LIBRARY UNIVERSITY OF NORTH CAROLINA Alcove Shelf UNIVERSITY OF N.C. AT CHAPEL HILL 00011546124 CHEMICAL AnJ^^y PEOBLEMS AND REACTIONS, TO ACCOMPANT STOCKHARDT'S ELEMENTS OF CPIEMISTRY. BY JOSIAH P.'^OOKE, Jr. #H:?-L:i3fary il;^ Vnhimliy of North Ca o;;;.. CAMBEIDGE: PUBLISHED BY JOHN BAETLETT, ISoofeseller to tje Sinfbcrsitj). 1857. C A 31 B E I D G E : ELECTROTTPED AND PRINTED BY METCALF AND COMPANY. PREFACE. This book has been prepared solely for the use of the undergraduates of Harvard College. It contains a collection of chemical problems and reactions, with references to the sec- tions of Stockhardt's Elements of Chemistry, and also a few chapters on the chemical nomenclature and the use of chemical symbols, subjects which are not sufficiently developed in that text-book for the purposes of college instruction. In writing chemical symbols the author has adopted a uniform system throughout the volume, which, as he hopes, will be found to be at once expressive and clear. The problems and reactions cover the Inorganic portion of Stockhardt's Elements ; the problems have only been extended to the section on the Heavy Metals. Beyond this, the reactions alone have been given, as it was supposed that, before reaching this section, the student will easily be able to propose problems for himself. In solving many of the problems it will be found convenient to use logarithmic tables of four places, which, with several other tables, will be found at the end of the volume. The student is advised to remove the tables of logarithms, and paste them for use on a card. The difficulty of insuring complete accu- racy in the printing of chemical formulge can be known only to those who have had to see a book of this kind through the press. Several errors have been already discovered, and corrected, but others unquestionably exist. Cambeidge, May I5th, 1857. Digitized by the Internet Archive in 2012 with funding from University of North Carolina at Chapel Hill http://archive.org/details/chemicalproblemsOOcook NOMENCLATURE OF CHEMISTRY. Origin of the Nomenclature. — Previous to the year 1787 the names given by chemists or alchemists to substances were not conformed to any general rules. Many of these old names, such as Oil of Vitriol, Calomel, Corrosive Sublimate, Red-Precipitate, Saltpetre, Liver of Sulphur, Cream of Tartar, Glauber^s and Epsom Salts, are stUl retained in common use. As chemical science advanced, and the number of known substances iucreased, it became important to adopt a scientific nomenclature. The admirable system now in use is due al- most entirely to Lavoisier, who reported to the French Acad- emy on the subject, in the name of a committee, in 1787. This system, now known as the Lavoisierian nomenclature, was generally adopted by scientific men soon after its pub- lication, and has not been materially modified since. In it the name of a substance is made to indicate the composi- tion. Names of the Elements. — The names of the elements are the only ones which are now independent of any rule. Those which were known before the adoption of the nomenclature, 1 2 NOMENCLATURE OF CHEMISTRY. such, as Sulphur, Phosphorus, Iron, Lead, retain their old names. Several of the more recently discovered elements have been named in allusion to some prominent property or some circumstance connected with their history; as, Oxygen from o^vs, yei/mc* (acid-generator) ; Hydrogen, from vbap, yewaca (water-generator) ; Chlorine, from -xkapos (green) ; Iodine, from labrjs (violet) ; Bromine, from ^papos (fetid odor), &c. The names of the newly discovered metals have a common termination, um, as Platinum, Potassium, Sodium ; and the names of a class of the metalloids terminate in ine, as Chlo- rine, Bromine, &c. ; but except in these respects the names of the elements are entirely arbitrary. Classification of Compounds. — There are three orders of chemical compounds : — 1st, Binary Compounds, consisting of two elements, or of the representatives of two elements ; 2d, Ternary Compounds, consisting of three elements, or of their representative ; and Sd, Quaternary Compounds, consisting of four elements, or their representatives. There are some chem- ical compounds containing more than four elements ; but in most cases two or more of these elements are representatives, i. e. occupy the place, of only one, as will be explained farther on. Binary compounds are subdivided into two classes, Elec- tro-Positive Binaries, or Bases, and Electro-Negative Binaries, or Acids. Each of these classes is distinguished by a peculiar set of properties, or at least this is the case with the promi- nent members of either class ; but the two classes merge so gradually into each other, that it is impossible to draw a line of demarcation between them ; and there is a large class of intermediate compounds, which either partake of the proper- NOMENCLATURE OF CHEMISTRY. 3 ties of both, or are entirely indifferent. Indeed, the binary compounds may best be regarded as forming a continuous series of substances, varying in their properties from those of strong acids on the one hand to those of strong bases on the otlier, and with every possible grade of qualities between the two extremes. In this series each binary may be considered as an electro-positive compound or base towards all those which precede it, and as an electro-negative compound or acid to- wards all those which follow it. Ternary compounds are gen- erally, at least in Inorganic Chemistry, composed of two binaries, i. e. of an acid and a base, and are then called Salts. The quaternary compounds are generally composed of two salts, and are called Double Salts. Names of Binaries. — The most important binaries, as well as those which have been the best studied, are the compounds of oxygen with the other elements. To these the generic term Oxide has been applied. The electro-positive binaries are called simply Oxides of the elements of which they con- sist. Thus we have Oxide of Hydrogen, consisting of oxygen and hydrogen. Oxide of Potassium, " " " potassium. Oxide of Sodium, " " sodium. When the name of the metal ends in iim, the name of the compound with oxygen is frequently formed by changing this termination into a, with such other modifications of the termi- nal letters as euphony may require. Thus we use, instead of Oxide of Sodium, Soda. Oxide of Potassium, Potassa. 4 NOMENCLATURE OF CHEMISTRT. Oxide of Calcium, Calcia (or Lime).* Oxide of Barium, Baryta. Oxide of Strontium, Strontia. Oxide of Magnesium, Magnesia. Oxide of Aluminum, Alumina. The two names are in all cases synonymous. Generally oxygen combines with an element in more than one propor- tion ; then, in order to distinguish between the different oxides of the same element, we use various Latin and Greek prefixes, such as suh, proto, sesqui, deuto, hyper. This is well illustrated by the names of the different oxides of mercury and man- ganese, which are as follows. Composition. Names. Suho:side of Mercury Protoxide of Mercury Protoxide of Manganese Sesquioxide of Manganese £li/pero:side of Manganese The electro-negative binary compounds of oxygen (the acids) are named on a different principle. These are called different kmds of acids. If the element forms but one acid with oxy- gen, the name is formed by adding to the name of the acid the termination ic, with such changes of the final letters as eu- phony may require. Thus, carbon and oxygen form Carbonic * The common name Lime is much more frequently used than either of its scientific synonymes, Oxide of Calcium, or Calcia. Indeed, the last has never been in general use. Mercury. Oxygen. 100 4 100 8 Manganese. Oxygen. 27.6 8 27.6 12 27.6 16 Arsenic. Oxygen. 75 24 75 40 Phosphorus. Oxygen. 