of an oxidizing agent, such as potassium permanganato or potassium bichromate, contains in 1 litre 1 gram-equivalent of oxygen, available for oxidation. Since exactly 16 grams of oxygeu unite with or oxidize 2.016 grams of hydrogen or their equivalent of other substance, then the normal solution contains exactly 8 grams of oxygen available for oxidation. A typical oxidation by potassium permanganate in the presence of sulphuric acid is represented by-- 2KMnO4 + 4H2SO4 + 511,C20. = 2KIISO4 + 2MnSO, + 8H2O + 10CO, The weight of permanganate in grams is 2(39:10 + 54.93 + 64:00) = 316.06. The number 316 is used as being more accurate than the experimental observations themselves. It will be observed on dissecting the equation thus- 2KMnO4 + 4H2SO4 = 2K HSO4 + 2MnSO4 + 37,0 +50 50 + 5C,H,04 = 1000, + 51,0 that only 5 of the 8 atoms of oxygen in the permanganate are available for oxidation. Hence the 316 grams of permanganate contain 80 grams of available oxygen. Thereforo 31:6 grams of permanganate contain 8 grams of available oxygen, so that for oxidation in the presence of acid a normal solution contains 31.6 grams of potassium permanganate. From equations expressing its oxidizing ·84 EXERCISES IN CHEMICAL CALCULATION CONTAP. action upon oxalic acid, ferrous salts, etc., the amounts of oxalic acid, ferrous salts, etc., in any given solution may be calculated. Normal Iodine Solution.-Iodine, dissolved in potassium iodide solution, is used for certain estimations. Since iodine is monovalent in the reactions concerned, & normal solution contains 126.92 grams of iodine per litre. Its action is really that of an oxidizing agent, and its estimation depends upon the reaction represented by 2Na2S2O3 + 14 = Na S408 + 2Nal In this reaction 1 molecule of sodium thiosulphate is equivalent to 1 atom of iodine. Therefore normal sodium thiosulphate will contain (46:00 + 64.00 + 48:00) = 158 grams of Na2S20g per litre. It must be noted that this normal solution contains two equivalents of sodium per litre, so that in other reactions it is possible for a solution containing only half the above amount of thiosulphate to be normal. Normal Solutions of Reducing Agents.-Such solutions are those of stannous chloride, titanous chloride, oxalic acid, etc. They are normal when 1 litre of the solution is completely oxidized by the equivalent of oxygen (8 grams), or of chlorine (35:46 grams), e.g. normal 119.0 stannous chloride could be made by dissolving 2 = 59.5 grams of tin in hydrochloric acid and making up to 1 litre. [SnCl2 + Cl, Snci.] Decinormal and Centinormal Solutions, etc.-Normal solutions are, as a rule, too concentrated for convenience. Weaker solutions are then employed, and, if one-tentb Similarly, one 10. normal, are designated decinormal (1).

·VOLUMETRIO ANALYSIS 85 hundredth normal (106), half normal and other 2 strengths are used. It is not always convenient to make up solutions which are exactly normal or decinormal, and in such cases, when the concentration of the solution has been determiued, it may be cxpressed by means of a factor, e.g. à 0 997 N. sulphuric acid contains 0 997 x 49:04 grains of sulphuric acid per litre. If a volume of, say, 10:45 C.C. of this solution has been taken, it is equivalent to 10:45 X 0.997 10:42 c.c. of normal acid. EXAMPLES.—A solution of copper sulphate was electrolyzed, producing copper and sulphuric acid. The weight of the copper obtained was 0·1463 gram, and the sulphuric acid produced neutralized 15-62 c.c. of an alkali. What was the normality of the alkali? 1 gram-atom (63.57 grams) of copper is equivalent to 1 gram-molecule of sulphuric acid, which is equivalent to 2000 c.c. of N. alkali. :: 0:1463 gram of copper is equivalent to 2000 X 0.1463 C.C. N. alkali = 4.603. 63.57 Heuce the 15.62 c.c. of the alkali are equivalent to 4:603 c.c. N. alkali or 1.00 0.2947 i.e, the normality of the alkali (the l'actor) is 0-2947. 11.382 grams of bleaching powder were suspended in a litre of water. 25.00 c.c. of this suspension, on treatment with dilute hydrochloric acid and potassium iodido liberated iodine, which required 24:37 c.c. of 0.1007 N. Na,S,Og. What is the percentage of availablo chlorine in the bleaching powder ? 1 c.c. N. Nags,Og is equivalent to 1 c.c. N. I., which 35-46 is equivalent to 1 c.c. N. CI, i.e. to 1000 gr. CL.

·86 EXERCISES IN CHEMICAL CALCULATION COHAP, :: 1 c.c. of 0.1007 N. NagSgOg is equivalent to 35-46 x 0.1007 gr. CI, 1000 .: 24:37 c.c. of 0.1007 N. NagsOg are equivalent to 35:46 x 0.1007 x 21:37 gr. Ci, 1000 which is therefore the amount of chlorine in the 25* 0.c. of bleaching powder solution. In the litre there were therefore 35-46 1000 25 x 1000 x 0.1007 Ⓡ 24 37 grams of chlorine, and the percentage of available chlorine in the bleaching powder is therefore 1000 35-46 100 Х x 0.1007 X 24-37 X 30.57. 25 1000 11:382 1. A quantity of ammonium chloride is boilod in an open vessel with 100 c.c. of normal sodium hydroxido until no further chemical chavgo takes place; it is then found that the excess of sodium hydroxide requires 10 c.c. of normal sulphuric acid to neutralize it. How much ammonium chloride was uscd ? (H = 1, N = 14,0 = 16, S = 32, Cl = 35-6, Na = 23.) (1st “Conjoint,” July, 1909.) 2. What weight of sodium hydroxide wonld be required to noutralize 100 c.c. of a normal solution of sulphuric acid ? What volume of carbon dioxide, measured at 15° C, and 760 mm., would be required to convert completely the above amount of sodium hydroxide into sodiam bicarbonate ? (1st “Conjoint," January, 1909.) 3. 10 c.o. of a solution of ammonium chlorido having a density of 1.011 gave on the addition of silver nitrate solution 0.9491 gram už silver chloride; calculate the amount of NH.CI in 100 grams and in 100 c.c. of the solution. How would you determine the amount of ammonis in 10 0.0. of tho solution, ·VOLUMETRIO ANALYSIS 87 N and how many c.c. of acid ought it to noutralize? (II = 1, 10 N = 14, Cl = 35.5, Ag = 108.) (London Inter. Hons., Extl., 1904.) 4. What amount of KMnO, must be dissolved in a litre in ordor that 1 c.c. of the solution may yield (a) 1 milligram of oxygen, (b) 1 c.o. of oxygen at N.T.P. in acid solution ? (Laboratory Paper, Manc.) 5. If 100 c.c. of a solution of sulphurous acid just docolourizo 10 c.c. of a docinormal solution of iodine, what volume of sulphur dioxide docs tho solution contain ? (Laboratory Paper, Manc.) 6. To neutralize 10 c.c. of diluto hydrochloric acid, 40 C.C. of normal sodium hydroxide were required. In another 10 c.c. of the same acid, 1 gram of precipitated chalk (CaCO3) was dis. solved, and it was then found that 20.5 c.c. of normal sodium hydroxide were required to neutralize the excess of hydrochloric acid. Calculate the percentage of pure calcium carbonate in the precipitated chalk. (H = 1, C = 12,0 = 16, Na CI 35.5, Ca = 40.) (1st “Conjoint,” January, 1910.) 7. 5.5502 grams of fuming sulphuric acid wore dissolved in water and made up to 250 C.C. 50 c.c. of this solution N roquirod 52:05 c.c. of alkali. 2 Express tho composition of the original acid in porcentages by weight of hydrogen sulphate and sulphar trioxide. 8. A solution contained sodium hydroxide and sodium peroxide. 100.0 C.C. wero neutralized by 21.26 c.o. of N. N sulphuric acid, and decolourized 16.6 c.c. of 10 permanganato in the presenco of sulphuric acid. How much sodium was present in 100 .c. of solation, (1) as hydroxide, and (2) as peroxide ? (Leeds Univ., B.Sc., 1911.) 9. On treatment with a slight excess of water, 1 gram of a commorcial sample of calcium carbide gave 290 0.c. of moist gas at 13° C. and 752 mm. pressure. (Pressure of aqueous Vapour 11 mm.) 23, ·88 EXERCISES IN CHEMICAL CALCULATION COHAP. IX. In order to neutralize the residue 29 c.c. of normal hydrooblorio acid were required. (H 1, 0 = 12, Ca = 40. 1 litre of hydrogen at 0° C. and 760 mm. weighs 0.09 gram.) Calculate from the above data two independent values for the percentage of calcium carbide in the sample and explain the discrepancy in the results obtainod. (London Intor Hons., Extl., 1907.) 10. An orango-coloured gas was found to be absorbed by water, the solution then containing nitrous and hydrochlorio acids. When 223 c.c. of the gas wero absorbod by water, onetenth of this solution, when titrated with pormanganate in the cold, required 0·0632 gram KMnO,; another tenth of the solution requirod 10 c.c. of decinormal AgNO, to precipitato tbe chlorine. On heating potassium in it the gis became colourless, and on cooling contracted to half its original volumo. Find the formula of the gas. (1st Year Hong., Manc.) 11. Air containing chlorino and vapour of hypochlorvus N acid was passed for some time through an excess of sodium 10 arsenito solution. The resulting solution was divided into two equal parts, the arsenite in one being determined by titration with iodine, the chloriile in the other boing determined by titration with silver nitrate. The arsonite oxidized being represented N by 9.15 c.c. of iodine, and tho chlorido produced by 7.27 c.c. 10 N of silvor nitrate, calculate what percentage of the total 10 chlorine in the air is present (a) in the free state, (b) as lypo chlorous acid ? (R. L. Taylor, J.C.S., 1910, 97, 2541.)

·CHAPTER X ORGANIC ANALYSIS Empirical Formulæ.-The first step in the full quantitative analysis of a pure organic substance is the determination of the proportions of the constituent elements separately. Carbon and hydrogen are estimated by weighing separately the carbon dioxide and water produced when a weighed quantity of the substance is burnt. Thus 0:1836 gram of a substance produced on combustion 01934 gram of carbon dioxide and 0:0693 gram of water. Since carbon 12 00 dioxide contains of its weight of carbon, the carbon 44:00 present in the original substance was 14 x 0.1934 = 0:05275 gram. Similarly the hydrogen present in the original substance was » x 0.0693 = 0·00770 gram.* Nitrogen is estimated from a measurement of the volume of free nitrogen produced when a weighed amount of the substance is burut, special precautions being taken to ensure that the nitrogen is all liberated as such, Certain classes of nitrogenous substances may be so treated that their nitrogen is converted into ammonia, and estimated by passing into a measured amount of standard a instead of 18 2 1 2.016 . The use of or simply a instead of introduces an error of 9 18.016 8 parts in 1110. This may be neglected, since the analytical figures are liable to much greater errors.

·90 EXERCISES IN CHEMICAL CALCULATION [CHAP. acid solution (Kjeldahl); in other cases the nitrogen may be converted completely into nitric oxide and measured as buch (Lunge's" nitrometer" method). Sulphur is converted into barium sulphate, and weighed as such. The halogens are converted into the corresponding silver halides. In each case the calculation of the percentage composition from the analytical results presents no difficulty that has not been met in carlier chapters; from the percentage composition the empirical' formula is deduced as shown in Chapter VI. In many of the exercises which follow it will be found unnecessary to express the analytical results as percentages, as the following example shows :- A body contains only carbon, hydrogen, and oxygen. 0-2427 gram of it gives on combustion 0-3553 gram of carbon dioxide and 0:1461 gram of water. What is its simplest (empirical) formula? In the 0-2427 gram are 0 3553 x 4 = 0.0969 gram of carbon and 0.1461 X } = 0·0162 gram of hydrogen Hence the total burnt, minus the hydrogen, minus the carbon, is the amount of oxygen in the substance, i.e. 0.1296 gram. These are the relative quantities of carbon, hydrogen, and oxygen present; hence, by dividing each by the corresponding atomic weight, the relative numbers of atoms are found (Chapter VI.), i.e.- 0.0969 00162 0.1296 = 0.00807 : 0.0162 : 0-0081 12 1 16 1 2 1 Hence the simplest formula is CH,O. . Moleonlar Formulæ.The molecular formula is, of course, either the same as the empirical formula or some ·x.) ORGANIO ANALYSIS 91 simple multiple of it. Molecular formulæ are deduced by one or other of the following methods for determining molecular weights :- (I.) Measurement of the vapour density of the substance or of some substance made from it by a simple process. Thus acetic acid has the empirical formula CH,0; its vapour density shows a molecular weight approximately 60; hence the molecular forinula is C2H4O2, which is twice the empirical formula. Oxalic acid has the empirical formula CHOZ. It decomposes partially on vaporization, so that a vapour density determination is valueless. But methyl oxalate, by analysis and vapour density determination, is shown to have the molecular formula C2H604, or C2(CH3)204; and since it is known to have been formed from oxalic acid by the replacement of hydrogen (H) by methyl (CH3), the molecular formula of oxalic acid is probably C,H,04, twice the empirical formula. (II.) Depression of freezing point and elevation of boiling point methods (see Chapter XII.). (III.) Special method for an acid. If the basicity of the acid (the number of atoms of hydrogen in the molecule which can be replaced by sodium on treatment with sodium carbonate) is known, then the silver salt is made and the proportion of silver in it determined by igniting in air and weighing the residue. The molecular weight of the acid is then given by-- Weight of silver salt used Weigut of silver produced molecular weight of acid - n + n. 108 n. 108 where n is the basicity. Thus lactic acid has the ewpirical formula CH,0. It % ·92 EXERCISES IN CHEMICAL CALCULATION (CHAP is monobasic, and its silver salt contains 54:8 per cento silver. Hence 100.0 molecular weight of lactic acid 1 - 108 54:8 108 :: Molecular weight of acid = 90, which corresponds with the formula C3H60s, three times the empirical formula. (IV.) Special method for a base. Organic bases, such, for example, as methylamine, CH3NH2, form salts in the same way as ammonia, so that methylamine forms mcthylamine hydrochloride, CH3NH2.HCl. Then, just as ammonjuin chloride forms an easily prepared double salt- (NII.CI). PtCl4 80 methylamine hydrochloride forms a double salt-- (CHI.NHA. (2).PtCl, Such salts, on ignition, leave a residue of pure platinum. The molecular weight of the base is obtained in a similar way to the determination in the previous example, i.e.- Weight of platinum salt Weight of platinum forined 2 x molecular wt. of basc + 2(1 + 35.5) + 195 + 142 195 Constitutional Formulæ.- When the molecular formula has been determined, other data may enable the constitutional formula to be deduced; for these a text-book on organic chemistry must be consulted. They receive illustration in many of the following examples. 1. A substance containing carbon, bydrogon, and oxygen gave on combustion the following results : substance, 0.32; CO3, 0:44; H,0, 0-36. What is the simplest formula for the substance, and by what tests would you determine its nature ? (Inter. B.Sc., Manc.)

·ORGANIC ANALYSIS 93 2. If 0-360 gram of an organic compound which contains only carbon, hydrogen, and oxygen yields on combustion 0.528 gram of carbon dioxide and 0.216 gram of water, what is ita empirical formula? (= 12.) (1st "Conjoint," Sept., 1909.) 3. A substance containing only carbon, hydrogen, and oxygen was found by analysis to contain C : 52.15, H = 13:1 per cont.; & vapour donsity determination gavo A = 23. Determine tho molecular formula of the substance. (Inter. B.Sc., Manc., 1909.) 4. 0-37 gram of a substance containing carbon, hydrogen, and oxygon gave on combustion 0:66 gram of CO, and 0.27 gram of water. Further, 0.123 gram of the substance displaced 39•6 c.c. of air at 17° C. and 760 mm, in a determination of its vapour density by the Victor Moyer method. Ascortain from those data tho molecular formula of the substanco. (A litre of hydrogen at N.T.P. weighs 0:09 gram.) (Inter. B.Sc., Manc., 1906.) 5. A liquid containing C, H, and O gave on analysis the following rosults: 01746 gram gavo 0.3492 CO, and 0.1426 gram II,O. The vapour density of the substance was found to be 22. (II = 1.) Find tho formula of the liquid. (General Organic, Manc.) 6. The analysis of an organic substance gave the following results : 0.2115 gram on complete combustion gave 0:4655 gram of carbon dioxido and 0.2533 gram of water, Tho vapour density of the substance was 29.7. Calculate the molecular formula, and givo the names of the substances possessing this formula. (London Inter., Intl., 1908.) 7. 0·1027 gram of a substance gave on combustion 0.2178 CO, and 0.0303 H,0. 0·1130 gram gave 0.0643 AgBr. What is the simplest possible empirical formula of tho substance ? (General Organic, Mano ) 8. A substanco gavo tho following results on annlysis :-- (a) 0·1836 gram gave by combustion 0.1934 gram of Co and 0·0693 gram of 11,0. (6) 0.3213 gram gave 0-3616 gram of AgBr.

·94 EXERCISES IN OHEMICAL CALCULATION COHAP. What is the simplest formula for this substance ? (Bd. of Ed., Stnge I., Organic, 1905.) 9. A dibasio acid containing only carbon, hydrogen, and oxygen was found by analysis to contain 0 267 por cont., H 2.2 por cont. A determination of the vapour density of its dimethyl ester gavo A = 59. What is the acid ? (Inter., Manc., 1910.) 10. A substancu gave the following rosnlts on analysis :- 0-2500 gave 0.6027 CO, and 0.339 H,O 0.2000 gave 33.4 c.o. of nitrogen at 20° and 750 mm. Its vapoar density compared with hydrogen was 37. Calculate the composition of the substance and give the constitutional formulæ of the possible substances of this composition. (Bd. of Ed., Stage II., Organic, 1907.) 11. A ketoxime, on analysis, was found to possess the following composition :- Carbon, 55.2; hydrogon, 10%; nitrogen, 16.1 por cent. What is the formula of this compound and from what ketono is it derived ? (Bd. of Ed., Stage II., Organic, 1910.) 12. A substance gave the following results on analysis :- (a) 0:1836 gram gave, on combustion, 0-2241 gram of CO2 and 0.0420 gram of II,O. (b) 0.3212 gram gavo 0.3736 gram of AgBr. What is the simplest possible formula for the substance ? (General Organic Class, Manc.) 13. Two liquids, one boiling at 83° C. and the other at 57° C., gave, on analysis, the same results, viz. 0.2376 gram gave, on combustion, 0.2112 CO, and 0.0864 H,0; 0:121 gram gayo, when heated with silver nitrate and nitric acid, 0.351 AgCl. Calculate the simplest formula for these substances. How do you account for two substancos having the same composition, and what do you rogard as the true formula of each compound ? (0 = 12, Cl = 355, Ag = 108.) (London Inter. Hons., Extl., 1904.) 14. A substance gave the following results on analysis : 0-197 gram gavo 0-293 gram Co, and 0-150 gram H.O.

·ORGANIC ANALYSIS 95 By the Kjeldahl method 0.59 gram required 10 c.c. of N. H,80, for neutralization of the ammonia. The original substance, when boiled with caustio soda, evolved ammonia, and the dry residue, heated with soda lime, gave off methane. What is the substance ? (Leeds Univ., Genl. Organic, 1909.) 15. Calculate the formula of the aliphatio dicarboxylio acid which gave the following results on analysis :- 0.1340 gave 0:1760 CO, and 0.0540 1,0. (Lond. Scholarships, 1910.) 16. 0.2543 gram of a liquid bydrocarbon gave, on combustion, 0·8513 gram CO, and 0.1991 gram H,0; a vapour density determination gave D = 46. Find its empirical formula. The hydrocarbon, on oxidation, gave an acid of the empirical formula C,H,O.. Find the structural formula of the hydrocarbon. (Organic Class, Manc.) 17. The combustion of 0-246 gram of a liquid hydrocarbon yiolded 0·792 gram of carbon dioxide and 0.270 gram of water. The hydrocarbon absorbod bromine readily, and yielded a compound containing 79.6 per cent of bromino. From theso data, what conclusions do you draw as to the formula of tho hydrocarbon ? (C = 12, 0= 16, Br = 80.) (London Inter., Veterinary Science, 1909.) 18. 0.2 gran of an anhydrous acid gave, on combustion, 0·040 gram H,0 and 0:195 gram CO.,. The acid is found to be dibasic, and 0.5 gram of the silver salt leaves, on ignition, 0.355 gram silver. What is the simplest formula for the acid, and what is its molecular formula ? (Organic Class, Manc.) 19. A mono-acid organic base gave the following results on analysis = 0.100 gram gave 0.2882 gram CO, and 0.0756 gram water. 0:200 gram gave 21.8 c.c. nitrogen at 15° and 760 mm. 0:400 gram of the platinichloride loft, on ignition, 0.125 gram of platinum. What is the molecular formula of the base ? 20. Certain white crystals having, in aqueous solution, y strongly acid reaction were found to consist of H = 46 ·96 EXERCTSES IN CHEMICAL CALCULATION (ONAP. C = 19.05, 0 = 76.19 per cent. ; what is the simplest formula which would express the results of the analysis ? Is this the formula adopted for the compound? If not, why not? (London B.Sc. Pass, Extl., 1904.) 21. A hydrocarbon of the formula C2H, yields, on oxidation, a colourless crystalline dibasic acid, containing 57.8 per cent. of carbon, 3:6 per cent. of hydrogen. The silver salt contains 56.8 per cent. of silver. On heating the acid it is convertod into a compound of the formula C,H,O,; on distillation with lime, benzene is formed. Discuss the structure of the original hydrocarbon. (B.Sc., Manc.) 22. The monocyclic carbon compound A containing only carbon, hydrogen, and oxygen gave the following numbers on analysis : 0:1120 gave 0.2638 CO, and 0.0720 H,0. When heated with ethyl alcohol and 5 por cent. sulphuric acid, A was converted into a substance containing C = 68.46, H = 8:07. When A was treated with bromine in chloroform solution, bromine was absorbed, hydrogen bromide evolved, and a compound produced containing Br = 41.78 per cent. With hydroxylamino hydrochloride A gave a substance containing N = 19.84 percent. (H = 1, C = 12, N = 14, 0 = 16, 80). What information can you deduce from thoso data regard. ing the presence of any particular groupings in A, and what formula can you suggest which would be in accord with the observed facts ? (London Hong., Extl., 1909.) 23. A substance of the formula CH CH,--CO CA OH, CH,- -CH on oxidation with potassium permanganate gavo a product which was analyzed with the following rosalts :- 0:111 gave 0·2142 C0, and 0·0654 H,O. N 0.2019 roquired 12.85 с.c. of sodium hydroxide for neutralization in the cold and a further 11.9 c.c. on warming. Br = 10 ·ORGANIO ANALYSIS 97 What structural formula would you assign to this oxidation product ? (London Hons., Extl., 1908, adapted.) 24. The ammonium salt of a non-nitrogenous organic acid gave the following results on analysis : 0.2856 gram gave 0.3102 gram CO2 and 0.180 gram H2O, and 0.1562 gram gave 22.1 c.o. of dry nitrogen at 13° C. and 779 mm. prossure. Find the simplest formule for the salt and for the acid. (H = 1, 0 = 12, N = 14, 0 = 16; 1 litre of nitrogen at N.T.P. weighs 1-25 gram.) (London Hons., Extl., 1907.) 25. A ketone having the formula CH, - CO (CII, MC CH CH, CCI ca was reduced by means of sodium in absolute alcoholic solation; 0·1499 gram of the resulting compound (A) gave, on combustion, 0-3835 gram 00, and 0.1559 gram H20. Whon treated with benzoyl chloride A was converted into a substance B, which contained 73:75 per cent. of carbon and 8.85 per cont. of hydrogen. What are the most probable formule for A and B? Explain the reactions by which they were produced ? (London Hons., Extl., 1907.) 26. An acid, derived from p-xylene, gave the following results on analysis : 0.162 gram gavo 0-4015 CO, and 0.1430 HO. When treated with bronine, it was converted into a bromo-derivative containing 36•2 per cent. of bromine, and this, when digosted with alcoholic potash, yielded an acid which gave the following results on analysis: 0-17 gave 0.4276 CO, and 0.1309 H,0. Assign constitutional formalm to these acids, stating your reasons for their selection. (B.Sc. Hone., Mano., 1905.) 27. A dibasic acid gave, on analysis, the following results : 0·1740 gram gave 0.3388 gram CO, and 0-1008 gram H.O. When it was treated with bromine it yielded a dibromo derivative which gavo, on analysis, C = 26.4, H = 3-2, Br = 50•3. Oxidation with permanganate converted the ·98 EXERCISES IN CHEMICAL CALCULATION (ORAP. X. dibasio acid into dimothyl malonio acid and oxalic acid. Wbat is the constitution of the acid ? (B.Sc. Hons., Mano., 1906.) 28. An organic acid gave the following results on analysis : 0·1792 gave 0.3826 CO, and 0.1370 4,0. The silver salt contained 4380 per cent. of silver, and the oxime yielded 8.8 per cent. of nitrogen. What constitutional formulæ may the acid have, and how would you distinguish between them? (B.Sc. Hons., Manc., 1907.) 29. A cyclic hydroxy-monobasic acid yielded on analysis the following results: 0·1800 gave 0.3854 CO, and 0.1346 11,0. When this acid was treated with dehydrating agents, water was eliminated and an unsaturated acid formed which gave, on analysis, C = 66-7, H = 7.9. Oxidation converted this unsaturated acid into y-acetobutyric acid, CII,.CO.CH,.CH,.CII,.CO,H. Discuss the constitutions of the hydroxy-acid and the unsaturated acid. (B.Sc. IIons., Manc., 1911.) 30. An organic acid gave the following results on anolysis : 0.1531 gram gave 0•3883 gran of carbon dioxido and 0.1410 gram of water. Tho silver salt gave 41.1 per cent. of silver, and when the acid was heated with bromino and afterwards with alcoholic potash it gave an acid containing 70.1 per cent. of carbon and 9.1 per cent. of hydrogen. The latter on heating with sulphuric acid was converted into an aromatic acil which yioldeil m-xyleno on distillation with limo. Suggest formulæ for the threo acids. (Iust. Chem., July, 1908.) 31. The ethyl ester of a dibasic acid, to which a normal carbon chain is assigned, has the composition: C = 61.11; 9.26 ; 0 = 29.63 por cent. Whon heatod at 120° with: sodium and a trace of alcohol, this ester is converted into a substanco A of the composition: C = 63.53; H = 8.23; O = 28.24 per cent. From A, when boiled with dilute sul. phuric acid, a second substance B is obtained of the composition: 0= 73.47; H= 10-20; 0 = 16.33 per cent., the oxime of which contains N = 12:39 por cent. What is the probable constitution of B? (B.Sc. Hous., Lond., 1911.) H = ·CIIAPTER XI TIIERMOCHEMISTRY THERMOCHEMISTRY is the study of the heat effects accompanying chemical reaction. The union or reaction of two or more substances is accompanied by the liberation or absorption of a definite amount of energy, usually in the form of heat. From this circumstance, chemic reactions and compounds are distinguished as exothermie and endothermic. Exothermic reactions are accompanied by evolution, endothermic by absorption, of heat. Exothermic compounds are formed from their elements with evolution, endothernic compounds with absorption, of heat. Units. Quantity of heat is measured in calories, calorio being defined as the amount of heat required to raise 1 gram of water through 1° C.* This unit is inconveniently small for chemical purposes so that it is customary to use tho“ large " calorie which is one thousand times as great as the small calorie, and is the amount of heat required to raise 1 kilogram of water through 1° C. To distinguish between the two units, gram calories are usually written cal.; kilogram calories, Cal. There is also the unit of heat (K) used by Ostwald which is the quantity * This amount is not exactly tho same for all intorvals of one degros. A more precise definition stipulates that the water shall be warmed from 15-5° to 16-50. Variation in the magnitude of the calorie with the particular portion of the temperature scalo which is being considered is too small to enter into ordinary calculation.

·100 EXERCISES IN CHEMICAL CALOULATION COHAP. 1 of heat required to raise 1 gram of water from 0° to 100° O., and is very nearly equal to 100 small calories. Ostwald also uses the joule, which is 10,000,000 ergs, 1 and equal to small calories, as the unit of heat. 4:18 Thus 1 calorie = 4:18 joules, and 1 Calorie = 4:18 kilojoules (kj:). Nomenclature.-- When by suitable means a determination has been made of the heat change in a given chemical reaction, the result is represented in a thermochemical equation, e.g.- Zn + 0 = ZnO + 86-4 Cal. i.e. 65.4 grams of zinc combine with 16 of oxygen to form 81'4 grams of zinc oxide, the heat evolution being 86.4 large calories.* The quantity 86-4 is called tho heat of roaction between zinc and oxygen, the heat of combustion of zinc or the heat of formation of zinc oxide. The above equation is often abbreviated to (Zn, 0) = +86-4. Laws.-The basis of thermochemistry is an application of the doctrine of the conservation of energy to chemical phenomena, e.g. Lavoisier and Laplace arrived at the conclusion that the decomposition of a compound into its constituents absorbs exactly the same amount of beat as is evolved during its formation from those constituents. Thus the decomposition of zinc oxide is represented as an endothermic reaction by the thermochemical equation -- ZnO = Zn +0 - 86.4 Cal. 80 that the members of a thermochemical equation may be transposed in the same way as the members of an algebraic equation. • Thermochemical equations are generally reduced to the simplest possible form, the above expression, for instance, being preferred to the more precise chemical equation 2Zn + 0, = 2200. Thus Zo + 0 = ZnO + 86.4 Cal. indicates the total thermal change due to the union of ·THERMOCHEMISTRY 101 G. H. Hess, in 1840, extended the conception by formulating a law which states that the total heat effect of a chemical action is the same whether the reaction proceeds in one stage or in several, so long as the initial and final states are the saine. For instance- H + Cl = HCl + 22 Cal. If, however, the hydrogen chloride is produced by means of the following processes :- H + Br = HBr + 13.5 Cal. HBr + Cl = HCl + Br + 8.5 Cal, it will be noted that tho suin of these two equations- H + Cl = 1[Cl + 22 Cal. representing the net thermal and chemical effect, is identical with the above. The truth of this law for all cases in which it bas been tested leads to its application in determining the heat evolved in reactions which cannot be conducted under conditions suitable for measurement, For instance, from the data- C+02 = CO2 + 97.0 Cal. CO + 0 = C0, + 68.2 Cal. it is deduced, by subtracting the second equation from the first, that- C+0= CO + 28.8 Cal. Influence of Physical State.--It must be remembered that since heat changes also accompany processes which are not chemical (6.g. melting, boiling), it is necessary to know the conditions under which the reaction has been carried out. The magnitude of the error which inay be 65-4 grams of zino with 16 grams of oxygen. No attempt is made to indicate what part of the heat effect is due to the dissociation of oxygen molecules, and what part is due to the union of zino atoms with oxygen atoms, or the volatilization of the zipo or the condensation of zinc oxide, ·102 EXERCISES IN CHEMICAL CALCULATION COHAP incurred by neglect of this is illustrated by the following equations : H2+0=H2O (ice) + 69.8 Cal.) difference=1.4 Cal. H2+0=H20 (liquid) +6864 gás at 1000 =9.7 and 760 mm. Hg+0=1,0 ( ) The differences represent respectively the latent heat of fusion and the latent heat of vaporization of the grammolecular weight of water. In thermochemical equations the reagents are assumed to be in the state of aggregation in which they would exist at ordinary temperature and pressure, unless the contrary is stated. Reactions in Aqueous Solution.--The thermal change due to the solution of substances in water must often be known for thermochemical calculations dealing with reactions in solution. A difficulty in dealing with this heat of solution is that it depends on the concentration of the resulting solution, as shown by the following numbers obtained by Thomsen for sulphuric acid dissolved in water :- H2SO4, H,0 6.38 Cal H2SO4, 2H,0 9.42 H2SO4, 311,0 11•14 HI,SO4, 19H2O 16.26 H,SO4, 199H2O 17:06 H2SO4, 799H,0 17.64 H2SO4, 1599H20 17.86 Obviously, if the solution is sufficiently dilute, the heat effect of further dilution is negligible, and the heat of solution of sulphuric acid at great dilution is expressed by the equation- HSO, + ag H2SO4aq + 17.9 Cal ... 2 ... ...