32 24 32 40 NOMENCLATURE OF CHEMISTRY. Acid. When the element forms two acids, by combining with different amounts of oxygen, the termination ie is reserved for that containing the most oxygen, while the termination ous is given to the other. We have, for example, ArsenioMs Acid Arsem'c Acid Phosphorotw Acid Phosphorzc Acid . If oxygen combines with an element in more than two pro- portions, to form acids, the names are formed with the Greek prefix Jiypo, indicating a less, or the Latin prefix per, indicating a greater, amount of oxygen than that contained in the acids to whose names they are prefixed. The acid compounds of sulphur and oxygen are Sulphur. Oxygen. IIyposVi\]Amvous Acid 16 8 Sulphuroits Acid 16 16 Hypo&\}\^\mxic Acid 16 20 Sulphur^c Acid 16 24 The acid compounds of chlorine and oxygen are Chlorine. Oxj'gen. IIypoQh\.OYous Acid 35.5 8 Chlovous Acid Jlypochloric Acid Chloric Acid PerdiAoTic Acid Very frequently the higher degrees of oxidation of an ele- 1* 35.5 24 35.5 .32 35.5 40 35.5 56 Nitrogen. 14 Oxygi 8 14 16 14 24 14 32 14 40 b NOMENCLATURE OF CHEMISTRY. ment are acids, when the lower degrees are bases, or indifferent compounds. This is the case with the oxides of manganese. Besides the three ah-eady mentioned, there are also Manganese. Oxygen. Manganic Acid 27.6 24 Permangamc Acid 27.6 28 The different oxides of nitrogen are, in like manner. Protoxide of Nitrogen Deutoxide of Nitrogen Nitro^er (Cuprum) Cu = 31.7 Silicon Si = 21.3 Didymium D Silver (Argentum) Ag = lOS.l Erbium E Sodium (Katrimn) Na = 23 Fluorine Fl = 18.9 Strontium Sr = 43.8 Glucinum G = 4.7 Sulphur S = 16 Gold (Aurum) Au = 197 Tantalum Ta = 184 Eydrogen H = 1 Tellurium Te = 64.2 Iodine I = 127.1 Terbium Tb Iridium Ir = 99 Thorium Th = 59.6 Iron (Ferrum) Fe = 2S Tin ( Stannum) Sn = 59 Lanthanium La Titanium Ti = 25 Lead (Plumbum) Pb = 103.7 Tungsten ( 'Wolfram) W = 95 Lithium Li = 6.5 Uranium U = 60 Magnesium Mg = 12.2 Vanadium V = 68.6 Manganese Mn = 27.6 Yttrium Y Mercury (Hydrargyrum^ Hg = 100 Zinc Zn = 32.6 Molybdenum Mo = 46 Zirconium Zr = 22.4 CHEMICAL SYMBOLS. 15 The student will do well to notice in the foregoing list the symbols of those elements whose Latin names commence with letters differing from the initial letters of the English names, since they are not so easily remembered as the others. Chemical Equivalents. — The chemical symbols not only stand for the names of the elements, but also for a fixed pro- portional weight of each. These weights are given in the above table, opposite to the symbols. They have only rela- tive values ; if one is in pounds, all the rest are in pounds ; and if one is in ounces, all the rest are in ounces. We may leave the standard indefinite, and express the weight in parts ; then Al stands for 13.7 parts of aluminum ; Sb stands for 129 parts of antimony, &c. The weight of an element indicated by its symbol is called one equivalent, and it is a law of chemistry that elements always combine by equivalents ; that is, one equivalent of one combines "with one equivalent of an- other, or else several equivalents of one combine with one or with several equivalents of another. As stands for 75 parts of Arsenic, or one equivalent. BK " 68.5 " Barium, « " H " 1 " Hydrogen, « « " 8 « Oxygen, " « In order to express two, three, or more equivalents of an element, we place a figure just below the symbol at its right hand; thus, 02 stands for 16 parts of Oxygen, or two equivalents. 03 " 24 " " three equivalents. These figures merely multiply the symbols beneath which 16 CHEMICAL SYMBOLS. they stand, and must not be confounded with algebraic powers, which are sometimes written in a similar way. We sometimes place the figure, though larger, before the symbol ; 2 means exactly the same thing as Oj. Symbols of Compounds. — In order to form the symbol of a compound, Ave write the symbols of the elements of which it consists one after the other, indicating by means of figures the number of equivalents of each which have entered into com- bination. Thus, H is the symbol of water, a compound con- sisting of one equivalent or one part of hydrogen, and of one equivalent or eight parts of oxygen ; S O3 is the symbol of sulphuric acid, a compound consisting of one equivalent or six- teen parts of sulphur, and of tliree equivalents or twenty-four parts of oxygen ; C12 Hn On is the s^bol of common sugar, a compound consisting of twelve equivalents of carbon, eleven equivalents of hydrogen, and eleven equivalents of oxygen. Binary Compounds. — The symbols of binary compounds are formed by writing the symbols of the two elements to- gether, taking care to place the symbol of the metal, or of the most electro-positive element, first. The binary symbol thus obtained represents always one equivalent of the compound. The weight of this equivalent is evidently the sum of the weights of the equivalents of the elements entering into the compound. 14 + 40 = 54. N O5 stands for one equiv. or 54 parts of Nitric Acid. 16 + 24 = 40. S O3 " " 40 « Sulphuric Acid. CHEMICAL SYMBOLS. 17 6 + 16 = 22. C O2 stands for one equiv. or 22 parts of Carbonic Acid. 1 + 8 := 9. II « « 9 « Water. 28 + 8 = 36. FeO " « 36 " Oxide of Iron. 20 + 8 = 28. CaO " " 28 « Lime. 23 + 8 = 31. Na " " 31 « Oxide of Sodium. 6 + 32 = 38. CS2 " " 38 " Sulphocarbonic Acid. 75+ 48 = 123. As S3 " 123 " Sulphoarsenious Acid. 28 + 16 = 44. FeS " " 44 « Sulphide of Iron. 39 + 16 = 55. KS " " 55 " Sulphide of Potassium. In order to express two or more equivalents of a binary, we place a figure immediately before the symbol, like an algebraic coefficient. A figure so placed always multiplies the whole hinary. 3 S O3 stands for 3 equiv. or 120 parts of Sulphuric Acid. 5PbO " 5 " 560 « Oxide of Lead. Ternary Compounds. — The symbol of a ternary compound is formed by writing together the symbols of the two binaries of which it consists, separated by a comma, taking care to place the most electro-positive binary, the base, first. If the salt" is composed of more than one equivalent of either base or acid, then the number of equivalents must be indicated by 2* 18 CHEMICAL SYMBOLS. coefficients. The ternary symbol thus obtained always stands for one equivalent of the compound, and the weight of this equivalent is evidently the sum of the weights of the equiva- lents of the elements entering into it. 1 + 8 + 16 + 24 = 49. H O, S O3 stands for 1 equiv. or 49 parts of Sulphate of Water (common Sulphuric Acid). 1 + 8 + 14 4- 40 = 63. H O, N O5 stands for 1 equiv. or 63 parts of Nitrate of Water (common Nitric Acid). 39 + 8 + 6 + 16 = 69. K O, C O2 stands for 1 equiv. or 69 parts of Carbonate of Potassa. 23 + 8 + 16 -(- 24 = 71. Na 0, S O3 stands for 1 equiv. or 71 parts of Sulphate of Soda. 39 + 8 + 27 + 24 = 98. K O, Cr O3 stands for 1 equiv. or 98 parts of Neutral Chromate of Potassa. 39 + 8 + 2 (27 + 24) = 149. K 0, 2 Cr O3 stands for 1 equiv. or 149 parts of Bichromate of Potassa. 39 + 8 + 3 (27 + 24) = 200. K 0, 3 Cr O3 stands for 1 equiv. or 200 parts of Trichromate of Potassa. 104 + 8 4- 27 4- 24 = 163. PbO, CrOs stands for 1 equiv. or 163 parts of Neutral Chro- mate of Lead. 2 (104 + 8) + 27 4- 24 = 275. 2 PbO, Cr03 stands for 1 equiv. or 275 parts of Basic Chromate of Lead. In order to express two or more equivalents of a ternary, we enclose the symbol in parentheses, and place before the whole the required figure. Thus, CHEMICAL SYMBOLS. 