·XI.) THERMOCHEMISTRY 103 This is quite different from the following :- H2SO4 + H2O H2S0.4,0 + 6-4 Cal. which refers to the formation of a definite hydrato of sulphuric acid. The symbol aq is a contraction for aqua, and means an indefinitely large amount of water; this is practically the same in thermal result as a moderately large excess of water. It is to be noted that aq may frequently enter into specific reactions besides those concerned with simple dilution, e.g.- K + aq = KOHaq + H + 481 Cal. where the thermal effcct consists of the heat of reaction of potassium with water and the heat of solution of the resulting potassium hydroxide. If the equation is to be used algebraically it should be considered as- K + H,0 + ag = KOHaq + H + 48:1 Cal. EXAMPLE,- Given K + aq = KOHaq + H + 481 K (i.) 12+ 0 = 11,0 + 684 K (ii.) KOIL + aq = KOHaq + 133 K (iii.) l'ind the leat of formation of KOH from its elements (K, O, H). (i.) inay be written- K + H2O + aq = KOHaq + II + 481 K Also H, + 0 = H,0 + 684 K By addition K +0+ II + aq = KOHaq + 1165 K . (iv.) (iii.) is- KOH + aq = KOHaq + 133 K By subtracting (ii.) from (iv.) K+0+H - KOH = 1032 K or K + 0 + H = KOH + 1032 K ... .I.

·104 EXERCISES IN CHEMICAL CALCULATION COHAP. or EXAMPLE. - Given CuSO4ag + Fe = FeSO4aq + Cu + 37240 cal. ... Fe + H2SO4aq=FeSO aq + H2 + 24840 cal. (ii.) H, + 0 =H2O + 68360 cal. .. (ii.) CuO+H2SO4qq=CuSO,aq + 18800 cal. (iv.) Find the heat of formation of CuO. Equation (iv.) is CuO + H2SO4aq CuSO4aq + H20 + 18800 By substitution from equations (ii.), (i.), and (iii.), for H2SO4aq, CuSO4aq, and H20- CuO + (FeSO4aq + H, - Fe + 24840) (FeSOcaq + Cu Fe +37240) + (H) + 0 - 68360) + 18800 Whence by cancelling- CuO = Cu + () -37160 Cu + 0 = CuO + 37160 cal. 1. Given 21 + Cl, = 2ICI + 11650 cal. ICI + Cl, = ICI, + 15660 cal. Find the hoat of formation of IC1, from its oleinonts. 2. On dissolving 100 grams of auhydrous copper sulphate in water, the heat evolved amounted to 9900 cals., whilst the sanio weight of crystallized coppor sulphate (Cu$0,5H,0) gavo an absorption of heat amounting to 1100 cals. : determine the hoat of transformation of the gram-molecule of Cuso, into Cu80,55,0. (B.Sc. Hone., Manc., 1905.) 3. Anhydrous magnesium chloride is to be propared froni magnesia, chlorine, and carbon. The beat of combination of magnesium and oxygen, of magnesium and chlorine, and of carbon and oxygon are 145860, 151000, and 28600 respectively in gram-calories per [gram] molecule. Calculate the balance of gram-calories given out or absorbed. (Inst. Chem., Inter., July, 1910.) 4. Find the heats of formation of the following hydrocarbons from their respective boats of combustion, givon that ·21.] THERMOCHEMISTRY 105 19 [C, 0] =96960 cal. and [H., 0] = 68360 cal. Methane, 211930; etbane, 370440; ethylene, 333350; acetylene, 310059; benzene (vapour), 799350 cals. per gram-mol. 6. Calculate the molecular heat of reduction of ethyleno to ethane, given the following data :- Molecular heat of combustion of hydrogen 64:49 Calorios. ethane 369.0 ethyleno 332.2 (Hons., Manc., 1910.) 6. From tho following data calculate the thermal effect produced by the decomposition of ozone by hydrogen peroxide : (H,0) = 68,000 cals. (,,0) = 45,000 cale. (0.2, 0) - 30,000 cals. (Hons., Vict., 1894.) 7. Calculate the heat of formation of formic acid from the following data - Ficat of combustion of carbon (C, 0,] = 96,960 сals. hydrogen (H., 0] 68,360 formic acid [CH,O.,,O] 65,900 (2nd Year Hons., Mnuc., 1910.) 8. Find tho hoat of formation of aldehydo, (i.) gaseous, (ii.) liquid, from the following data :- (20,1,0, 50,) liquid = 551 Cals. (20,11,0, 502) gasoous = 564 (0, 0,) 97 (2H,, 0.2) to water = 136-7, (B.Sc. Hons., Manc., 1908.) 9. Calculate the heat of combination of calcium and oxygen from the following equations :- CaO + 2HCl = CaCl, + 1,0 + 52.7 Col. Ca + Cl, = CaCl, + 169.8 Cal. H, + 0 = H2O + 57.9 H + Cl = HCl + 22:0 (Inst. Chem., July, 1907.) 99 ·106 EXERCISES IN CHEMICAL CALCULATION COHAP. 10. Given [Na, ag] = 43,450 cal. [H, Cl] = 22,000 [HCI, aq] = 17,314. [NaOHaq, HClag] 13,745 [NaCl, ag] =-1180 Find the heat of formation of sodium chloride from its elements. 11. From the following figures of Thomson- [Fe, 2HClaq] = 21,320 cal. [FeClaq, CI] = 27,770 [FoCl,, aq] = 17,900 [TeCl3, aq] = 31,680 [H, Cl] = 22,000 , [HCI, (q) = 17,314 find the heats of formation of forric and ferrous chloridos from their elements. 12. From the following data calculato the honts of forma tion of gaseous hydrochloric and hydriodic acids respectivoly from thoir olemonts : II, + Cl, (gas) + aq = 2HClaq + 78,600 cals. HCI (gas) + aq = HClaq + 17,300 cale. 2H/aq + Cl, (gas) = 2HClaq +1, +52,400 cale. Al (gas) + aq = Hlaq + 19,200 cals, (B.Sc. Hons., Manc.) 13. From the followiug data calculate tho hcat of formation of solid potassium chloride from its elements :- (i.) KOHaq + HClaq = KClaq + 1,0 + 137.5 k (ii.) H, + Cl2 + aq = 2HClaq +- 786 k (iii.) 2H, + 0, = 24,0 + 1368 k (iv.) 2K + 2H,0 + aq 2KOHaq +H2+ 962 k (v.) KCl + aq = KClaq - 4+k (B.Sc. Hons., Vict., 1902.) 14. The heat of formation of hypochlorous acid has boon determined by two indopendent methods by Thomson ; his 9 .

·THERMOCHEMISTRY 107 rosults are givon below. From both sets calculato the heat of formation of an aqueous solution of the acid (H, CI, O, aq). A. B. (2NaOHaq, Cl,) =24,647, (HClOaq, 2H/aq) = 51,435, (NaOHaq, HClOaq)= 0,976, (H, I, aq) = 13,171, (NaOHaq, HClaq) = 13,740, (II, CI, aq) = 39,315, ( HO) 68,357. (London Hons., Extl., 1907.) 15. In order to determine the heat of formation of ainmonia, Thomsen passed chlorino into its aqueous solution, and found for the reaction [4NH307, 301] an evolution of 119,613 hent units; ho also found for- [H, Cl] = 22,001 [NII,aq, HClaq) = 12,270 [HCl, aq] : 17,314 [NII, aq] = 8,435 but afterwards by tho combustion of ammonia he found- [2NH,, 30] 181,296 11,0) = 68,357 ... . Calculate the heat of formation of ammonia gas from each set of duta, and compare and contrast the reliability and probable sources of error in each method. (London Hons., Extl., 1905.) 16. Find the heat of formation of hydrogen sulphate from its elements from the following data - (4,80,119, 11,0, C1,) = 11,50 aq + 20Claq +- 73,900 cal. (i.) (II, Cl, aq) HClaq + 39,300 cal. (ii.) (12, 0) = 4,0 + 68,400 cal. (iii.) (SO,, aq) = H,80,aq +7700 cal. (iv.) (8,0) = 80, + 69,900 cal (H,90,, aq) = H,SO aq + 17,000 cal. (vi.) 17. From the following data, based on the use of seminormal solutions throughout, calculate the heat which should be liberated in the production, in aquoous solution, of one gram-moleculo of ammonium acetate. 250 0.0. of caustio soda ... ... ...

·108 EXERCISES IN CHEMICAL CALCULATION [OHAP. mixed with an equal volume of hydrochloric acid liberate 1710 calorios; the samo volumes of caustic soda and acetio acid liberate 1020 calories, and of ammonia and hydrochlorio acid 1530 calorios. For the purpose of the calculation it is assumed that, in the solutions employed, tho caustio soda and hydrochloric acid are fully dissociated. (B.Sc., Lond., 1910.) 18. 0.5599 gram of peat gave on combustion with excess of oxygen an evolution of 2521 calories of heat. Calculate (i.) the number of calorios 'evolved by the combustion of one gram of peat; (ii.) the number of B.T.U.'s evolved by the combustion of 1 lb. of peat. (B.T.U. means British Thermal Uuit; it is the amount of heat required to raise the tow peratura of 1 lb. of water through 1° Fahr.) 19. A gaseous hydrocarbon was annlszod by explosion with tho following result: 20 c.c. of hydrocarbon with 80 c.c. of oxygen gave 50 c.c. of residual gas, of which 40 c.c. wero Co, and 10 c.c. 02. When 2-86 litres of this bydrocarbon (measured over water at 13° C. and 751 mm.) wero burnt in a Junker calorimeter, 2.0 litres of water wero hcatod from an average temperaturo of 4' to an averago temporature of 22', and 5 c.o. of water were condensed from the products of combustion. Calculate tho molecular hoat of combustion (calories evolved by burning 22-4 litres) when the hydrogon of the hydrocarbon is burnt to steam and the carbon to carbon dioxide, Latent hoat of stcam 540 calories. Pressure of aqueous vapour at 13° = 11 mm. (B.Sc. Hons., Manc.) 20. Given SO, (dissolved) + Cl2 + H,0 = H,SO. (dissolved) + 2HCl (dissolved) + 73,907 cal. HI, + 0 = H,0 (liquid) + 68,360 H + Cl = HCI + 22,000 FICI + water = HCl (dissolved) + 17,314 8 + 0, = SO, + 71,080 80,+ water =SO, (dissolved) + 7700 80+ water = IL,SO. (dissolved) + 39,170 ·IL] THERMOCHEMISTRY 109 Find the thermal effoot of the following reactions : So, (dissolved) + 0 = H,80. (dissolved) SO, +0+ water = H SO, (dissolved) 8 + 30+ water = H,80, (dissolved) and 8 +0,= SO, Heat of Reaction at Constant Pressure and Constant Volume. -Up to this point heat values have been expressed for reactions occurring at constant pressure, namely, atmospheric. If a chemical reaction at constant pressure results in the production of gas of different volume from that which was present initially, then work is done by the system if expansion has occurred, or on the system if contraction has occurred. Hence the heat evolution is respectively less or greater than that which would occur had no work been done. The difference is frequently significant, especially in the case of the determination of heats of combustion. In the Berthelot bomb the reaction is performed at constant volume; in some other forms of calorimeter at constant pressure. It is useful, therefore, to be able to transform each value of the thermal effect of a reaction into the other. This is dono as follows: The production of 1 grain-molecule of a gas at constant pressure necessitates the use of pv units of work (see text-book). Now, po = RT, so the work done = RT, which in calories = 2T. Hence the production of n grammolecules of gas involves work done by the gas equal to 2nT calories, where n may be positive or negative or zero. EXAMPLE The combustion of methane at constant pressure- CH4 + 202 = CO2 + 21,0 + 211.93 Cal. at 15° C. (initial and final temperatures) has resulted in the absorption of + 2 x 1 x 288 calories, i.e. + 0.576 Cal., on account of the decrease in volume of 1 gram- ·110 EXERCISES IN CHEMICAL OALCULATION (ORAP. molecule of gas (the water is assumed to be entirely condensed). Hence the heat of combustion of methane at constant volume 211.93 -0.576 = 211:35 Cal. 21. Find the heat of combustion at constant volume of ethane, ethylene, and acetylene, given that their heats of combustion at constant pressure are 370440, 333350, and 310059 calories respectively at 15° C. 22. Find the heat of combustion of bonzene vapour at constant volume, given that its heat of combustion at constant pressure is 801.2 Cal. at 120° C. Calorific Power. The calorific power of a substance is the number of units of heat which unit amount of it (1 gram, 1 kilogram, 1 pound, 1 cubic foot, as the case may be) will produce on complete combastion. This term is used rather than “heat of combustion " for combustible substances which are not pure chemical compounds and whose oxidation cannot be represented in a single equation. As water is generally a product of the combustion of such & substance, a distinction is made between the gross calorific power, which incluiles the heat of condensation of the water, and the nett calorific power which corresponds to the formation of water vapour. Calorific Intensity.--The calorific intensity of a reaction is the temperatme attained in it on the assumptions (i.) that the reaction is complete before any heat is lost by radiation from the flame; (ii.) that the specific heats of the products of the reaction are known. The calorific intensity of a reaction, in degrees Centigrade, total heat evolved heat required to raise the products through 1° 0. 23. Calculate the gross calorific power of a cubic foot (at 0° and 760 mm.) of dry gas containing Co, = 13-2; CO 15:3; H2 = 19.5; CHA 3.8; N = 48.2 per cent., assuming the following molecular heats of combustion :-- ·THERMOCHEMISTRY 111 12 70 > H, + 0 = 11,0 (water) = 48•4 kilogram-centigrade units. CO +0= CO, = 68.0 CH,+ 20% = CO + 2H,O = 213-0 [N.B. 1 cubic foot = 28.3 litres.] (Fuel, Manc., 1905.) 24. A regenerative furnaco holds 10 crucibles, each contuining 100 lbs. of steel. Assume- (1) That the total lont in cach lb. of stool when melted = 300 calorios. (2) That the furnaco is fed by gas with a calorific power of 480 calories per cubic foot. (3) That only 5 por cent. of the heat generated is used for melting the steel. Required: The number of cubic feet of gas for molting 1 tou (= 2000 lbs.) of stcel. (llons., Mano., Metallurgy, 1907.) 25. Calculato tho temperaturo of the hydrogen flame burning in (a) oxygen, (b) dry air. Hoat of combustion of 1 cubic metre of hydrogen is 2613 Calories. The inean specific heat of wator vapour between 0' and tº 0-34 + 0.00015t, and of oxygen and nitrogen 0.303 + 0.000027t. (Fuel Class. Manc., 1908.) 26. Calculate the calorific intensity of carbon burning in oxygen and in air respectively if (C,02) = 96960 specific heat of carbon dioxide = 0.2164 specific hent of nitrogen = 0·2438 Composition of air = 79 volumes of nitrogen and 21 volumes of oxygen. (London Pass B.Sc., Extl., 1903.)