19 3 (Na 0, S Og) stands for three equivalents of Sulphate of Soda. 5 (2 Pb O, Cr O3) stands for five equivalents of Basic Chromate of Lead. Neutral Salts. — The larger number of inorganic acids com- bine most readily with one equivalent of base, and the salts so formed will be called neutral salts. If the salts contain more equivalents of acid or base than one, they are called acid or basic salts respectively. There are, however, some acids which combine most readily with two or three equivalents of base, in the same way that the others combine with one. Such acids are called bibasic or tribasic acids, in order to distinguish them from the rest, which are frequently called monobasic. Of bibasic and tribasic acids, the most important in inorganic chemistry is Phosphoric Acid. This Is known in three differ- ent conditions. In the first of these it is monobasic, in the second bibasic, and in the third tribasic, the last being the ordinary condition. The three conditions are designated by the symbol aP O5 ; ^P O5 ; ^P O5. The acid eP O5 forms neutral salts when combined with three equivalents of base, the acid bP Og when combined with two, and the acid ^P O5 when com- bined with only one. There are three compounds of the acid and water corresponding to the three conditions, which are rep- resented in symbols by H O, ^P Oj ; 2 H 0, tP Og ; 3 H O, ,P O5. In these compounds we can substitute for the equivalents of water equivalents of other bases, either in whole or in part, forming such compounds as Na O, ^P Og ; 2 Na O, bP Og ; 3]S-aO,,P05 ; [HO, 2 NaO] ^P O5 ; [2 H 0, Na 0] ^P Og. The equivalents of water may even be replaced by different bases, as in the compounds [Na O, Pb 0] bP Og ; [H 0, Na 0, 20 CHEMICAL STJIBOLS. K 0], eP O5 ; [K 0, 2 Mg 0] ,T O5. All the above are sym- bols of neutral salts. As protoxide bases combine most readily with one equivalent of acid, so sesquioxide bases combine most readily mth three equivalents. A neutral salt of a sesquioxide base is therefore one which contains for every equivalent of base three equiva- lents of a monobasic acid, or one equivalent of a tribasic acid, and for every two equivalents of base three equivalents of a bibasic acid. Hence, Fca O3 , 3 S O3 ; 2 Fcg O3 , 3 tP O5 ; P62 O3 , cP O5 are all symbols of neutral salts. On the other hand, Fea O3 , S O3 ; 4 AI2 O3 , 3 ,P O5 ; 2 Al^ O3 , cP O5 are sym- bols of basic salts. It will be noticed, on examining the above symbols, that neutral salts of monobasic acids contain as many equivalents of acid as there are equivalents of oxygen in the base, and neutral salts of bibasic and tribasic acids one half and one third as many, respectively. This rule must be kept in mind when writing the symbols of salts. Compound Radicals. — There is a large class of substances which, although compound, nevertheless act in chemical changes exactly as if they were simple, frequently replacing the ele- ments themselves. Such substances are termed compound radicals. Many of these radicals have, like the elements, received arbitrary names, such as Cyanogen, Ammonium, Ethyle, Acetyle, &c. ; and, moreover, the first letter or letters of these names are frequently used as their symbols. It is best, however, to write out the symbols of the elements form- ing these compounds, and enclose the whole in brackets ; thus, [N H4] stands for Ammonium, [Cj N] for Cyanogen, [C4 H5] for Ethyle, [C4 H3] for Acetyle. The oxides of the compound CHEMICAL SYMBOLS. 21 radicals, like those of the elements, may be divided into acids and bases, and their symbols are Avritten exactly like those of other binaries. Thus, [C4 H3] O3 is the symbol of Acetic Acid. [N H4] O " " Oxide of Ammonium. [C4 H5] " " Oxide of Ethyle. So, also, with the salts. [N H4] O, [C4 H3] O3 is the symbol of Acetate of Ammonia. [C4 Hs] 0, [C4 H3] O3 " " Acetate ofOxide of Ethyle. Water of Crystallization. — Besides the water of constitution, which frequently forms a part or the whole of the base of a salt, most salts combine with water as a whole. This water is held in combination by a comparatively feeble affinity, and may be generally driven off by exposing the salt to the tem- perature of 100° C, and sometimes escapes at the ordinary temperature of the air, the crystals of the salt in all cases fallmg into powder. Its presence is essential to the crystalhne condition of many salts, and hence the name "Water of Crys- tallization. The presence of water of crystallization in a salt is expressed in symbols, by writing after the symbol of the salt, and separated from it by a period, the number of equiva- lents of water. Thus, Fe O, S O3 . 7 H is the symbol of Crystallized Sulphate of the Oxide of Iron (Green Vitriol). H 0, 2 Na O, eP O5 . 24 H O is the symbol of Crystallized Phosphate of Soda. The same salt, when crystallized at different temperatures, not unfrequently combines with different amounts of water of crys- , tallization, the less amounts corresponding to the higher tem- 22 CHEMICAL SraiBOLS. peratures. Thus, the Sulphate of Manganese may be crystal- lized with three different amounts of water of crystallization. Mn O, S O3 . 7 H O when crystallized below 6° Centigrade. Mn 0, S O3 . 5 H O " « between 7° and 20°. Mn O, S O3 . 4 H O « « between 20° and 30°. The crystalline forms of these three compounds are entirely different from each other, proving that the form depends, in part at least, on the amount of water which the salt contains. The symbols of other ternary compounds are written like those of the oxygen salts, and therefore require no further explanation. Below are a few of these symbols, together with those of the corresponding oxygen salts, which may serve as examples. K O, C O2 = Carbonate of Oxide of Potassium. K S, C S2 = Sulphocarbonate of Sulphide of Potassium. K O, As O3 = Arsenite of Oxide of Potassium. K S, As S3 = Sulphoarsenite of Sulphide of Potassium. 3 Na CI, Sb CI3 = Double Chloride of Antimony and Sodium. [NH4] CljPtCla = Double Chloride of Platinum and Potassium. K I, Pt I2 = Double Iodide of Platinum and Potassium. Quaternaries. — The symbols of the double salts are formed by writing together the symbols of the two salts of which they consist, separated by a period. Thus, K O, S O3 . iMg O, S O3 . 6 H O = Double Sulphate of Mag- nesia and Potassa. K O, S O3 . AI2 O3, 3 S O3 . 24 H == Double Sulphate of Alu- mina and Potassa (Alum). When the salts contain water of crystallization, the amount of this water expressed in equivalents is written after the symbol of the salt, as already explained. CHEMICAL EEACTIONS. The various chemical changes to which all matter is more or less liable are termed, in the language of chemistiy, reac- tions, and the agents which cause these changes, reagents. In every chemical reaction we must distinguish between the sub- stances which are involved in the change and those which are produced by it. The first will be termed the factors, and the last the products, of the reaction. As matter is indestructible, it follows that The sum of the weights of the products of any reaction must always be equal to the sum of the weights of the factors. This statement seems at first sight to be contradicted by experience, since wood and many other combustible substances are apparently consumed by burning. In all such cases, how- ever, the apparent annihilation of the substance arises from the fact that the products of the change are invisible gases ; and when these are collected, their weight is found to be equal, not only to that of the substance, but also, in addition, to the weight of the oxygen from the air consumed in the process. As the products and factors of every chemical change must be equal, it follows that A chemical reaction may always be repre- 24 CHEMICAL REACTIONS. sented in an equation hy writing the symbols of the factors in the first member, and those of the products in the second. The I'eaction of sulphuric acid on common salt may be represented by the following equation : 23 + 35 + 1+8 + 16+24 = 23 + 8 + 16+24 + 1 + 35 = 107. NaCl + ^0, SOs = Mi 0,SOs +HC1. The correctness of this may be proved by adding together the equivalents of both sides, when the sums will be found to be equal. In like manner, the reaction of a solution of common phosphate of soda on a solution of chloride of calcium may be represented by the equation 1+8 + 2 (23 + 8)+ 32+40 + 3(20+35) = IIO,2I^aO, ,FOs-^SCaCl-\-Aq* = 3 (20 + 8) + 32+40 + 2(23 + 35) + 1 + 85 = 308. 3CaO, eP05 + 2iVa CI -\- JI CI -\- Aq. So, also, the reaction of hydrochloric acid on chalk, which may be proved like the other two : CaO,CO^-{- !£ CI -\-Aq= CaCl-]-II0-{-Aq-\-COa. Although the equation is the most concise, and therefore in most cases the best form of representing chemical reac- tions, it is nevertheless frequently advantageous, in studying comphcated changes, to adopt a more graphic method, by which the various steps of the process may be indicated. The reactions represented by the preceding equations may be writ- ten thus : — * The symbol Aq, for Aqua, merely indicates the condition of solution, and is not to be regarded in adding up the equivalents in order to prove the equa- tion. CHEMICAL REACTIONS. 