·CHAPTER XII SOAIE PRINCIPLES OF PHYSICAL CHEMISTRY THE examples of this chapter and of the previous one cover a great part of physical chemistry. The student who understands his text-book will find himself capable of solving most of them without further explanation; to attempt explanation would involve the writing of a textbook. One point, derived from experience of teaching students, may be mentioned. It is, as a rule, safer and more permanently useful to work from principles than to memorize formulæ. Just as in correcting a gas volume to NT.P., inost experienced chemists seem to remember the principles and apply them as best fit the particular case, so, for example, in deducing a molecular weight from an observation of the depression of a freezing point, most chemists use the principles, but do not remember the formula. The difference is that instead of putting the principles into a formula and translating that to the particular case in point, the direct method is used. A few of the following examples, more especially some of those in the sections on rate of diffusion of gases, solubility of gases, molecular weight determinations and electrochemistry, may be attempted by the junior student. He may choose those obtained from Intermediate B.Sc, papers, and some others.

·CHAP. XII.) PHYSICAL CHEMISTRY 113 Deviations from the Simple Gas Laws. 1. Guye found that the weight of a normal litre of the gas obtained by the decomposition of ammonia was 0-3799 gram. Taking tho weight of a normal litro of hydrogen as 0.0899 gmm and the weight of a normal litre of nitrogen as 1.2507 gram, then (p+ (1) On the assumption that the gases obey Boyle's Law, calculato the combining volumos of nitrogen and hydrogen at 0° C. and the atomic weight of nitrogen. (2) If, owing to deviations from Boyle's Law, the litre of nitrogen at 0° C. bocamo 1.0004 litres when diluted with the 3 volumes of hydrogen, and tho litre of hydrogen at 0° C. bccame 0.9999 litro whon diluted with one-third of its volume of nitrogon, at atmospherio pressure, what effect would theso deviations have on the determination ? (Hons., Manc., 1910.) 2. Calcnlate the critical temperature and critical pressure of nitric oxide from the following equation : 0.00257 ) (0 – 0-001156) = RT. (Inst. ('hem. Phys., 1910.) Rate of Diffusion of Gases. 3. If 50 c.c. of by,rogon take 10 minutes to diffuso out of a vessel, how long will 20 c.c. of nitrogen tako to diffuse under the same conditions? (H = 1, N = 14.) (Bd. of Ed., Stage I., Inorg., 1910.) 4. If 50 volumes of hydrogen take 85 seconds to diffuse from a vessel, how long would it take 35 volumes of nitrous oxide to do so under the same conditions ? (London Pre. So., Vet., 1905.) 5. Tho relative rates of diffusion of oxygen, hydrogen, and nitrogen are respectively 0.96, 3.80, and 1.015. What dodaotion may be drawn from those figuros concerning the donsity of nitrogen? (London Inter., Intl., 1908.) 6. Soret found that when chlorine mixed with oxygen, diffused through a small aperture into oxygon, the proportion E ·114 EXERCISES IN CHEMICAL CALCULATION [CHAP of the diffused chlorine to that remaining in the Vossel in 45 minutes was reprosented by the number 0.2270; whilst in & similar experiment made with ozonized oxygen the proportion of the diffuscd ozone to that remaining in the vessel in 45 minutes was 0.2708. From these numbers determine the density of ozone (a) with chlorine as unity, and (b) with hydrogen as unity. (London Inter., Veterinary Science, 1909.) 7. If, in Soret's apparatus, 100 c.c. of HCl gas had been mixed with oxygon and 18 c.c. had diffused in half an hour, what molecular weight would yon assign to an acid gas of which 12 cc. (out of 100 c.c.) diffused undor the same conditions ? (1st Year Hons., Manc.) 8. Two vessels, A and B, contaiuing respectively chlorine and a mixture of oxygen with another gas, aro connected by means of a short capillary tube. The mixture contains 10 por cont. of oxygen. Alter diffusion for a limited time, tho chlorine in A is absorbed by potash, and the rosidual gos contains 11.05 per cent, of oxygen. What is the density of the second (Fions., 1903, Manc.) Solubility of Gases. 9. Air, containing 20.9 por cent. of oxygen, 79.0 per cent. of nitrogen, and 0:1 per cent. of carbon dioxido, is bubbled through water for half an hour. Calculate the composition of the mixture of gases which would bo obtained on boiling the solution, given the following coefficients of solubility at the temperature of the water: oxygen = 0.040, nitrogen = 0.020, carbon dioxide = 1.79. 10. A natural water, on boiling, gave a mixture of gases of the following composition : oxygen = 20.9 per cont., nitrogon 39 5 per ceni.g carbon di'xido = 39.6 per cent. Assuming that the water had dissolved these gases from the air with wbich it had previously been in contact, calculate tho composition of that air. Use the coefficients of solubility giver in the previous question. 11. One litre of a mixture of oxygen and carbon dioxide gas?

·ITT.) MOLECULAR WEIGHTS 115 in equal proportions by volume is shakon with 1 litro of water at 0° C. until equilibrium is established. The initial pressure is 760 mm. Calculate the final pressure and the composition of the rosidual gas. (The coefficients of absorption at 0° are : oxygon, 0.01; carbon dioxide, 1.79.) (Inst. Chem., Inter., July, 1906.) 12. Wator is shaken up with its own bulk of a mixture of 1 volume of nitrogen and 3 volumos of nitric oxide. Supposing the temperaturo and pressure to remain normal and constant throughont the experiment, calculate the composition of the residual gas. Bunsen's coefficient for nitrogon is 0.0235, and for nitric oxide, 0.0738. (Inst. Chem. Phys., July, 1910.) Molecular Weight Determinations-Osmotic Pressure-Freezing and Boiling Points and Capillarity. 13. Admitting the analogy between gaseous and osmotic pressures, calculate the osmotic pressure of a 1 per cent. cane sugar solution at 0°. (Leeds Univ., Phys. Chem.) 14. Calculate the osmotic pressuro of an aqueous solution containing 5.00 grams of dissolved substance por litre given that the solution freezes at -0.25º C. The molecular lowering of the freezing point of water may be takon as 18.5°. (Locds Univ., Physical Chem.). 15. 3.29 grams of a substance dissolved in 100 granis of wator (K = 19) gave a doprossion of the freezing point of 2:04° C. Find the molecular weight of the substance. (London Inter., Intl., 1908.) 16. Successive quantities of 0.317, 0-394, and 0-5152 gram of a substanco woro dissolved in 18:054 grams of benzene. The depressions of freezing point were 0-278°, 0-348°, and 0-452° rospectively. What was the molecular weight of the substance ? The molecular lowering of the freezing point of benzeno is 50°. (Inst. Chom., July, 1902.) 17. Tho electrical conductivity of a 5 per cent. solution of chromium trioxide is not increased by further dilution, and its ·116 EXERCISES IN CHEMICAL CALCULATION COHAP, freezing point is – 1.34º. What conclusions can you draw as to the condition in which the oxide exists in solution ? (K = 19.) (Hons., Mano., 1907.) 18. A current of dried air was passed through a solution of 3.458 grams of a substance in 100 grams of ethyl alcohol, and then passed through ethyl alcohol alone. The loss in weight of the former liquid was 0.9675 gram, and of tho aloohol 0.0255 gram. Find tho molecular weight of the dissolved substanco. (B.Sc. lIons., Manc., 1907.) 19. 2.012 grams of resorcinol are dissolved in 100 grams of ether. The riso in the boiling point is 0-399º. Find the molecular weight of the resorcinol, being given that the boiling point of ether is 34.8°, and the latent hoat of vaporisation 84.5 calories. (Inst. Chem. Phys., 1910.) 20. Calculate the latent heat of fusion of tin from the following data : melting point of tin 232° 0. The depression of the freezing point for 1 gram-atom of silver per 100 gram-atoms of tin is 3.0° C. (London Hons., Intl., 1907.) 21. Show from the following data that the molecules of formic acid in the liquid state are associated, but that they gradually dissociate as the temperature riges. Radius of cn pillary = 0.01425 cm. Temperaturo 16.8 46.4 79.8° 0. Height of liquid 4.442 4.205 Density 1.207 1.170 1.129 (Inst. Chom. Phys., 1909.) 22. On heating nitrocamphor (formula C.H.NO,) only water is eliminated, and the other product gave the following rosults on analysis :- 0-3391 gram gavo 21.8 c.c. of moist nitrogen at 19° and 767 mm. 0.4332 28.2 21° 760 0.1650 0-3858 gram of carbon dioxide and 0.1138 gram of water. The molecular weight was dotermined by the elevation of the boiling point of benzene. .. 3.90 cm. 9 > 1 2 ·10.) VELOCITY OF REACTION 117 2.8487 grs. raised the briling point of 11.10 grs, of benzene 2.011° 1.8842 1.214° 1.9012 1.191° 59 Molecular elevation of boiling point for benzone = 26.7. What is the probable formnla of the product? (Lowry, J.C.S., 1898, 73, 997.) Velocity of Reaction, 23. Tho spontanoous change of radiuni omanation, freshly drawn, is monomolccular reaction. The rosults obtained wero Time, 0 7 11 14 16.5 20 25 36 45 6+ 75 Volume, 0.102 0.062 0.044 0.033 0025 0.019 0.016 0.007 0.003 0 002 0.000 Show bow these figures fit the order of the reaction. (London Hons., Intl., 1907). 24. Diazoamido benzene (C.H..NH.N.C.H;) is converted into amidoazobenzeno (C.A.N.C.H.NH) under the influence of auilino hydrochlorido. The former compound liberates two-thirds of its nitrogen when boiled with dilute acids ; the Intter gives no nitrogen under these conditions. The progress of the reaction was therefore followed by withdrawing from time to time a weighed portion of tho reaction mixture and determining the volume of nitrogen evolved. From tho following results of a series of moasurements show that the reaction is monomolecular. The original mixture consisted of 0.8475 gram of aniline hydrochloride, 4.925 grams of ·118 EXERCISES IN CHEMICAL CALOULATION COHAP. diazoamidobenzene and 50 grams of aniline, which acted as solvent Time in hours). Weight of sample. O.o. of moist Ns. Bar. Temp. 18.11 2-645 22.8 750 14:0° 66.25 5.255 21.8 748 12 50 115.2 5.940 12.9 753 8.5° (Inst. Chom., Phys., July, 1911.) 25. In an investigation of the reaction betwoen hydrogen peroxide and hydriodic acid tho following results were obtained : Experimont I. Concentration of III, 0·02; of I1,02, 0.02 II. 0.01 III. 0.005; 0-005 The times taken for the completion of ono-fourth of the reaction were respectively: 21-7, 43-2, and 974 minutes. What is the order of the reaction ? (B.Sc. Hons., Manc.) 26. The transformation of yolydroxybutyric acid into its lactono is a reversible action of the first order, as represented by the equation, CH,OH.(CH2)2 - CO.IIeJH, - (CHI) - CO + HNO 0-01; 20 log Assuming that only the acid is prosout in a solution at tho commencement of the ronction, show that the course of the reaction will be represented by the formula - А =const. А. (1 + ada where A is the conceutration of tho acid at the commencement of the roaction, x is the amount of lactone formod after the time t, and a is the ratio of the concentrations of the lactono and acid when cquilibrium is attained. Verify tho formula for the following valucs of t and *, when A is 18.23 and a is 0.373. t = 80 100 160 7.08 8.11 10:35 (Inst. Chem. Phys., Jan., 1911.)

·XII.) VELOCITY OF REACTION 119 27. The concentrations of a compound undergoing complete change were 5'72, 3-23, 1.96, and at the times 10, 30, and 60 hours from the commencement of the reaction. Ascertain the apparent order of the reaction, the probable initial concentration of the compound, and the velocity constant. (Hons., Manc., 1911.) 28. The rate of roaction between iodine and potassium phenyl propiolate in equivalent quantities was determined (James and Sudborough, J.C.S., 1907, 1042) by measuring tho rate of disappearanco of the iodino. 25 c.c. of the solution Was withdrawn after suitable intervals froin tho reaction N mixture and titratod with sodium thiosulphate. 10 Time in houre, 0 20 24 Titration, 24.9 10:39 9.25 8.15 6:36 28 42 1 What is the order of the reaction, and what is the velocity constant ? 29. When electrolytio gas was circulated over porcelain heated to 450° C. the following observations were made of the pressuro of the gas remaining (dry) at intervals of 24 hours :- Time, days 0 2 3 4 5 Pressuro in mm. 465 232 116 58 29 11.5 What information do these figures afford as to the order of the change occurring ? (B.Sc. Hons., Manc., 1907.) 30. In the reaction between equivalent quantitics of potassium persulphato and hydrogen peroxido the rate of disappoarance of tho hydrogen peroxide was measured (Friond, J.C.S., 1906, 1094). The numbors given below reprosent N polassium permanganato equivalent to the pur- 50 oxide remaining in solution. What is the order of the reaction ? 0.0. of ·120 EXERCISES IN CHEMICAL CALCULATION Couar. Time in hours. C.o. of permanganate 0 14:06 1 13 19 2.5 12.05 3.5 11.35 10.36 7 9:16 31. Potassium persulphute and potassium iodide interact with the liberation of iodine. 25 c.c. of a solution which was si normal with respect to both tho persulphate and iodide, wero titrated from time to time with ido normal sodium thiosulphate. From the followiny rosults show that the reaction is bimolecular, t being the time of titration and the number of c.c. of thiosulphuto used. t 9 16 32 50 4:52 7.80 14:19 20.05 (Inet. Chem. Phys., June, 1909.) C Equilibrium. 32. Amyl bromide dissociates according to the equation- CII Br = C,H1,0 + A Br The donsity of amyl bromide vapour comparod with air = 5:2 at temperatures up to 150° C. At 215° its density is 4:12 compared with air at the same temperature. Find the percentage dissociation at 215°. 83. The density of sulphur vapour varios gradually with the temporaturo, some of the values obtained being as follows: Temperature (°C.) 468 524 606 900-1100 Density (Air = 1) 7.81 7.09 4.73 2.23 What information do those numbers give as to tho nature of gaseous molecules of sulphur? 34. From the following numbers obtained by Natanson (Wied. Ann., 1885, 24, 465), for the oquilibrium N,0. 2NO, at 21°, calculate for each pressure (i.) the percentage ·201.) EQUILIBRIUM 121 ... dissociation of N,0, (ii.) the partial pressures of N20, and NO, respectively, (iii.) the equilibrium constant. Pressuro (mm.) 43.58 101.64 230.6 492.1 617.6 Density (Air = 1) 2.049 2-283 2-501 2.684 2.722 The vapour density of NO, = 1.589. (Air = 1.) 35. If nitrogen poroxide is onc-half dissociated undor cortain conditions, what proportion will remain undissociated when the volume is increasod to ton times the original value while the temperature remains the same? (Hons., Manc., 1910.) 36. In four cxporimonts in which one gram-molecule of acotic acid was allowed to react with the following different quantities (o grau-molecules) of ethyl alcohol until equilibrium was established, Berthelot and l'éan de Suint Gilles found that the following amounts of ethyl acetate (y grama-molecules) were produced : (i.) (iii.) (iv.) 0.5 1.0 2.0 12.0 y 0.42 0.66 0.83 0.98 Show that these results aro in accord with the law of Guldborg and Waage. (B.Sc. Hone., Vict., 1902.) 37. 1 litre of iodine vapour at 1092º and 760 mm. was found to weigh 2.25 grams. Calculate the proportion of monatomic molecules. 38. 2.447 grams of chloral hydrate dissolved in 100 grams of acetic acid lowered its freezing point 0.755º. Assuming K = 39 in the caso of acetic acid, calculate the degree of dissociation of tho chloral bydrate. (B.Sc. Ilons., Vict., 1903.) 39. Bodonstein (Zeitschr. Physikal. Chem., 1897, 22, 16) found that at 443° the dissociation of hydrogen iodide, according to the equation 2H1 = II+ In, was 21.98 per cent. For experiments at the same temperature starting with varying amounts of hydrogen and iodine, the amount of hydrogen iodide present when equilibrium was attained is given by the following numbers:

·122 EXERCISES IN CHEMICAL CALCULATION [CHAP. Iodine (0.c. vapou:). Hydrogen (0.0.) Hydrogen iodideo.c.) Initial. Initial. Found. 2.94 8:10 5.66 5.30 7.94 9.52 9.27 8.07 13:34 14:44 8.12 14.82 27.53 8.02 15.40 33.10 7.89 15.12 Calculate from the first two columns, the amounts of bydrogen iodide predicted by the law of mass action, and compare them with the experimental numbers. 40. At 11° the distribution coefficient of iodine botween carbon tetrachloride and water is 79. A solution of iodine in tenth-normal potassium iodido was shakon up with carbon tetrachloride. After equilibrium had beon attained 1 litre of the carbon tetrachloride phase contained 0·02022 grammolecule of iodino, whereas the same volumo of the aqueous phase contained 0-02118 granı-molecule of iudine. Calculato the equilibrium constaut of the reaction KI +1, 2KIZ. (Inst. Chem. Phys., January, 1911.) 41. The following data (Richardson, J.C.S., 1887, 51, 401) show the variation in density of uitrogen peroxide with change of temperature, produced by the dissociation 2NO, = 2N0 +0, Volume Temperature. (corrected to N.T.P.). Weight. 184° C. 143.5 C.C. 279° C. 116.6 0.226 494' 0. 58.6 0.093 620° C. 19.6 0.027 What is the percentage of the total nitrogen peroxide which is dissociated at each temperature ? 42. For the system 200 =CO, + C, tho following results wero foaud (Rheud and Wheeler, J.U.S., 1910, 97, 2178): The equilibrium mixture contained Temperature. Carbon dioxide. Carbon monoxido. 850° C. 6:23 per cent. 93.77 per cent. 10002 0. 0:59 99.41 1200° 0. 0.06 99 94 0.287 gram.

·xn.) ELECTROCHEMISTRY 123 Calculate the constant in the expression 19500 T O CO + loge Coug whore T is the absolute tom porature, Cw and Cw, are the concentrations of carbon monoxido and carbon dioxide respectivoly. Electrochemistry. 43. An electric current is passed simultaneously through the following solutions : hydrochloric acid, ferrous sulphate, ferrio sulphato, and silver potassium cyanido. If 5-2 litres of hydrogon [at N.T.P.) were cvolved from the hydrochloric acid solution, how much metal would be deposited in the case of the iron and silver salts ? (Fo=56, Ag=108.) (B.Sc., Manc.) 44. An electric current is passed through a series of colls containing respectively acidified water, and aqucous solutione of common salt, copper sulphate, silver nitrate, and potassium acetato. What is the nature and amount of substance liberated in each cell if the current is allowed to pass until 50 c.c. [at N.T.P.] of oxygen have been liberated in the first cell? Na = 23, Cu = 63-5, Ag = 108, K = 39. (Bd. of Ed., Stage III., Inorg., 1906.) 45. A current of electricity is passed through thrco cells in sories. The first is provided with carbon electrodes, and the electrolyte is a 10 per cent, solution of hydrogen chloride. The second has & copper cathode and a carbon anode, and the electrolyte is a solution of copper sulphate slightly acidified with sulphuric acid. The third has a silver anode and a coppor cathode, the electrolyte being a solution of silver nitrate, slightly acidified with nitric acid. State what reactions occur in ench cell, and calculate the weight of the products at the anodo and cathode of each cell per ampere-hour from the following data: ono coulomb of electricity liborates 0.0000105 gram of hydrogen. (Inst. Chem., Min., July, 1910.) 46. Aluminium is manufactured by electrolysing a mixture of alumina, alaminium fuoride, and sodium fluorido. The ·124 EXERCISES IN CHEMICAL CALCULATION [OHAP. heat of combination of A1,0, is 392 Cal., of AIF, is 249 Oal., of NaF is 102-6 Cal., all being stated in large calories por gram-molecule. Current is passed from one electrode to the other at a pressure of 6 volts. Stato what reactions occur and why; and calculate that fraction of the energy which is used in producing aluminium, and also state what use is made of the balance. (Ono large calorie = 4200 joules; one gramequivalent requires 96,540 coulombs.) (Inst. Chem., Min., March, 1909.) 47. A current which deposited 0.101 gram of silver in a silver voltameter was passed through a solution of silver nitrato. After tho experiment the cathode liquid gave 1.6347 gram of AgCl and an equal volume of the solution beforo the orporimont gave 1.7463 gram of Agol. Calculate the transport numbers of tho ions in the solution. (2nd Year Hons. Class, Manc., 1909, adapted.) 48. From the electrolysis of hydrochloric acid in a coll with a cadmium anode the following results were obtained : Change in concentration of chlorine at anode and cathode respectively + 0.00545 gram; silver deposited in voltameter connected in series with the cell, 0.0986 gram. Calculate the transport numbers of hydrogen and chlorine. (Cl = 35.46, Ag = 107.9.) (London Hons., Extl., 1909.) 49. A carrent which deposited 0.047 gram of copper in a voltameter containing copper sulphate, was passed through a solution of silver nitrate using silver electrodes. The liquid from the cathodo contained, before tho experiment, 1'315 gram of Ag, and after the experiment 1.231 gram of Ag. Calculate the transport numbers of the ions in the solution. (Ag = 107.9, Ou = 63-4.) (2nd Year Hons. Class, Manc., 1911, adapted.) 50. A solution of potassium chlorate is electrolysed, and during electrolysis 0.3514 gram of silver is deposited in a silver voltamoter in series. 41.079 grams of solution gavo before electrolysis 1.0576 gram of potassium sulphate. The snode solution after clectrolysis weighed 69.954 grams and gave 1'3802 gram of potassium sulphate, The anode was ·XII.) ELECTROCHEMISTRY 125 cadmium, and was not attacked by chlorate ions. Find the transport numbor of the cation. (Inst. Chem., June, 1908.) 51. Calculate the transport number of the cation from the following experimental data :-- N A corrent of 0.9 milliampero was passed through a 100 solation of silver nitrate between silver electrodes for 6 brs. 52 min. Aftor the experiment 11:42 c.c. (tho total anodo N solution) required 6.90 c.c. of HCl for complete pre- 25.56 cipitation. (Electrochemical equiv. of silver = 0.001119.) (Leeds Univ., Electrochem., 1905 adapted.) 52. Caloulato the degree of ionisation in each of the following solutions of acetic acid, find also the dissociation constant of the acid. (Temp. = 25.) . V (litros) ... 128 256 Molec. conductivity 16.99 23.82 36+ (Inst. Chem. Phys., Jan., 1911.) 50 53. A gram-molecular weight of an acid is coutained in 30:15, 00:3, and 120.6 litres of solution respectively. The obsorved resistances for these concentrations are (i.) 92.9 ohms, N (ii.) 135.4 obms, (iii.) 200 ohms. Tho resistance of potassium chlorido is 67 ohms in the samo cell, and the molecular conductivity for potassium chlorido of the same strength is 129.7. Calculate the value of K for each concentration. (M co for the acid is 350.) (B.Sc. Hons., Vict., 1900.) 54. Calculate from the following data the volume of the solution which contains 1 gram-moleculo of silver chlorido at 13.8° : ratio of known resistance to that of the cell and contents, 1:53:02, the known resistance being 2140 ohms; limiting ionic conductivities of Ag and Cl respectively, 45.8 and 56:4; cell constant, k, 111.7. (B.Sc. Hons., Lond., 1910.)

·126 EXERCISES IN CHEMICAL CALCULATION CONTAP. 55. Given that the specifio conduotanco of water at 18° is 0.038 X 10-8 mbos, and that the ionic conductivities at infinite dilution of hydrion and hydroxidion Are 318 and 174 respectively, calculate the degree of ionisation of water. (Inst. Chem. Phys., July, 1911.) 56. Calculate the percentago hydrolysis of sodium acetato in tenth-normal solution at 25° from the following data, 8g8uming that the salt is completely dissociated : Dissociation constant of acetic acid = 0.000018 Ionic product for water = 1.21 x 10-14 (Inst. Chem. Phys., March, 1909.) 57. A 5 per cont. solution of cane sugar has a rotation of 20 = +66.7° in a 2-dm. tubo. After complete inversion the rotation is – 19.7°. If the solution is 0:01 N with respect to hydrogen chloride, the Anglo of rotation decreases by 69.2° in 20 minutes. What is the velocity constant of inversion ? What would be the valuo of the velocity constant if tho hydrogen chloride were replaced by 0:1 N lactio acid, the dissociation constant of which is 1.4 x 10-4. The bydrogen chloride may bo taken as completely dissociated in 0.01 N solution, [N.B.--~ is in litros in dissociation formula.] (Inst. Chem. Phys., Jan., 1910.) 58. Tho velocity constant of inversion of cane sugar by N 4 acotic acid at 25° is 0.75 X 10-3. Find the valac of the N constant when the acid solution is also with respect to 40 sodium acetate, being given that the dissociation constant of acetic acid is 0.000018, and that the sodium acetate is dissociated to the extent of 86 per cent. (N.B.-u is in litros in dissociation formula. (Inst. Chem. Phys., June, 1909.) 59. At a given tomporature a litro of saturated silvor bromate solution contains 0·0081 gram-molecules of salt; 0-0085 gram-molecules of silver nitrate are then added. Calculate the now solubility of the bromato, assuming both salts to be completely dissociated in solution. (Inst. Ohem., Juno, 1908.)

·XII) ELECTROCHEMISTRY 127 60. A lead accumulator with an electrolyte containing 1 gram-molecule H,80 per litre is joined in opposition to a similar coll with an electrolyte containing 0:01 gram-molecule acid per litro. Assuming all the acid to be dissociated, calculate the E.M.F. which will be given by the combination. (Uso data in No. 61.) (IIons., Manc., Electrochem., 1911.) 61. A coll is arranged by having as one electrodo a platinum plnte surrounded by an atmosphere of hydrogen over a normal solution of an alkali (80 per cont. dissociated), and as the other eloctrode an atmosphero of hydrogen over a normal solution of acid (80 por cont, dissociated). Calculate the E.M.F. given by this cell, noglecting tho P.). at the boundary of tho solutions, and stato what change accompanies the discharge. 1 faraday = 96,510 coulombs. 1 joule = 0.24 cal. conc. F* x conc. OH' = 0.56 x 10-14 gram equivs. per litro logis = 0.1343 log, .. logio 1:15 = 0.060. (Hons., Manc., Electrochem., 1911.) 62. Calculate the E.M.F. of the Daniell cell, given that- Zu + CuSO, = ZnSO, + Cu + 50,110 cals. I volt-coulomb = 0.24 cal. 1 faraday = 96,540 coulombs. Temp. coefficient = +0.000034 volts per degree. (Fons., Manc., Electrochem , 1910.) 63. Find the E.M.F. of an oxyhydrogen cell from the following data: H, + 0 = 11,0 + 67,520 cals. 1 joule = 0.24 cal. 1 faraday = 96,540 coulombs. Temp. coefficient of cell = 0.0014 volts per degrce. (3rd Year Manc., 1910). 64. Calculato the hont of formation of learl chlorido at 498° from the following data The E.M.F. of the combination Cl,/ molten PLC), Pb is givon by the formula, e = 1.263 - 0.000679 (1 - 498) volts. F = 96,540 coulombs. 1 jonlo = 0.2391 cal. (Inst. Chem. Phys., March, 1909.)

·128 EXERCISES IN CHEMICAL CALCULATION COMAP. Refractivities, (n 1) 65. The value of for othyl alcohol at 20° is d 0-4528, for water at the same temperature 0.3324. Tho density of a mixture of alcoliol and water at 20° was found to be 0.8558, and its rofractive indox 1.3616. Calculate the percentage of alcohol present. (Hons., Vict., 1899.) 66. Calculate the rofractive power (using Gladstone and Dale's formula) of sodium chloride from the following data :- 22:2219 grams of an aqueous solution of sodium chloride contained 2.1336 grams of NaCl. The density of the solution at 19° was 1.06803 and its refractive index for sodium light was 1.31978. At the samo temperature 1 gram of water occupies 1:001564 c.c. and has a refractive index of 1.33300. (Inst. Chom. Phys., Jan., 1910.) Miscellaneous. 67. The vapour pressuro of water at 99.8° is 754.58 mm., and at 100.2° it is 763•45 mm. Calculate tho latont heat of vaporisation for 1 gram of water at 100°. Regnault's value for 100° is 537 calories; how do you account for the differenco between the calculated and oxperimental valuca ? (Inst. Chem. Phys., July, 1910.) 68. If the ignition temperature of electrolytic gas is 585°, what adiabatic compression would cause ignition, if the gas were originally at 15° C. and 1 atmosphere pressure? Assume for the purpose of this calculation that the ignition temperaturo is not altered by variation of pressure. (y = 1:41.) (For this and the following calculation, compare K. G. Falk, J. Amer. Chem. Soc., 1906, 28, 1517; 1907, 29, 1536; and H. B. Dixon, Presidential Address to the Loudon Chemical Society, J.C.S., 1910 and 1911.) 69. A column of gas of composition represonted by H, + 30, and of length 540 mm. was compressed adiabatically into a length of 40 mm. without ignition ; into a length of 36 mm, with ignition resulting. The cross-section of the ·MISCELLANEOUS 129 column being constant, show that the ignition tomperuture of the mixture (at the pressure of the experiment) lies between 550° and 578°. Assume y (ratio of the specific heats at constant pressure and constant volume) 1:40, and the initial temperature of the gases 17•v and 15° rospectively. (H. B. Dixon, Juur. Chem. Soc., 1911, 99, 590.) 70. Calculate the approximate molecular hoat of solution of anilino (CIN) in water, given the following data :- 1 part by weight of anilino dissolves in 31.5 parts by weight of water at 12.5° and in 32:1 parts at 16'. logio 3.15 = 0.4983. 10g10 3.21 = 0·3065. log. X = 2.303 log10c. (Hons., Manc., 1911.) 71. Tho melting point of acetic acid is 16:0° C. ; its heat UT of fusion is 46.42 calorios ; and the value of 0.0242?. dp Calculate the change in volumo when 1 gram of solid acetic acid is molted. (Londou llons., Intl., 1908.) 72. The mean distance betwcon two nodus for a tube filled with dry CO, was found to be 4.528 cm., when filled with dry air under the samo conditions the distance was found C') to be 5.948 cm. If the ratio for air is 1.41 and its density 0, C, is 14:45, calculato the ratio for CO ☺ log P (B.Sc. Huns., Manc.) 73. The vapour prossure of a substauco varies with temperature according to the equationb T where p is the vapour pressure, T the absolute temperature, and a aud b constants, b being connected with the heat of reaction per grain-molecule of gus produced (q), by the equation- 9 = log, 10.RO R being the gas constant (log, 10 = 2:30). в ·130 EXERCISES IN CHEMICAL CALCULATION [CHAP, XIL Caven and Sand (J.C.S., 1911, 1359) found the following relations for sodium carbonate monohydrate :- log P . 30000 = 10.825 T and for sodium bicarbonate (2NaHCO) 3340.0 11.8185 T log P Caloulate q in each of these cases and compare tho second value with that obtained from the following thermochemical data- [Na, H, C, 0,] = 228,380 cal. [Nang, 0, 0] = 271,970 » [H,,0) = 58,060 [C, 0] = 97,000 13 ·ANSWERS CHAPTER II 3 22 99 9 1. 1 silver : 0.149 sulphur in each caso (Borzolius 1 silver : 0.14852 sulphur 0.14834 0.14854 (Stag) 0.14851 0.14849 2. 1 silver : 0.32841 chlorino, 0•32844 0.32843 0.32850 0.32846 0.32848 0•32842 3. 39.1620 4. Load 74.62 per cent 39.1605 Chlorine 25.46 39.1560 100.08 39:1527 39.1550 39.150 39.147 39.156 6. Ratio of nitrogen in (i.) to nitrogen in (ü.), combined with the same weight of oxygen is 2.000. 6. 1st contains 3:0 A to 1.0 B. 2nd 4.0 A to 1:0 B. 3rd 6.00 A to 1.00 B. ..

·132 EXERCISES IN CHEMICAL CALCULATION 7. Ratio of arsonic in A to arsenic in B combined with samo weight of iron is 2-66: 1.35 = 2 (approx.). 8. Ratio of sulphur in A to sulphur in B combined with same weight of oxygen is 1 : 1497 = 2:3 (approx.). Ratio of iron in A to iron' in B combined with saine weight of oxygen is 1 : 1.499 = 2:3 (approx.). Ratio of carbon in A to carbon in B combined with sano weight of oxygen is 1 : 2.000. 9. Ratio of copper oxide in copper sulphato to cupper oxide in basic sulphato combined with the sumo weight of sulphuric anhydride is 1:04: 4:00 = 1:4 (approx.). 10. Ratio of iron in first compound to iron in pyrites combined with the same weight of sulphur is 1.99:1= 2:1 (approx.). 11. Ratio of copper in black oxilo to copper in other oxide combined with the same weight of oxygen is 1 : 2:00. Iodino 12. l'otassium = 3-58 (from potassium iodido). Iodino Potassium (combined with saine weight of chlorine) =3.58. 13. 1 grum of hydrogen combines with 3:00 or 0:00 grams of carbon, either of which quantities will combine with 8.00 grams of oxygen, which in turu combines directly with 1 gram of hydrogen. 14. Sulphur, 39.9. Oxygen, 60-1. 15. 12:1. 16. 108. 17. 107.9. 18. 10:34 + 0.02. 19. 7.94; 1:00; 31:57; 15.89; 35:17. 20. 32.0; 64.2. . CHAPTER III 1. 34:0 с.C., 340 c.c. 2. 471 mm. 3. 3:0 atmospheres. 4. 22:78 c.c., 9:18 c.c., 0.45 c.c. 5. 65.5 c.c. 6. 10 c.c. 7. 265 C.C., 322 c.c., 449 c.c., 549 c.o. 8. The former. 9. 240° 0. 10. The former. 11. 273 c.o. 12. 0•67 atinosphoru. 13. The formor.

·ANSWERS 133 13702 9.2 um. water vapour. 14. 370:5 mm. oxygen. 370-5 mm. nitrogen, 15. 150 c.c. 16. 15.0 с.o. (Oxygen, 20:5 per cent. 17. Water vapour, 2:0 por cent. (Nitrogen, 77.5 por cent. av 731 mm. total : 542 mm. hydrogen, 77 mm. oxygen, 112 mm. nitrogen. CHAPTER IV 1. 249.76 grams. 2. 99.404 c.c. 3. 1.0519, 1.0525, 1.05273, 1.0514, 1.05157, 1.0517. 4. 65 c.o. 5. 0.0046 gram, 0.0043 gram. 6. 134.873 grams. 7. 2:19 +0.03, 2.14 003, 2:17 +0:04. 8. 2:02 grams. 9. 140° C. 10. 126. 11. 96.7. 12. 38.7. 13. 37.8. 14. 52.0. 15. 75. 16. 46, 23, 15:3. N20,2NO, NO + 02. 18. 12. CHAPTER V > 1. 65-4. 2. 29.5, 118.0. 3, 8.67, 52-0. Heravalont. 4. M.O(80), MO(80), M,0(240), MO,(160). Accept 80, i.e. MO (atomic woight of mictal 61). 5. 26.6, MCI 6. 29.7, 119. 7. 58 6. 8. 519 c.c. 9. 214, which is molecular weight of GeCl.. 10. 3•67, 11•0. 11. 11.04, 10.95. 12. 1973. 13, 107.89, 107.93. 14. 127:55, 127.55, 127:53, 127:53, 127:53. 15. 35.41. 16. 35.48. 17. 12:00, 12:06. 18. 18.1. 19. 39.8. 20. 2001. Molecular wt. approx. 198, i.e. mercury vapour is monatomic, and its exact molecular wt. is 2001. 21. 1:006; or if CO, is absorbed by potash in drying tubes, 1.008. 22. 18.017 grams. 1.009.

·134 EXERCISES IN CHEMICAL CALCULATION CHAPTER VI 1. K, 56-6; 0, 8-7; 0, 34.7. 2. H, 5.93; 0, 94.07. 3. Fe, 34:42; 0, 65.68. 4. N, 21:40; 0, 24-44; C1, 54.16. 5. Na, 32-8; AI, 12.9; F, 54-3. 6. N, 6-31; H, 1.82; Pt, 43.96 ; 01 47.91. 7. C, 31.98; H, 6:72; N, 18.67; 0, 42-64. 8. C, 55-35; H, 7.75; 0, 36.90. 9. C, 39.98; H, 2.80'; 0, 26-65; S, 17-81 ; Na, 12.77. 10. Cu0, 71:96 ; CO., 1990; HEO, 8:15. 11. A1,03 39.5; SiO2, 46:6; H,0, (const.) 7:0; 1,0, (hyd.) 7.0. 12. H,0, 62.96; Na,CO., 37.04. 13. K, 17:2; Na, 5:1; Co, 13:0; NO2, 60-8; 1,0, 3.9. 14. 29.2 per cent. gold from analysis an:l from forniuln. 15. (C311,50),Ba; (30·15 per cent. BA), (C,H,O2).Ba requires 30•42 por cent. 1.6. FOO, 17. Cillig 18. NagAlF6 19. NiCO3, K_CO., 411,0. 20. CuSN.,0,11g. (CuSO., 2NH3, H,0.) 21. 3B.O, A1,03, 6SiOg. 22. 0.209 gram. 23. Cal, A1,03, 6SiO2, 611,0. 24. Th(NU) (0,0.).7H2O. 25. Pb(NO) (FeCN), 58,0. 26. K,Cr((:20.).(OH), or K,Cr(C,0,1,OH. 27. 2K,S,06 31,0. 28. Hg(NO2)2, 3KNO,, 11,0. 29. 28.0. 30. [Fc, Al],03, [Fe, Mg]0. 31. [Ni, Fe, Mg], (AsO.)., 8H,0. 32. NA,O, A1,03, 3Si0, 211,0. 33. FeWO., or (Fe, Mn, Ca, Mg] WO. 34. 2Ca0.28i02-[Al, B],03.H.0, or Ca[Al, B]OH.SiO.. 35. 341,03.3N2,0.6SiO,2NaCl. 36. 2nP,0,01 37. Ca,Si[O, F.]... CHAPTER VII 1. 94'5 grans. 2. 6.1 grams of KCl, 3.9 grams of oxygen, 5.6 grams of C&O, 4.4 grams 3. 74.6 lbs. 4. 296 grams per litre. 5. 6.004 in CO, 8.002 in CO of CO2 ·ANSWERS 135 7. 48.0 per cont. 6. 8.09 grams, 7067 grams. 8. 29 per cont. 9. 63.0. (M20.; MOX 10. Atomic weight of motal caloulated fron nitrate or either oxide is 207. 11. 29.4 per cent. Feo. 706 per cent. Fo,Os. 12. 43.7. 13. 103.5, 207. 14. 3TICI + 2AuBrg = 2Au + TiCl2 + 2T1Brg. 15. K,Cr,0, + 6Hg + 14HCI = 2CrCl, + 30g,Cl, + 2KCI + 7H,O. 16. 2.9 per cont. 17. 60.0 per cent. KC1, 30-0 per cent. KBr, 10.0 per cent. KI. 18. Add 113 grains NaOH and 1:1 grains Na Coz per gallon. (Ca is precipitated as CaCO3, Mg as My(OH)2-) 19. 8.16 grams, 29.19 grams. 20. 38 grams, 158 grams. 21. 0.933 grain. 22. 89.41 per cent. 23. 2.54 grans (if the gasos remain in contact with the solution for some minutes). 24. 0-200 gram. 25. 800 lbs. 26. KI 0.313 gram. NaCl 0:687 gram. 27. KCl 73.8 per cent. KI 26.2 per cent. 28. Caso, 43.2 por cont. CaCO, 50.8 per cent. 29. NaCl 42.5 per cont. KCl 57.5 per ceut. 30. BaO 81'1 per cont. Bal, 18.6 per cent. (a) 11.8, (12. 31. 0.0198 gram, 0132 granu, 32. 11.1 litres, 0.509 litro, 0.716 litre, 0.509 litre. 33. 274 c.c., 16:3 c.c., 110 c.c. 34. 0.226 grain. 35. 4:00 grams, 267 c.o. 36. 18.7 litres of hydrogen, 9:3 litres of oxygen. 37. 33.3 grams, 11•4 litres. 38. 497 glans, 002 grains. 39. 209 c.c. 40. (a) 16.3 c.c., (6) 32-6 c.o., (c) 10:3 c.c. 41. 5.6 litros in each caso. 42. 630 mm. 43. 0-47 Na. 1•20 grams NaCl. 44. 43.9 grams. 16.8 litros. 45. 367 grams, 3800 litres. 46. 18.7 litres. 47. 22-4 litres, 22-4 litres, 11•2 litres. 48. 35.5 grams, 8.7 litros. 49. 30 grains S per 100 cu. ft. S per litro 0-48 o.o. H.S por litro. 50. 33.5 kilos. of sulphur. 0.295 gram, 0.432 grain. gram 0-69 mgr.

·136 EXERCISES IN CHEMICAL CALCULATION 51. 6:32 grams per litre. 52. 131.0 grams, 120.1 grams, 322 c.o. 53. 58.6 por cent., 670 C.C., 720 0.c. 54. 5.6 litres (assuming no SO, formed). 14.9 litres. 55. H.S: 10-6 grams, 7.0 litros. SO,: 200 grams, 7-0 litros. 56. 35.2 litres. 57. 27.5 litres. 58. 95.1 per ceut. 59. H.S. 60. 2 +3 litres at N.T.P. 37.0 grams. 61. 61 per cent. CaCO3. 39 per cent. MgCO3. 62. 3:2. 63. 90 cubic metres air. 10.5 kilograms wator. 12.4 cubic metres CO2 64. 155 thousand cubic feet. CHAPTER VIII 1. 20 c.c. CO. 2. 5:0 litres oxygen ; 10-8 litres COg. 3. 22:4 litres. 4. 456 litres. 5. (a) Production of so, involves no change in volume, but the small amount of SO, pro luced in such a combustion would cause a corresponding small contraction. (6) 20 c.c. contraction. (c) Total contraction on 3 litres of 1 litre, or resultant volume is 2 litres. 6. 29.8 litres. 7. 68 grams; 4.8 litros. 8. 60 vols. CO2; 55 vols. oxygen. 9. 60 vols. CO2; 44 vols, oxygen. 10. CO2, 807 C.C. ; E1,0, 1571 c.o.; SO2, 2.7 c.0. ; No 5554 c.C. 11. 44.8 litres; 104 litres. 12. 327° 0. 13. 2.0027 vols. 14. 2·005 vols, hydrogen: 1 vol. oxygon; Density 5.96. 15. CIO. 60 с.o. 16. NO. 17. CH 18. C,02 19, 79 vols. nitrogen ; 38 vols, hydrogen assuming tho original gases to be measured saturated with water vapour, 20. 36 c.c. oxygen ; 64 0.c. nitrogen.

·ANSWERS 137 21. 90 1.0. Co; 10 c... CH. 22. 40 vols, oxygen; 40 vols. N,0; 20 vols. nitrogen: 23. 38.2 por cent. N,0; 49.4 per cent. N0; 12-4 per cont. nitrogen, 24. 60 vols. CO; 40 vols. acetylene. 25. 30 0.0. CS2 ; 70 0.0. O,Ng. 26. 2 c... C,H, ; 6 c.c. hydrogen ; 12 c.c. CO. 27. 12.5 per cent. hydrogen ; 7-5 per cent. OH, ; 800 per cont, nitrogen. 28. 22-4 per cent. 29. 12.6 per cent., C,H,; 41:9 per cent. CO; 43:6 por cent hydrogen; 1.9 por cent. nitrogen. 30. 0.28 per cent. CO,; 1.22 per cent. CH.. 31. CO.,, 13; 00, 48.0; C,H, 2:7; CH,, 2:3; hydrogen, 43°4; nitrogen, 2-3 per cent. 32. CO2, 0.7; C,H., eto., 4.2; CO, 7.0; CII, 40:0; hydrogen, 48.0 por cent. 33. CH., 38.6 c.c.; C,H, 9-0 c.c.; nitrogen, 5.9 c.o. 34. Có, 4 c..; CH, 4 c..; C,H, 2 c.. 35. CS.,, 40 c.c. ; CO, 20 c.c. ; 10 per cent. 36. CH,, 0:43 per cent.; nitrogen, 0.04 por cent. 37. CO., 64:7 per cent.; N,O, 20.5 per cent. ; CO, 6-3 per cent. ; nitrogen, 8.5 per cent. 38. Oxygen, 17.65 c.c.; ozone, 18.59 c.c. ; CO, 39.96 o o.; CO2, 9•67 c.c. 39. (a) Possible. (b) Impossible. No. CHAPTER IX 1. 4:815 grams. 2. 4.00 grams; 2.36 litros. 3. 3.54 grams in 100 C.C. ; 3.50 grams in 100 grams ; 6.62 c.c. 4. 3.95 grams; 5.64 grams. 6. 11.2 0.0. gaseous SO, at N.T.P. 6. 97.5 per cent. 7. SO, 66-6; H, SO, 3304 por cent. 8. 0:451 gram sodium as hydroxide. 0.038 peroxide. 9. 77.7 por cont. ; 92.8 per cent. 10. NOCI. 11. (a) 74 per cent. (6) 26 per cent.