25 (2.) HO,2XaO,„ PO^ Chemical reactions may be classed under three divisions. First, those reactions in which a compound is decomposed, and divides into simpler compounds or into elements. E. g. when oxide of mercury is heated, it is decomposed into oxygen gas and metallic mercury. Thus, Hg0 = ^y + O. Again, when Chlorate of Potassa is heated, it is resolved into oxygen gas and chloride of potassium. Thus, K O, CI O5 == K CI + 6 O. 3 26 CHEMICAL REACTIONS. So, also, when sulphate of lead is heated, it is resolved into anhydrous sulphuric acid and oxide of lead. Thus, PbO,S03= PbO+SOs. Such reactions as these Avill be called analytical, and the process analysis. Second, those reactions in which the elements are united to form compounds, or compounds of a lower order to form those of a higher. E. g. when hydrogen and carbon bum in the air, they combine with oxygen to form water or carbonic acid. Thus, + 0=HO; C + 02=C02. Again, when anhydrous sulphuric acid combines with lime to form sulphate of lime. Thus, CaO + S03=CaO,S03. Reactions like these will be called sy7ithetical, and the process synthesis. Third, those reactions in which one element displaces an- other. E. G. when sodium takes the place of hydrogen in water, or zinc the place of hydrogen in dilute sulphuric acid. Thus, ^6> + Na = iVa(9 + H. Zn-^HO,SO^-\-Aq=ZHO,SO,-\-Aq^U.. This division includes also those reactions in which there is a mutual interchange of elements between two compounds. E. g. when a solution of chloride of barium is added to a solution of sulphate of soda, the sodium and barium change places, and we have formed an insoluble precipitate of sulphate CHEMICAL REACTIONS. 27 of baryta and chloride of sodium (common salt), which remains in solution. Thus, Ba a-\-Na 0, S 0^-^- Aq = Bs, 0,^0, -\- Fa Cl+Aq. Reactions like these will be called metathetical, and the process metathesis* Of the three classes of chemical reactions, the last is by far the most important ; indeed, the larger number of reactions described in an elementary treatise on chemistry are examples of metathesis. All metathetical reactions can be illustrated very elegantly with the aid of mechanical diagrams, as fol- lows : — (4) (5.) They are easily made by printing with stencils on the larger piece of pasteboard, A B, Fig. 4, the symbols of the elements not disturbed in the reaction, and on the smaller piece, a b, the symbols of the elements which exchange places. The smaller piece having been fastened to the larger by means of an eye- let at O, the reaction is represented by merely turning it half round. (See Fig. 5.) If the symbols of the interchang- ing elements are not symmetrical on all sides, it is of course necessary to make them reversible, by printing each on a sep- arate small square of pasteboard, fastened by an eyelet to the * From the Greek /aerar I'^Tj/xt, to displace or to transpose. 28 CHEMICAL REACTIONS. top or bottom of the revolving piece a b, since otherwise the letters would be inverted when the diagram is turned. This method of illustration may, with a little ingenuity, be extended to some of the most complicated cases of chemical change. The most important condition of chemical action is, that the particles of the substances involved in the change should be indued with freedom of motion. This condition is generally fulfilled, both in nature and in our laboratories, by bringing the substances together in solution, either in water or in some other fluid. When substances are brought together in solution, there are two circumstances which, more than any others, determine the nature and extent of the resulting change. First, If hy an interchange of analogous elements an insoluble coynpound may he formed, this compound always separates from the fluid as a precipitate. As this circumstance is by far the most important of all in determining chemical reactions, it requires full illustration. Ba 0, N 0: HO, SO, 1 , . HO,NOs\ . |+^^=Ba6,S0:|-^?- In these examples, and in general throughout the volume, the symbols of the substances, when in solution, are printed in italic letters, and the solid precipitate in Roman letters. The symbol Aq, as already stated, stands for an indefinite amount of water, in which the substances are supposed to be dissolved. It is obvious, from the above examples, that, in order to ascer- CHEMICAL REACTIONS. 29 tain whether two salts will react on each other, when brought together in solution, so as to form a precipitate, it is only neces- sary to Avrite the symbol of one under that of the other, and interchange the symbols of the metallic elements. If either compound whose symbols are thus formed is insoluble in the menstruum present, a reaction will take place, and the insoluble compound will be precipitated. At the end of the volume will be found a table, reprinted from the English edition of Fre- senius's Qualitative Analysis, by means of which the student can easily ascertain from inspection the solubility of any of the more frequently occurring binary compounds or salts, and thus will be able to solve the following problems. Problem 1. If chloride of barium and sulphate of soda are mixed together in solution, wiU there be a reaction ; and if so, what will be formed ? Problem 2. If chloride of sodium and nitrate of silver are mixed together in solution, will there be a reaction, &c. ? Problem 3. If sulphide of hydrogen and nitrate of lead are mixed together in solution, will there be a reaction ? Problem 4. If sulphide of hydrogen and sulphate of zinc are mixed together in solution, will there be a reaction ? In solving the last problem, it must be noticed that the fluid which would result from a reaction would be a weak acid, in which many substances are soluble which would be insoluble in pure water, as may be seen from the table. Problem 5. If chloride of sodium and sulphate of copper are mixed together in solution, will there be a reaction ? Problem 6. If sulphuric acid and borate of soda are mixed together in solution, will there be a reaction ? It will be found that, by an interchange of metallic elements 3* 30 CHEMICAL REACTIONS. in the last two examples, no insoluble compound will be formed, and hence the conclusion follows from our data, that there will be no precipitate. We must not, however, conclude from this that there will be no reaction, since, as can easily be seen, it does not necessarily follow, because the possible formation of an insoluble compound always determines a reaction, that the re- verse is equally true, and that no reaction can take place unless an insoluble compound is formed. Indeed, in the last two ex- amples, we are able, from incidental phenomena, to determine satisfactorily that a change does result ; thus, in Problem 5, when the solutions are mixed, the blue color of sulphate of copper changes into the green color of chloride of copper ; and in Problem 6, if sulphuric acid is not added in excess, the claret color to which blue litmus-paper turns in the mixed solution proves that it is boracic acid, and not sulphuric acid, wlaich is in a free state ; nevertheless, in most similar cases it is impossible to determine, unless an insoluble compound is formed, whether any reaction has taken place. Second. The circumstance which, next to insolubility, is most important in determining metathetical reactions, is vola- tility, and it may be laid down as a general principle, that, If hy an interchange of analogous elements between two substances in solution, a substance can be formed, which is volatile at the temperature at which the experiment is conducted, such an inter- change cdioays takes place, and the volatile product is set free. In order to illustrate this principle, a few examples may be adduced. 1. If diluted sulphuric acid is poured upon granulated zinc, a brisk evolution of hydrogen gas ensues, and sulphate of oxide of zinc is retained in solution. Thus, CHEMICAL REACTIONS. 31 Zn 1 jH HO,SO^-{-Aq\ — \ZnO,SO,-\-Aq. In this example, and those that follow, the volatile or gaseous products are always printed with a full-face type. 2. If diluted sulphuric acid is poured upon protosulphide of iron, sulphide of hydrogen gas escapes, and sulphate of protox- ide of iron remains in solution. Thus, FeS ) (HS HO,SO^-{-Aq] — \Fe 0,SO^-\-Aq. It will be noticed from the last two reactions, that it is not essential that more than one of the factors of the reaction should be fluid, or in solution. 