·138 EXERCISES IN CUEMICAL CALCULATION CHAPTER X 1. CHO. 2. OH,O. 3. C,H160. 4. OgH 0.2 5. C,H,O. 6. C,0,0. (Propyl alcohol, isupropyl alcohol, methyl etliyl ether.) 7. OH,O,Br. 8. C,H,O,Br. 9. C,H,O,; oxalic acid. 10. C,H,,N.(CH3.CHZ.CH.CH.NH7; (CH3)2.CH.CH.,.NH,; (CH).C.NH; CHỊ.CH, CH.NII,; (C,II:),NII; CH CHOEL, H, CH, C.H. NH; CIL CH N CH, CHI, NH; CH, CII, 11. CHIA C:NOH from ruethyl ethyl kotona C, H. 12. C,H,0,Brz. 13. Empirical formula CH1,01. CHOI und CH,C1 CIICI 14. CH.CO.NH, acetamido. 15. CHOH.COOH malic acid. CIL.COOH CII,.C(OH) (COOH), isomalic acid. CII,OH.CH.(COOH)2 B-hydroxy isosuccinic acid. 16. C&H.CHE. 17. C.410 A doubly unsaturated aliphatio hydrocarbon. 18. CHO; C,H,O, 19, C,IIN. 20. CH.03. The compound is (COOH),2H,0. 21. 0-xylene C.H.(CH2), 22. Tho simplest formula is C311,0, but tho bromino conpound is C,H,Brog. Hence A is C,H,Oz. With alcohol and salphuric acid it gave 0,H,O,U,)02, thus indicating the presonoe of a bydroxyl group. Hydroxylamine gave a dioximo, indicating two ketonic groups. Hence, since there aru only CH, (11,6 ·ANSWERS 139 two atoms of oxygen in A, one group_CH.CO-must exist and behave both as a ketone and as-OH = C(OH), Dihydroresorcinol satisfies all the conditions. Its formula is represented as CHCO CH-CO CH, Сн, or CH, CH CII, CO CHI,-0-0II) OT) 23. COOH CII, CH_0 (Lactono of a-lıydroxy BB-dimethyl glutaric acid). CH, CHCO 24. Salt is C1,0-N, and acid C.H.07, or C.H.0(COOH)3, i.e, citric al or an isomer. 25. CL-C110H (A) (CH). CH, CHỐC,H, CH-CHOC,H, CI,CII-O-COC,II, (B) (CII,), CH, HOCH, CH-CHOC,H, CII,-CH. 26. CH.CH >CLI.COOII. CH-CH, CIL OF CH.CH O.COOH. CH-OH 27. aa-dimethyl glutaconic aoid, COOH.C(CH,) (3)..CH:CH.(COOH). 28. C,H,O, or C,H,O(COOH). A number of isomerio keto and aldehyde heptylic acids have the above formula:

·140 EXERCISES IN CHEMICAL CALCULATION 29. y-aceto butyric acid is CH,,Og. This is formed by oxidation of an aoid C_H100, thus :- CH, COOH CII, H000 0:0 from CH, CIL CH, CH, CH, CH, The cyclic hydroxy-monobasic acid is then either COOH CH, COOL CF, πος CH C(OII) HO.C CIL or CH, CH CH, CH, CH, CHZ CHI, Alternatively, the carboxyl group may bo exchanged in the last three formulæ for one of the hydrogen atoms in a CH, group. 30. The first two acids are represented by the molecular formulæ 0,710, and C,H,O,. m-xylene being 0.410, the third acid must be monocarboxylic, i.e. 0,41,02, a dimethyl benzoio acid. The first and socond are then boxn-hydro and tetrahydro-dimethyl benzoic acids. 31. CH-CH2-('H, CH-CH-CO CHAPTER XI 2. + 18,550 cal. 3. + 33,740 cal . 1. + 21,485 ral. 4. OH,: + 21,750. C2H.: + 28,560. C,H,: 2710. C,H,: - 47,779. 0,11,: - 12,510. 5. + 27.7 Calories. 6. + 53,000 cals, ·ANSWERS 141 7. + 99,420 cals. 8. (i.) + 48.7 Cals. (ii.) + 55.2 Calg. 9. t 131.0 Cal. 10. + 97,689 cal. 11. [Fe, Cl.]= + 82,048 cal. [Fe, Cl,]= + 96,038 cal. 12. [H, CI] = + 22,000 cals. [H, I] 6100 cals. 13. + 1055.5 k. 14. A. 29,973. B. 29,895. 15. + 26,707, + 11,889, 16. 192,700 cal. 17. 11,520 cals. 18. 4503 cals. 8105 B.T.U. 19. C.H. 371,600 cals. 20. 63,639. 71,939. 142,419. 103,249. 21. 369,000. 332,198. 309,195. 22. 801.6 Cal. 23. 40.2 kgr. cent. units. 24. 25,000 cub. ft. 25. 3600°. 1950º. 26. 10,180°. 2755º. CHAPTER XII 1. 1 volume nitrogon with 3.003 volumes hydrogen, in cach caso. N = 14:01 (H = 1:008). Tho data do not pernit a distinction to bo drawn. 2. 71.2 atraospheres. - 93° C. 3. 15 minutes. 4. 279 seconds. 5. Tho deusity of nitrogen, by comparison of its rate of diffusion with that of (a) hydrogen, is 14'1, (1) oxygen, is 14.3. 6. (a) 0.703. (b) 24.7. (Alumic weights used : CI = 3546, II = 1.008.) 7. 82. 8. 20. 9. 32 per cent. oxygen, 61 per cent. nitrogen, 7 per cont. CO2 10. 20-7 per cent, oxygen, 78.4 per cent. nitrogen, 0.9 per ccut. CO 11. 365 mm. oxygen, 136 mm. CO.,. Total = 501 mm. 12. 196 mm. nitrogen, 564 mm. NO. From 1 volume of the mizod guscs, 0.939 volumo remains unabsorbed.

·142 EXERCISES IN CHEMICAL CALCULATION 13. 50 cm. mercury. 14. 0:30 atmosphere. 15. 30.6. 16, 315. 17. H + H + Cr,0,". 18. 60. 19. 113. 20. 1700 calories per gram-atom. 21. Molecular weight from first two detorminations 164, from last two = 144. The molecular weight of formio acid is 46. 22. C2H,,N,Os 1 0. 23. 0.0309, 0.0332, 0.0350, 0.0370, 0.0365, 0·0322, 0.0323, 0.0340, 0.0267. C ( = 0.00564, 0.00631, 0.00598. 25. Second 26. Const. = 0.00414, 0.00410, 0.00410. 27. Second order. Initial concentration (moan) = 9.26. Velocity constant (mean) = 0.00670. 28. Second. 2.81 x 10-3, 2.83 x 10-5, 2.80 x 10-5, 2.79 logio 510 TL 24. log. X 10-3. 29. The numbers would be regarded as proving the reaction to be monomolecular, wore it to be proceeding in a homogeneous system. Since it is taking place on the surface of the porcelain, no deduction can be made as to the order of the reaction. (See Bone and Wheeler, Phil. Trans., 1906.) 30. First. 31. 0.000255, 0.000259, 0.000257, 0.000253. 32. 26.2 per cent. 33. Over the range 900°-1100° the molecules are Są; as tho temperaturo falls, a greater proportion of more complicated molecules is formed, until the density approaches, but does not rench that corresponding with Sg. 34. Per cent. diss. 55:09, 39.20, 27.07, 18.40, 16.75 Partial (press. N20, 12.62, 44-39, 132-4, 339.1, 440-4 NO, 30.96, 57.25, 98.2, 153.0, 177.2 C*NO, KE Oy 75.9, 73.8, 72.9, 69.0, 71.3. 36. 14:6 per cent. Caster. Ometer 86. K Coeld.com (i.) 3.8, (ii.) 3.8, (iii.) 3.5, (iv.) 4.4. Alcohol ·ANSWERS 143 37. 1.5 per cent. of total number of molecules are monatomic. 38. 31 per cent. 39. 5.64, 9.49, 13:47, 14:93, 15:54, 15.40. 40. Oki. Cg = 0.000967., Скі, 41. 5 per cent., 12 per cent., 59 per cent., 98 per cent. 42. 20-01, 20-44, 20-65. 43. Iron 13.0, 8.7 grams; silver 50.1 grams. Negative Pole. Positive Pole. 44. First coll. 100 c.c. hydrogen. 50 c.c. oxygen. 100 c.c. hydrogen. 100 c.c. chlorine Second cell. 0.358 gr. NaOHI (less amount dissolved). 50 c.c, oxygen. Third cell, 0.283 gr. coppor. 20:438 gr. H.SO. 50-563 gr. IINO, Fourth cell. 0.964 gr. silver. 50 0.c. oxygen. $100 c.c. hyılrogen. 100 c.c. C.H. Fifth cell. 0-500 gr. KOH. 200 c.c. CO., Cathode. Anode. 45. First cell. 0.0378 gr. hydrogen. 1.330 gr. chlorine. 50-300 gr. oxygen. Second cell. 1.192 gr. copper. (1.840 gr. 11,50, Third cell. 4:047 gr. silver. 6:373 gr. AgNO, 46. 47.4 per cent. 47. Ag : 0.478. NO,: 0.522. 48. H:0.832. CI: 0.168. 49. Ag : 0:475. NO,: 0:525. 50. 0.574. 51. 0-455, 52. 0.0467, 0.0654. 1.79 X 10-8 (u in c.c.). 53. 72.3 x 10-4, 72.3 x 10-8, 72.2 x 10- (v in c.o.). 64. 103,000 litres. 55. 0.77 x 10-7 gram-molecule per litre. 56. 0·0082 per cent. 57. 0·0350, 0.0129. 68. 0.067 x 10-4. 59. 0.0049 gram-moleoule per litro 60, 0.109 + 0.00041. 61. 0•76 + 0.00281. 62. 1.090 + 0.000034t. 63. 1.075 - 0.00141. 84. 82-5 Cal. 65. 74.8. 66. 0.2708.

·144
EXERCISES IN CHEMICAL CALCULATION
67. 547 calories. It is assomod that steam behavos as a
perfect gas; its volume is, howover, less than that calculated,
hence the difference.
68. 43 atmospheros
70. 890 calorios
71. + 0.161 c.c.
72. 1.24.
73. 13,800 calories.
30,700 calories.
29,700 calorios.

·SOME CONVERSION NUMBERS
THE student will remenber the following after using them for a
short tiine. Ho will find them sufficient to enable him to perform
most of the conversions ho inay meet.
Lenyth.
1 metre = 39:37 inches.
1 inch = 2:54 cm.
Muss.
1 ounce (avoirdupois) = 28:35 grams.
1 gram
= 15:13 grains.
Volume and Jass.
1 gallon of water (at 62° F.) = 10 lbs. avoirdupois = 70,000 grains.
1 litro of hydrogen at N.T.P. weighs 0·0999 (or 0-09) grain.
1 litre of air at N.T.P. weighs 1.293 gram.
The gram-mulecular weight of a gas occupies at N.T.P. approxi.
mately 22 4 litres.
Temperature.
t° C. = (t + 32)º F.
Specific Gravity.
5 x Twaddell reading + 1000
Specific gravity =
1000
Logarithms.
log10 a x 2.3026. Log10
Logo a =
-
log. a * 0:4343.

·146 EXERCISES IN CHEMICAL CALCULATION VAPOUR PRESSURE OF WATER .C. mm. 0 4.6 1 4.9 2 5.3 3 5.7 4 6:1 5 6.5 6 7.0 7 7.5 8 8.0 9 8.6 °C mm, 10 9.2 11 9.8 12 10.5 13 11.2 1 11.9 15 12.7 16 13.6 17 14.5 18 15:4 19 16-4 °C. min 20 17.4 21 18.5 22 19.7 23 20.9 21 222 25 23.5 26 25.0 27 26.5 28 28.1 29 29.8 •c. mm, 30 31.6 40 55.0 50 92.2 60 149.2 70 233.8 80 955-5 90 526.0 100 760.0 DENSITY OF WATER 0.99987 93 07 ... 99 0.99862 43 23 02 0.99780 50 32 .. ILO . 07 00 1° 20 30 40 50 6° 7° 8° go 100 11° 12° 130 14° 150 16° 17° 18° 19° 20° 21° 220 23° 24° 25° 26° 27° 28° 29° 30° 31° 3:29 33° 34° 35° ... ... 1.00000 0.99999 97 93 88 81 73 63 52 40 27 13 0.99897 80 4 0.99081 54 26 0.99597 67 37 05 0.99173 40 06 ... .. ... ...

·... Ni .. ... ... .. . ... ... ... ... ... ... ... 1.. ... . ... ... ... INTERNATIONAL ATOMIC WEIGHTS 147 1912. 0 = 16. 0 = 16. Aluminium Al 27.1 Neodymium Nd 144.3 Antimony Sb 120.2 Neon No 20.2 Argon .. A 39.88 Nickel 58.68 Arsenio As 74.96 Niton (radium Barium Ba 137.37 emanation) Nt 222:4 Bismuth Bi 208.0 Nitrogen N 14.01 Boron B 11.0 Osmium 0.5 190.9 Bromine Br 79.92 Oxygen 0 16.00 Cadmium Cd 112-40 Palladium Pd 106.7 Cæsium Cs 132.81 Phosphorus P 31.04 Calcium Ca 40:07 Platinum Pt 195.2 Carbon С 12.00 Potassium K 39:10 Cerium Co 140.25 Praser.dymium Pr 140.6 Chlorine CI 35.46 Radium Ra 226.4 Chromium Cr 52.0 Rhodium Rh 102.9 Cobalt Co 58.97 Rubidium Rb 85.45 Columbiuna Cb 93-5 Ruthenium Ru 101.7 Copper .. Cu 63.57 Samarium Sa 1504 Dysprosium Dy 162.5 Scandium Sc 44:1 Erbium Er 167.7 Selenium 79.2 Europium Eu 152.0 Silicon Si 28.3 Fluorine F 19.0 Silver Ag 107.88 Gadolinium Gd 157.3 Sodium. Na 23-00 Gallium Ga 69.9 Strontium Sr 87.63 Germanium Ge 72-5 Sulphur... S 32:07 Glucinum Gl 9.1 Tantalum Ta 181.5 Gold Au 197.2 Tellurium Те 127.5 Helium Но 3.99 Terbium Tb 159-2 Hydrogen H 1.008 Thallium TI 204.0 Indium In 114.8 Thorium Th 232.4 Iodine I 126.92 Thulium Tm 168.5 Iridium Ir 193.1 Tin Sa 119-0 Iron Fe 55.84 Titanium Ti 48.1 Krypton Kr 82.92 Tungsten W 184.0 Lanthanum La 139.0 Uranium U 238.5 Lead Pb 207.10 Vanadium 51.0 Lithium Li 6.94 Xenon Xe 130-2 Lutecium Lu 174.0 Ytterbium (Neo- Magnesium Mg 24:32 ytterbium) Yb 172-0 Manganese MA 54.93 Yttrium Yt 89.0 Mercury Zino Hg 200:6 za 65.37 Molybdenum Mo. 90.0 Zirconium Zr 90-6 Se .. ... .. ... ... ... .. ... ... ... ... ... ... ... ... ... .. .. ... .. ... I..