8. If strong sulphuric acid is poured upon common salt, and the mixture slightly heated, chlorohydric acid gas is evolved, and bisulphate of soda remains dissolved in the excess of sul- phuric acid. Thus, Na a 7 f H CI } = HO,SOs . HO, ^6>3 1 — XNa 0, S O3 . HO, S 0,. 4. If strong sulphuric acid is poured upon nitre, and the temperature of the mixture slightly elevated, the vapor of nitric acid is given off, and bisulphate of potassa is formed. Thus, KO,NO, \ fH0,JV05 HO, 8 0^.H0,S0^\ — \K0, S O3 . HO, S O3. 5. If diluted nitric acid is poured upon chalk, or any other analogous carbonate, carbonic acid gas is set free, and a salt of nitric acid formed. Thus, CaO, CO, ) _ j H0-\-CO, HO,NO,^Aq\ — \ OaO,J^Os-\-Aq. 32 CHEMICAL KEACTIONS. It is very frequently the case that two, or even all of the three, classes of chemical reactions are combined, and going on simultaneously, in a single experiment. In the last example, for instance, the metathesis is succeeded by an analysis of one of the products, owing to the want of aflBnity between C O2 and H O. Again, the reaction of nitric acid on copper is an ex- ample where all three varieties of reactions are combined. Three equivalents of copper react on four equivalents of nitric acid, and the reaction may be conveniently studied in two parts. In the first part, the copper is oxidized by one of the equiva- lents of the acid ; here Ave have analysis accompanied by syn- thesis ; and in the second part the copper changes place with the hydrogen of the acid, a case of metathesis. 1st. 3Cu + ^0, iV^(95 = 3(CuO, H0)-f-]\O2. HO,NO,\, SGuO,NOs 1 , . Z"^- CuO, H0[ +^^~ l^O + ^Oj +^^- The whole reaction combined may be expressed thus : 3 Cu + 4 {HO, NO,) -{-Aq = 3(at 0,N0,) +^g + JVO^. Such mixed processes are very common in all complex cases of chemical change. StocJtiometrieal Problems. — The chemical symbols enable us not only to represent chemical changes, but also to calculate exactly the amounts of the substances required in any given process, as well as the amounts of the products wliich it will yield. The method of making such calculation can best be illustrated by examples. In these examples the weights and measures will be given according to the French decimal system, which is now very generally used in chemical laboratories, and CHEMICAL REACTIONS. 33 which, on account of the very simple relation between the units of measure and of weight, greatly facilitates stochiometrical calculations. The French measures and weights can, when required, be very easily reduced to the English standards, by means of a table at the end of the volume. Pi'oblem 1. We have given 10 kilogrammes of common salt, and it is required to calculate how much chlorohydric acid gas can be obtained from it by treating with sulphuric acid. The reaction is 23+35.5 = 58.5. 1+35.5 = 36.5. I^eiCl-{- ffO,S03-{-Aq = mO,SOs-\-Aq-\-U€L Hence one equivalent, or 58.5 parts, of common salt, will yield one equivalent, or 36.5 parts, of chlorohydric acid gas. There- fore the amount which 10 kilogrammes will yield can be cal- culated from the proportion 53.5 36.5 Na CI : H CI = 10 : X = 6.239 kilogrammes, Ans. Problem 2. It is required to calculate how much chlorine gas can be obtained from chlorohydi'ic acid with 6 grammes of hyperoxide of manganese. The equation representing the reaction is 28.6+16 44.6. 35.5. Mn02 -\- 2 H 01 -\- Aq = Ml a -\- 2 HOi^Aq + €1. Hence the amount of hyperoxide of manganese represented by Mn O2, or 44.6 parts, yields an amount of chlorine gas repre- sented by CI, or 35.5 parts, and we have the proportion, 44.6 35.5 Mn O2 : CI = 6 : a; = 4.775 grammes. Problem 3. It is required to calculate how much sulphuric 34 CHEBIICAL REACTIONS. acid and nitre must be used to make 250 gi'ammes of the strongest nitric acid. 101.2 98 63 K0,-^0,+2(JI0,S0,)=K0,S0s.II0,S0s-\-nO,NO,. Hence, 63 98 H0,I^0si2{II0,S O3) = 250 : x. 63 101.2 HO,NO^ : K 0, N Og = 250 : x. From the above examples we can deduce the following gen- eral rule for calculating from the amount of any given factor of a chemical reaction the amount of the products, or the reverse. Express the reaction in an equation : make then the proportion, As the symbol of the substance given is to the sym- bol of the substance required, so is the amount of the substance given to x, the amount of the substance required ; reduce the symbols to numbers, and calcidate the value of x. On account of the very great lightness, the amount of a gas is very much more frequently estimated by measure than by weight. At the end of the volume a table will be found giving the weight in grammes of one thousand cubic centimetres, or one litre, of each of the most important gases. With the aid of this table such problems as the following may be solved. Problem 4. How much chlorate of potassa must be used to obtain one litre of oxygen gas? One litre of oxygen gas weighs 1.43 grammes. K 0, CI O5 = K CI + 6 O. 48 122.7 6 : K 0, CI O5 = 1.43 : x. CHEMICAL REACTIONS. 35 Problem 5. How much zinc and sulphuric acid must be used to obtain 4 litres of hydrogen gas ? Four litres of hydrogen weigh 0.357 grammes. Zn -^HO,SOs-\-Aq= Zn 0, S 0^ -\- Aq -\- H. 1 32.6 H : Zn == 0.357 : a; = amount of zinc required. 1 49 H : HO, S03 = 0.357 : x = amount of sulphuric acid required. Problem 6. If ten grammes of water are decomposed by galvanism, how large a volume of mixed gases will they give ? ^0 = M+O. 9 1 JIO : H = 10 : x = i^- grammes of hydrogen.^ 9 8 ^0:0=10 : x = ^^- grammes of oxygen. J^ grammes of hydrogen occupy 12.^429 cubic centimetres. -^ grammes of oxygen occupy 6.214 " " The mixed gases occupy 18,644 " " And hence * water, when decomposed into its elements, expands 1864 times. The use of chemical symbols, both in expressing chemical reactions and in stochiometrical calculations, having been ex- plained, they will be used in the following pages to illustrate the text of Stockhardt's Elements of Chemistry. The reac- tions described in that work are represented in the form of equations, the first member containing always the factors, and * It must be remembered that one cubic centimetre of water weighs one sramme. 36 CHEMICAL KE ACTIONS. the second, the products of the process. It remams for the student to work out the reaction, and represent it in the manner explained on page 25. The equivalents of water which are set free or formed during a reaction, are not generally indicated, but are merged in the general symbol ^5'; and the student will frequently be obliged to supply these equivalents in order to work out the reaction. The symbols of solids are printed in Roman letters, those of fluids in italics, and those of gases in full-faced type. "When, however, the solid or gas is dis- solved in water, the symbol is printed in italics, followed by the general symbol Aq, or aq. Aq always stands for an indefi- nite and large amount of water ; ag, for an indefinite but small amount of water. The color of a substance, especially of pre- cipitates, is frequently printed above its symbol. The head- ings and figures, or letters at the side of the page, refer to the sections of the above-mentioned book. In order still further to illustrate the subject, a large number of problems have been added to the reactions, which the stu- dent is expected to solve. The method of solving these prob- lems can, in most cases, be deduced from the explanations already given, and from the sections of the text-book ; in all other cases the method is explained under the problem. Throughout the following pages, all gases and vapors are sup- posed to be measured at the temperature of 0° and when the barometer stands at 76 centimetres, unless some other tempera- ture or barometric pressure is expressly stated. All specific gravities are referred to water at its maximum density (at 4°). The temperatures are all given on the Centigrade scale, and the weights and measures according to the French decimal system. EEACTIONS AND PROBLEMS. WEIGHING AND MEASUEING. 10. Problems on French System of Weights and Measures. 1. Reduce by means of the table at the end of the book, — a. 30 inches to fractions of a metre. h. 76 centimetres to English inches. c. 36 feet to metres. d. 10 metres to feet and inches. 2. Reduce by means of the table at the end of the book, — a. 8 lbs. 6 oz. to grammes. h. 7640 grammes to English apothecaries' weight. c. 45 grains to grammes. 3. "What is the diameter, and what is the circumference, of the globe in French measure ? 4. What is the distance from Dunkirk in France to Barce- lona in Spain ? The latitude of Dunkirk =51° 3', that of Barcelona = 41° 22', and the two places are on the same meridian. 5. "Were our globe composed entirely of water at its great- est density, what would be its weight in kilogrammes ? 6. What is the weight of one cubic decimetre of water ? 7. Reduce by means of the table at the end of the book, — a. 4 pints to litres and cubic centimetres. h. 5 gallons to litres and cubic centimetres. 40 WEIGHING AND MEASURING. c. 5 litres to English measure. d. 4 cubic centimetres to English measure. 11-17. Problems on Specijlc Gravity. 1. The specific gravity of iron = 7.84. What is the weight of 1 cubic centimetre, 4 cubic centimetres, &c. of the metal m grammes ? 2. The specific gravity of alcohol = 0.81. "What is the weiglit of one litre in grammes ? of 45 cubic centimetres, &c. ? 3. The specific gravity of sulphuric acid = 1.85. K you wish to use in a chemical experiment 250 grammes, how much must you measure out ? 4. Knowing the specific gravity of any given substance, how can you calculate the weight corresponding to any given meas- ure, or the measure corresponding to any given weight ? Give a general algebraic formula for the purpose, representing spe- cific gravity, weight, and volume, by Sp. Gr., W., and V. 6. Determine the Sp. Gr. of absolute alcohol from the following data : — Grammes. Weight of bottle empty, 4.326 " " fiUed with water at 4°, 19.654 « « " alcohol at 0°, 16.741 6. Determine the Sp. Gr. of lead shot from the following data : — Grammes. Weight of bottle empty, 4.326 " " « filled with water at 4°, 19.654 « " shot, 15.456 " " bottle, shot, and water, 33.766 7. Determine the Sp. Gr. of iron from the following data : — Grammes. Weight of iron in air, 3.92 " " « water, 3.42 WEIGHING AND MEASUKING. 41 8. Determine the Sp. Gr. of copper from Grammes. Weight of copper in air, 10.000 « " « water, 8.864 9. Determine the Sp. Gr. of ash wood from Grammes. Weight of wood in air, 25.350 " a copper sinker, 11.000 " wood and sinker under water, 5.100 10. How much bulk must a hollow vessel of copper fill, weighing one kilogramme, which will just float in water ? 11. How much bulk must a hollow vessel of iron occupy, weighing ten kilogrammes, which sinks one half in water ? 12. An alloy of gold and silver weighs ten kilogrammes in the air, and 9.735 kilogrammes in water. What are the proportions of gold and silver ? Sp. Gr. of gold = 19.2, of silver = 10.5. Solution. — In the French system the volume of a solid in cubic centimetres w equals its weight in grammes divided by its Sp. Gr., or V = g — q^. Since one cubic centimetre of water weighs one gramme, the volume of a solid in cubic centimetres is equal to the weight of water it displaces in grammes. Hence the w weight of water displaced = g — ^. Put, then, x = weight of gold in alloy, 10 — X will equal weight of silver, y^ = weight of water displaced by gold, and -jjy^ = weight of water displaced by silver. Hence -^ + ~^ = 0.265. 13. An alloy of copper and silver weighs 37 kilogrammes in the air, and loses 3.666 kilogrammes when weighed in water. What are the proportions of silver and copper ? 22. Problems on Expansion of Liquids and Gases. 1. If 34.562 cubic centimetres of mercury at 0° are heated to 100°, what increase of volume do they undergo, and what is the increased volume ? Solution. — The small fraction of its volume by which one c. c. of a liquid 4* 42 WEIGHING AND MEASURING. or gas increases ■when heated from 0° to 1°, is called the coefficient ofeocpandon of that liquid or gas. The coefficient of expansion of mercury, for example, = 0.00018, that is, one c. c. of mercury at 0° becomes 1.00018 c. c. at 1°. If Tve assume that the expansion is proportional to the temperature, then one c. c. at Qo becomes 1.00036 at 2°, 1.0009 at 5°, and 1.018 at 100°. Hence, 84.562 c. c. of mercury would become, at 100°, (34.562 X 1.018) c. c. To make this solution general, let h = coefficient of expansion ; then (1 + ^) = increased volume of one c. c. when heated from 0^ to 1°, and (1 + < h) = increased volume of one c. c. when heated from 0° to <°, and V (1 + < ^) = increased volume of V c. c. when heated from 0° to t° ; representing by V the increased volume, we have Y'=Y {1 + toli), from which the increased volume of any liquid or gas may be calculated when the volume at 0°, the coefficient of expansion, and the temperature are knoAvn. It is not true, as we have assumed, that liquids expand twice as much for two degi'ees, three times as much for three degrees, &c., as they do for one; but, on the contrary, the rate of expansion slowly increases with the temperature. For example, one c. c. of mercury at 0° becomes 1.000179 at 1°, but one c. c. at 300o becomes 1.000194 at 331°. This difference of rate, however, is so small, that we can neglect it, except in the most refined experiments, more especially if we use not the coefficient observed at any particular temperature, but a mean coefficient obtained by observing the total amount of expansion between 0° and 100°, and then dividing the result by 100, by which we averrge the error. Again, experiments on the expansion of fluids are commonly con- ducted in glass vessels, which expand themselves by heat, and therefore cause the expansion to appear less than it is. If they expanded as much as the fluid, it is evident that the fluid would not appear to expand at all. They in fact expand much less than the fluids, but, nevertheless, sufficiently to make a material difference between the absolute expansion of a fluid, and its apparent expansion in glass vessels. The mean absolute coefficient of mercury between 0° and 100° is 1.000181. The apparent expansion in glass between 0° and 100° is 1.000156. In Table IV., at the end of the volume, will be found the mean coefficients of expansion in glass of some of the more important fluids. The rate of expansion of water varies so rapidly and so anomalously, that no use should be made of its coefficient except in experiments extending over 100°. When it is required to determine the amount of expansion between naiTower limits, use should be made of Table V., which gives the volume to which one cubic centimetre of water at 0° or at 4° increases, when heated to the tem- peratures at the side of the table. It also gives the Sp. Gr. of water at different temperatures, when either water at 0° or at 4° is taken as the unit. 2. What will be the volume of 5.346 c. c. of water at 0°, when heated to 100° ? 3. What will be the volume of 250 c. c. of oil of turpentine, at 0°, when heated to 50° ? WEIGHING AND MEASUKING. 43 4. What will be the volume of 35 c. c. of water at 0°, when heated to 4°, to 25°, to 40°, and to 84° ? 17. Problems on reducing Specific Gravities to the Standard Temperature. The specific gravities of both liquids and solids are supposed to be referred to water at 4°, its greatest density ; but in prac- tice we always use water at a much higher temperature, and it becomes therefore necessary to reduce the results to 4°, or in other words, to calculate what would have been the result had the temperature been 4° during the experiment. Hence an important class of problems like the following : — 1. The Sp. Gr. of zinc was found to be 7.1582 when the temperature of the water was 15°. What would have been the Sp. Gr. at 4° ? Solution. — Sp. Gr. of water at 4° : Sp. Gr. at 15° = Sp. Gr. of zinc at 15° : Sp. Gr. at 4°; or 1 : 0.9992647 = 7.1582 : x. 2. The Sp. Gr. of antimony was found to be 6.681 when the temperature of the water was 15°. What would have been the Sp. Gr. at 4° ? Ans. 6.677. 3. The Sp. Gr. of an alloy of zinc and antimony was found from the following data : — Grammes. Weight of the alloy, 4.4106 " Sp. Gr. bottle, 9.0560 " " " full of water at 4°, 19.0910 « bottle, alloy, and water at 14°.6, 22.8035 Ans. 6.375. 