·148 LOGARITHMS 0 1 2 3 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 0000 0043 0056 0129 0170 4 9 13 17 21 26 80 84 85 0219 0253 0294 0334 0374 + 8 12 16 20 24 28 82 87 11 0+14 01530192 0531 0569 4 8 12 15 19 28 27 81 35 0607 064509820719 0757 1 7 11 15 19 22 20 80 38 12 0792 1828 0904 0899 0934 0969 3 7 11 14 18 21 25 28 32 100+ 1038 1072 1101 3 7 10 14 17 20 24 27 31 13 1139 1173 206:12391271 3 7 10 13 10 20 23 20 30 1303 1333 1307 1399 1430 3 7 10 12 10 10 22 25 29 14 H01 19:13:23 13.13 3 6 9 12 15 18 21 24 28 1551101+ 1044 1073 1709 1732 3 0 9 1915 17 20 23 26 151761 1790 1518 1847 | 1873 1203 3 6 9 11 14 17 20 23 20 1931 1959 1957 2014 3 5 8 11 14 10 19 22 25 12011 -003 | 2005 2122 2148 3 5 8 11 14 16 19 22 24 21732201 227 22.13 2279 3 5 8 10 13 15 18 21 23 17 2304 233023352350 2405 3130 3 5 8 10 13 15 18 20 23 2135 2450 2304 2329 2 5 7 10 12 15 17 19 22 18 2353 2377, 2001, 2023 2018 2 5 7 12 14 10 19 21 20: 2093 2715 274 2765 2 5 7 9 11 14 16 18 21 192783 810 2833 970-878 2 4 7 9 11 13 10 18 20 2900 29.23 2915 2:07 2989 2 4 6 8 11 13 15 17 19 20 3010 3032130343073 3090 311 313 3110 311 3,01 2 1 6 8 11 13 15 17 19 :21 3.222 3243 3203 325+ 3304 3324 3340 3315 335 3404 2 4 6 8 10 12 14 16 18 22 3424 311131613183 3.302 339 33410 3579 3598 2 4 6 8 10 12 14 15 17 23 80173630 3653 3074 3002/3711 3720 3717 3700 37$! 2 4 6 7 9 11 13 15 17 24 3302 3520 38983330 3874 3392 3909 3:27 3915 3902 2 4 5 7 9/11 12 14 16 27397 39971014 4031 1049 1005 109 110 1116 41332 3 9 10 12 11 15 20 4150 4100 4183 4200 4210 42321310200 1281 4298 2 3 5 7 8 10 11 13 15 271304330 1346 43024378 139311094123 410 4430 2 3 5 6 S 9 11 19 1+ 291472 4487 1502 1518 1533 4318 4301 4379 310 4609 2 3 5 6 8 9 11 12 14 21634 4039 | 4051 4669 4083 10:18 1713 4728 47424757 1 3 4 6 7 9 10 12 13 30 1771 1750; 4800 481+15+3 41771871 4500 49001 3 4 0 7 9 10 11 13 311914 1923 1942 1933 1902 1983 1937 | 3011 5021 5038 1 3 4 6 7 8 10 11 12 32 5031 30330793092 31075119 5132 311 5150 5172 1 3 4 5 8 9 11 12 3351333195 5211 5221 5237 5250 7031 7270 520 530211 3 4 5 8 9 10 12 313313 525 5310 5333 5306 5375 030151033410 5424 1 3 4 5 6 8 9 10 11 35 5411 15 153 | 54055178, 34903302 7 9 10 11 36 5509 5573 5587 5599 5011 5123 5535 56173035 3070 1 2 4 5 6 7 8 10 11 375082 3694 5705 !5717 5729 57 10 5752 5703 57753780 1 2 3 5 6 7 8 9 10 35798 5800 5921 5832 3813 7535 350655775588 3899 1 2 3 5 6 7 8 9 10 30 5:11.59225:33 3911 5935 3960 29770859996010 1 2 3 4 5 7 S 9 10 10 50216031 0043 605360107.) 100 0107 011712 3 + 3 8 10 41 | 0125 0139 014951000170 6180 0191 0:01 0212 022: 1 2 3 4 5 7 8 9 12 0232 6213 02.33 6203 0274028162294 63040311 03:25 1 2 3 4 5 6 7 8 9 43 0333 0343 0355 630563756383 6395 040504156423 1 2 3 4 5 6 7 8 9 14 6139 0444 1454 0461 (474 6484 6493670303130522 1 2 3 4 3 6 78 15 0332 65+2 6551 | 0561 0710.380.59005390009 601 2 3 4 5 6 7 $ 9 40 1028 00370046 0656 66050075 00540093 0702 6712 1 2 3 4 5 0 7 7 8 47 0721 073067396749 | 6758 076767786785 6704 6803 1 2 8 4 5 5 6 7 8 450812 6821 6$300839 1848 0857 OSCA 6875 6884 6893 1 2 3 4 4 5 6 7 8 49 0902 6911 6920 6928 937 6946 09550964 09726981 1 2 3 4 4 5 6 7 8 50 000 0998 7007 70107024 7033 7012 7050 i 70597007 18 8 1 0 7 8 5 4 1 انری | ارور 7رة الزر The copyright of that portion of the above table which gives the logarithms of nuni bers from 1000 to 2000 Is the property of Plessis, Diacruillan and Conipany, Limitol.

·LOGARITHMS 1 149 0 1 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 1 e. 51 70707084 70:3 7101 7110 7118 7120 7133 7143 7152 1 3 3 1 5 6 7 :52 7100 7108 71777185 7193 7202 7410 7218 7226 7235 1 2 2 3 5 6 7 7 53 7213 7231 7259 7267 7275 7284 7202 7300 7308 7316 1 2 2 3 5 6 7 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 73996 1 2 2 3 5 6 ů 7 55 7404 74127419 7427 743571437451 7459 7466 7474 1 % 2 3 4 5 5 6 7 50 7452 7490 74977505 7513 7320 7528 7536 75437031 1 2 ! 31 1 5 5 6 7 577559 7566757475827589 7597 7004 7012 7619 70271 2 2 3 5 5 58 7634 7612 7649 7037 7604 7672 1679 7686 7691 7701 1 1 2 3 4 5 6 7 59 7709 7710 7723 7731 7738 7745 7752 7760 77077774 1 1 2 3 4 11 00) 7782 7789 7711803 7810 7818 7825 7832783978401 1 2 3 6 61 7853 7807868787378027389 78.967003 791079171 1 2 3 4 45 6 0) 62 7924 7:31 7038 7945 7952 7939 7900 7973 7980 79871 1 2 3 3 6 6 63 7993 8000 80078014 8021 028 8035 8041 30185057 1 1 2 8 3 4 5 5 6 64 8002 8069 8075 8082 8089 8096 S102 109 81168122 1 1 2 3 3 4 5 5 6 65 8129-130; 8143 8149 8150 >102 109 81708182 189 112 33 4 5 5 08195 S20292098215 8222 22235 8241 248 9201 1 1 2 3 3 4 0 67 8261 320732748280 287 293 299 300 831283191 1 2 3 3 4 5 5 6 68 8325 331 83388344 $351 357 1363 8370 8376 838 1 1 3 3 5 6 8388 8315 84018107 8414 8120 8120] 8132 5499 8431 1 2 3 3 5 0 70 8131 845718403 $170 84784828158 3194) 8300,001 1 1 5 b) 71 80138719 325 8531 8537 $513 9719 835785618707 1 132 3 72 8573 8579 53838591 8397 8003 002 3015 3021 8027 1 1 2 2 3 4 4 5 5 73 8033 89 1045 8651 8057 8003 86098075 8681 SONO I 1 2 2 3 5 5 74 8642 $4.98 3704 5710 8710972: $727 733 873 87411 1 ! ! 3 3 5 75 8731 87565702 sitis 877779 783 3791879750 1 > 3 3 1 5 70 SSOS511 SS20 8525 8531 $537 8842 $849885468391 1 2 2 3 3 u 77 8805 8871 8570 SS2 8887 883 88998:904 8910 9915 1 1 2 3 3 1 1 78 8921 8927 81328938 9:943 8:249 8934 SOGO $965 19:11 1 2 2 3 3 T9 $976 99289878993 SOOS 9001 900 9015 9020 9023 1 1 2 2 3 3 10:01 903690129047 90.53 0.55 903 90099907490711 3 3 1 3 S 9085 909090909101 9100 0119 117 9122 91291331 1 2 2 3 1 3 S2 913 91439149191.37 9159 :)165 170 1 91159150 91501 1 2 3 3 4 1 53 9191 219092010200 921292179222992329231 1 2 2 3 4 1 13 349213921892539389203 9209:2711 9270281 289 1 1 2 3 15929492999301 9309 9317:3_0239330.335 93101 1 2 3 4 1 86 1:315 9350 9355 9300 9305 :137079375 9350 9385 93.90 I ? 2 3 4 879395 9400 940394109115 9420 912591309435 91100 1 1 2 2 3 3 4 $89415 9150 94.35 910091670469 947194799481901 129 3 3 4 4 899491919) 9301 9500 9513 9318 323 9929533 95881011? 3 34 00:31205470352 03.07 95019576 :1571 907693SI 31 3 1 4 91 9500 9595 9600 2003 2000 190149191 9014 96289033 0 1 1 3 3 4 92 9638 961396479652 96579061 966690719675 SOLO 11 22 3 3 4 4 .3 9085 90899094 9600 9703 9705 97139717972:72701 3 S 4 + 91 9731 97369741 9745 9750 9754 97599713 970897730 1 1 2 3 3 4 4 95 0777 97829786 4791 9795 9800 9805 98019 9811980SU1 1 ? !! 3 3 4 4 900823 98279834 9836 9841 :18 13:1530 3834 99,39 9803 0 1 1 2 2 3 4 4 97 9808 98729877 9881 9880 9800 9804 999 9903 990 0 1 1 2 2 9 9 4 98 9012 9917 9921 9926 9930 934 99999913 99489952 0 1 1 2 2 3 3 4 4 99 0950 0901 9905 9900 99749978 9993 99979991 9990 0 1 1 2 2 3 3 3 4 1 1 + سا celes ·INDEX 36 Absolute temperature, 18 Densities of liquids and suide, Accuracy of figures, 1 23 Acid and alkali solutions, normal, relative, 23 82 Density of ait, 26 Air density of, 26 hydrogen, 26 Analysis, gravimetric, 55 water, 146 of gases, 70 Deviations from gas laws, 113 of minerals, 50 Diffusion of gases, 113 organic, 89 Dulong and Petit's generalization, Approximate and oxact laws, 19 Aqueous solutions, thermo-1 chemistry of roactions in, 102 Electrochemistry, 123 Atomic weights, approximate, 64 Empirical formulee, 89 deduction of, Endothermic reactions, 99 from molecular Equations, use of, for calculating weights, 32 reacting quantities, 6:1 exact, 34 Equilibrium, 120 tablo of, 147 Equivalents, 12 Avogadro's hypothesis, 28 law of, 10 Error of experiment, 4 Boiling points of solutions, 115 influence of, Boyle's law, 15 in special Calorie, 98, 99 Exact and approximate laws, 19 Calorific intensity, 110 Exothermio reactions, 99 power, 110 Capillarity, 115 Charles' law, 17 Fixed ratios, law of, 5 Conservation of mass, law of, 5 Formula of a pure gas, determi. Constancy of combiniug ratios, 5 nation of, 73 Constitutional formulæ of organic Formulæ, constitutional, 92 compounds, 92 empirical, 89 Conversion numbers, 131 of compounds, doduo- Corrections for temperature and tion of, 46 pressure in gases, 15 of minerals, 60 molecular, 90 Dalton's law of partial pressures, Freezing points of solutions, 115 20 Densities, application of, in Gas analysis, 70 weighing, 25 influence of water, of gases, 25 Vapour in, 75 casos, 61 ·INDEX 151 &, 73 'Gas, determination of formula of Logarithms, use of, 4 tablo of, 148 (tas laws, 15, 70 deviations from simple, Marriotte's law, 15 113 Minerals, formula of, 50 Gases, density of, and molecular isomorphous replaceweights, 30 ment in, 50 diffusion of, 113 Molocular formulæ, 90 reacting volumes of, 70 volumes of gases, 31 relationships of temperaweights, 28, 115 turo, pressure, and Multiple ratios, law of, 8 volume in, 15 solubility of, 114 Normal solutions for volumetrio Gay-Lussac's law (reacting analysis, 82 volumes of gases), 28 N.T.P. (Normal temperature and Gay-Lussac's law (tomperature pressure), 20 and volume of gases), 17 Numbers, meaning of, 1 Gram-moleculo, weight of, 31 volume of, 31 Gravimetric analysis, 55 Organio analysis, 89 Osmotic pressure, 115 Oxidizing solutions, normal, 83 Heat of reaction at constant pressure and constant voluine, 109 Partial pressure, 20 Hoat of solution, 102 Percentage compositions from leat, units of, 99 formula, deduction of, 43 Hess' law, 101 Physical chemistry, 112 Hydrogen, density of, 25 Physical state, influence of, on thernial effect, 101 Indirect analysis, 60 Platinum salts, use of, for deter- International atomic weights, mining inolecular weights, 92 tablo of, 133 Pressure, temperature and volume Iodino, use of, in volumetrio in gases, relations between, 15 analysis, 84 Pressure, partial, 20 Isomorphous minerals, 50 Vapour, 20 Law of Boyle (or Marriotte), 15 Reacting quantitios, calculation Charlos, 17 of, from equations, 54 conservation of mass, 5 volumos of gasos, 70 equivalents (reciprocal Reactions at constant pressure and proportions), 10 constant voluine, heat of, 109 fixod ratios (definite pro- Reactions successivo, 58 portions), 5 velocity of, 117 Law of Guy-Lussac (Reacting Reciprocal proportions, law of, 11 volumes of gases), 28 Roduring agopts, uormal solutious Law of Gay-Lussac (temperaturo of, 84 and volume of gases), 17 Recurring decimals, 2 Law of Hess, 101 Refractivitios, 128 Marriotte, 15 Relativa densities, 23 multiplo ratios, 8 reciprocal proportions, 11 Silvor salts, use of, for detormin. Laws, approximate and exact, 19 ing molecular weights, 91 of thormochemistry, 100 Simple ratio, meaning of, 12 Liquids, densities of, 23 Solids, densities of, 23 ·152 INDEX Solubility of gases, 114 Solutions, volumetric analysis of, 82 (normal, decinormal, etc.), 82 Specific heats of solids, 36 Successive reactions, 53 Vapour pressure of water, 20 table of, 146 Velocity of reaction, 117 Volume, pressure, and temperature in gases, relations between, 15 Volume-weight relationships of gases, 62 Temperature, pressure and volume in gases, relations between, 15 Thermochemistry, 99 Volumetric analysis of gases, 70 solutions, 82 Units of heat, 99 Water, density of, 146 Water, vapour pressure of, 20 table of, 146 Valency of elements, 95 Vapcur densities, 27 Weighings, correction to vacuum, THE END.

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