4. Find the Sp. Gr. of metallic zinc from the following data : — Grammes. Weight of the zinc, 12.4145 « bottle, 9.0560 " « full of water at 18°, 19.0790 " " zinc and water at 12°.4, 29.7663 Ans. 7.153. 44 WEIGHING AND MEASURING. 24. Problems on reducing Centigrade Degrees to Fahrenheit, and the reverse. 1. What do —40°, —20°, —15°, —9°, 0°, 4°, 10°, 13°, 15°, 80°, 100°, 150° Cent, correspond to on the Fahrenheit scale ? Ride. — Double the number of degrees, subtract one tenth of the whole, and add 32 if the degrees are above 0°, or sub- tract 32 if they are below. 2. What do —40°, —32°, —18°, —7°, 0°, 10°, 32°, 42°, 50°, 70°, 90°, 100°, 150°, 212°, 300°, 450° Fahr. correspond to on the Centigrade scale ? Rule. — Subtract 32 if above 0°, or add 32 if below, add one ninth to the result, and divide by two. 27. Problems on Expansion of Solids. 1. What will be the length of a rod of iron 424.56 metres long at 0°, when heated to 20° ? Solution. — The small fraction of its length by which a rod of iron, or of any other solid, one metre long, expands when heated from 0° to 1°, is called the coefficient of linear expansion of the solid. A bar of iron one metre long at 0° becomes 1.0000122 at 1°, and the small fraction 0.0000122 is the coefficient of linear expansion of iron. Eepresenting this coefficient by h, Ave have for the new length of the rod, at 1°, V = 1 (1 + X"), and at io^ V = l{\-\-t'k). We may assume, in the case of solids, that the expansion is proportional to the temperature, especially if we deduce our coefficient from experiments made between 0° and 100°, as described under examples on the expansion of fluids. Substituting for ?, ^, and t the values given in the problem, we have 11 = 424.56 (1 + 20 x 0.0000122). The coefficients of linear expan- sion for a number of solids are given in Table IV. at the end of the volume. 2. What will be the length of a rod of copper 2.365 metres long, at 0°, when heated to 100° ? 3. What will be the length of a rod of silver 0.760 metres long, at 12°, when heated to 20° ? Solution. — Denoting by I the unknown length of the rod at 0°, hy I' the known length at i°, and I" the required length at t'o, we have as above, l'=l{l + tk), and l"^Hl + fk). WEIGHING AND MEASURING. 45 By combining these, we obtain ^" = ^ (rrfi) = ^' [1 + ^ «' - + &c.]. All the terms of the quotient may be neglected after the first, because they contain powers of the already very small fraction Ic. Substituting the values in the above equation, we get l' =z 0.760 [1 + 0.000019 (20 — 12)]. 4. One of the large iron tubes of the Britannia Bridge over the Menai Straits is 143.253 metres long. What increase of length does it undergo between 8° and 20° Centigrade ? 5. What is the increased capacity of a globe of glass which holds exactly one litre, at 0°, when heated to 250° ? Solution. — If we consider for a moment the glass globe as forming the out- side shell of a solid globe of glass, it is evident that the increased capacity of the globe will be equal to the increased volume of this solid globe, which we have supposed for a moment to fill the interior. The problem, therefore, resolves itself into calculating the amount of cubic expansion of a solid glass globe having a volume equal to one litre, or 1000 c. c. The coefficient of linear expansion of glass is given in the table as 0.00001. The coefficient of cubic expansion is always three times * as great as the linear expansion ; in the case of glass, therefore, it is 0.00003. By this is meant that, if a rod of glass one metre long at 0° becomes 1.00001 long at 1°, then a cube of glass of one cubic centimetre at 0° becomes 1.00003 c. c. at 1°. Using, then, the equa- tion we obtain by substitution V = 1000 (1 + 250 X 0.00003) = 1007.5 c. c. 6. What is the increased capacity of a globe of glass which holds exactly 495 c. c. at 15°, when heated to 300° ? * Each edge of a cube of glass one centimetre long at 0°, would become (1 + h) c. m. long at 1°. The increased volume of the cube would be equal to (1 -)- Z;)3 = 1 + 3 ^- 4- 3 X-2 + 7j3; but as Ic is an exceedingly small fraction, Z;2 and Z;3 may be neglected in comparison, so that a cube of glass of one c. c, at 0° becomes {1 + Zh)^i 1°, which proves that the amount of cubic expan- sion is three times as great as the linear. NON-METALLIC ELEMENTS, or METALLOIDS. FIRST GEOUP : ORGANOGENS. Oxygen (0). 56. Hg b = ^ + ©. 1. How much oxygen can be obtained by heating 108 grammes, 250 grammes, 25 grammes, or 5 grammes, of red oxide of mercury ? 2. How many cubic centimetres of oxygen can be obtained from the amounts of oxide of mercury given in the last example ? 3. How much oxide of mercury would be required to yield one litre of oxygen by the process of 1 ? 4. How much mercury would remain after the experi- ment of last example ? 59. KO, CIO5 = KCl+6©. 1. How much oxygen can be obtained from 1 kilo- gi-amme, 50 grammes, or 5 grammes, of chlorate of po- tassa ? 2. How much chloride of potassium would remain after the oxygen in the last examples had been driven off? NON-METALLIC ELEMENTS, OR METALLOIDS. 47 3. How much cMorate of potassa would be required to make one litre of oxygen ? 63. C + 2 O = C O2. 1. How much carbonic acid gas will be formed by burn- ing 5 grammes of carbon ? 2. How many cubic centimetres of carbonic acid will be formed by burning 5 grammes of carbon ? 3. How much oxygen will be consumed in the last two examples ? 4. How many cubic centimetres of carbonic acid will be formed from one litre of oxygen, by burning in it carbon ? 64. S + 2 O = S O2. 1. How much sulphurous acid gas wiU be formed by burning 10 grammes of sulphur ? 2. How much oxygen will be consumed in the last example ? 3. How many cubic centimetres of sulphurous acid wiU be formed by burning sulphur in one litre of oxygen ? 4. Assuming that one litre of oxygen yields exactly one litre of sulphurous acid, what is the Sp. Gr. of S Og gas ? 65. P + 50 = P05. 1. How much phosphoric acid can be formed from 48 grammes of phosphorus ? 2. How much phosphorus will exactly consume one litre of oxygen ? 3. How much phosphorus is required in order to make 250 grammes of phosphoric acid ? 48 NON-METALLIC ELEMENTS, OR METALLOIDS. 67. NaH-© = NaO. 1. How much oxide of sodium can be made from 28.75 grammes of sodium ? 2. How much oxide of sodium can be made with one litre of oxygen ? and how mucli sodium will be consumed in the experiment ? 68. 3 Fe + 4 O = Fe O, Fea O3. 71. 2NaO-^POs-^Aq= HO, 2NaO,PO,-{- Aq. The ciystallized salt = [HO, 2NaO] PO5 . 24HO. 72. Ca0^aO^^Aq= CaO, CO2 + Aq. 73. CaO-{- SOi-\-Aq= Ca 0,S0^-\- Aq 1. How much lime would be required to neutralize 20 grammes of sulphurous acid ? 2. How much lime would be required to neutralize the sulphurous acid obtained by burning 5 grammes of sulphur ? 3. How much lime would be required to neutrahze one litre of sulphurous acid gas ? 79. 3 Mn Oi = Mn O, Mng O3 + 2 O. 1. How much oxygen can be obtained from one kilo- gramme of hyperoxide of manganese by heating ? 2. How much of the oxide must be used in order to obtain 30 grammes of oxygen ? IVInOa + HO, ^S C>3 = MnO, S O3 4- HO + O. 1. How much oxygen can be obtained from one kilogr. of hyperoxide of manganese by the last process ? and how much sulphuric acid wiU be required to decompose it ? 2. "What per cent of the whole amount of oxygen in the mineral is obtained by the two processes ? and by how much does the second exceed the first ? NON-METALLIC ELEMENTS, OR METALLOIDS. 49 Hydrogen (H). 81. Na + ^g = Na 0, HO^Aq-\- H. 1. How much hydrogen will be set free by 5.25 grammes of sodium? How many cubic centimetres will be set free? 82. 3 Fe + 4 M © == Fe O, Fcj O3 + 4 H. 1. How much hydrogen would be obtained in the last experiment by the decomposition of 39 grammes of water ? 2. By how much would the weight of the tube increase in last example ? 83. Zn + HO, S 0^^ Aq = Zn 0, S 0^-^ Aq -\-ll. 1. How much sulphuric acid and how much zinc must be used in order to make 2 grammes of hydrogen ? How much in order to make one litre ? 2. How much hydrogen can be obtained with one kilo- gramme of zinc ? 3. How much with one kilogramme of sulphuric acid ? Decomposed by Galvanism. 55. iro = H + ©. 1. How many cubic centimetres of hydrogen would be obtained by the decomposition of one cubic centimetre (one gramme) of water ? How many cubic centimetres of oxygen ? How many of mixed gases ? 85. A solid immersed in a liquid or a gas is buoyed up by a force equal to the weight of the liquid or gas which it displaces. The excess of the buoyancy over its own weight is called its ascensional force. 1. What would be the ascensional force of a small balloon filled with one htre of hydrogen gas, when the balloon itself weighs five centigrammes ? 5 50 NON-METALLIC ELEMENTS, OR METALLOIDS. 2. What would be the ascensional force of a spherical balloon seven metres in diameter, two thirds filled with hydrogen, when the balloon and attachments weigh twenty kilogrammes ? 87. n-i-o = iro. 1. How much water would be formed by burning one thousand litres of hydrogen? and how much oxygen would be consumed in the process ? 2. How much vapor of water would be formed in Ex- ample 1 ? 93. Problems on the Barometer. 1. When the surface of the column of mercury in a barometer stands at 76 centimetres above the mercury in the basin, with what weight is the atmosphere pressing on every square centimetre of surface ? Sp. Gr. of mercury = 13.596. 2. To what difference of pressure does a difference of one centimetre in the barometric column correspond ? 3. When the water barometer stands at ten metres, what is the pressure of the air if the temperature is 4° ? 4. How high would an alcohol barometer, and how high a sulphuric-acid barometer, stand under the same circumstances, disregarding in each case the tension of the vapor ? Sp. Gr. of alcohol = 0.8095 ; Sp. Gr. of sulphuric acid = 1.85. 94. Prohlems on the Compression and Expansion of Gases. Mariotte's Law. — It is an estabhshed principle of sci- ence that I'he volume of a given weight of gas is inversely as the pressure to which it is exposed ; that is, the greater the pressure, the smaller is the volume ; and the less the NON-METALLIC ELEMENTS, OR METALLOIDS. 51 pressure, the larger is the volume. This may be illustrated by an India-rubber bag holding one litre of air, or of any other gas. This is exposed to a pressure, under the ordi- nary conditions of the atmosphere, of a little over one kilogramme on every square centimetre of surface. If this pressure is doubled, the volume of the bag will be reduced to one half; if trebled, to one third, &c. On the other hand, if the pressure is reduced to one half, the volume will double ; if to one third, the volume will treble, &c.* The principle is expressed in mathematical language by the proportion - H' : H=: V : V; (1.) where H and H' are the heights of the barometer which measure the pressure to which the gas is exposed under the two conditions of volume V and V. Since the density of a given weight of gas is inversely as the volume, or D' : D = V : V, it follows that H' : H = D' : D, (2.) or the density of a gas is proportional to the pressure to which it is exposed. Moreover, since the weight of a given volume of gas is proportional to its density, and its density, as just proved, proportional to the pressure, it follows that The weight of a given volume of gas is directly as the pressure to which it is exposed, or H' : H = W : W. (3.) These three proportions are very important, and will be constantly referred to in the following pages. The student must be careful to notice that in (1) the weight of gas is supposed to be constant and the volume to vary, and in (g) the volume is supposed to be constant and the weight to vary. * We suppose the bag to have no elasticity. 52 NON-METALLIC ELEMENTS, OR METALLOIDS. The variations in the pressure of the atmosphere, amount- ing at times to one tenth of the whole, necessarily cause equally great changes in the volume of gases which are the objects of chemical experiment. Hence, in order to compare together different volumes of gas, it is essential that they should have been measured when exposed to the same pressure. A standard pressure has therefore been agreed upon, that measured by 76 centimetres of 3iiercury, to which the volume of gases measured under any other pressure should be reduced. Hence a number of problems like the following : — 1. A volume of hydrogen gas was measured and found to be equal to 250 c. c. The height of the barometer, observed at the same time, was 74.2 centim. What would have been the volume if observed when the barom- eter stood at 76 centim. ? Solution. — Proportion (1) gives, by substituting the data of the prob- lem, 74.2 : 76 = 250 : V = Ans. 2. A volume of nitrogen gas measured 756 c. c. when the barometer stood at 77.4 centim. What would it have measured if the barometer had stood at 76 centim. ? 3. A volume of air standing in a bell-glass over a mer- cury pneumatic trough measured 568 c. c. The barome- ter at the time stood at 75.4 centim., and the surface of the mercury in the bell was found, by measurement, to be 6.5 centim. above the surface of the mercury in the trough. What would have been the volume had the air been ex- posed to the pressure of 76 centim. ? Solution. — It can easily be seen, that the pressure of the air on the surface of the mercury in the pneumatic trough, measured by the height of the barometer at the time (75.4 centim.), was balanced first by the column of mercury in the bell, and secondly by the tension * of the confined air. Hence, the pressure * to which the air was exposed was * The tension of a gas is the force with which it tends to expand, and, when the gas is at rest, must evidently be exactly equal to the pressure to which it is exposed. NON-METALLIC ELEMENTS, OR METALLOIDS. 53 equal to the height of the barometer less the height of the mercury in the bell, or 75.4 — 6.5 = 68.9 centim. We have then the proportion 68.9 : 76 = 568 : V = Ans. 4. A volume of air standing in a tall bell-glass over a mercury pneumatic trough measured 78 c. c. The barom- eter at the time stood at 74.6 centim., and the mercury in the bell at 57.4 centim. above the mercury in the trough. "What would have been the volume had the pressure been 76 centim. ? 5. What would be the answers to the last two problems, had the pneumatic trough been filled with water instead of mercury ? 97. Problems on Expansion of Gases hy Heat. 1. What will be the volume of 250 cubic centimetres of air at 0° when heated to 300° ? Solution. — It has been found by Eegnault and others, that the per- manent gases expand so nearly equally for the same increase of tempera- ture, that the differences may be entirely disregarded except in the most refined investigations; and it has also been found, that their rate of expansion does not materially vary from the lowest to the highest tem- peratures at which experiments have been made. The coefficient of expansion for air, as determined by Eegnault, is equal to 0.00366, and we can therefore calculate the volume, V, of a gas at any temperature, from its volume, V, at 0°, by means of the equation, already explained, V' = V(l + iX 0.00366); (1.) or if we know the volume Y' for a tempei-atiire t, we can calculate the volume V" for another temperature i\ by means of the equation V" = V' (! + («' — <) 0.00366). (2.) By transposing, we can obtain from equation (1) l + tX 0.00366' ^ ' As the volume of a gas varies very considerably with the temperature, it is important, in comparing together different measurements, that we should adopt a standard temperature, as we have adopted a standard pressure. The temperature which has been agreed upon is 0° ; but as it would be inconvenient, and often impossible, to make our measurements at this temperature, it becomes necessary to calculate, by means of equa- 5* 54 NON-METALLIC ELEMENTS, OR METALLOIDS. tion (3), from a volume Y' measured at t°, what would be the volume at 0°. This is called technically reducing the volume to 0°. There can be obtained also from equations (1) and (2) the equations _ V'' — V , _ V" — V ' ~ V X 0.00366 ' ^^-^ ^^^ * ~ * "^ V X 0.00366 ' ^^'^ by means of which we can calculate the change of temperature when we know the change of volume. Kepresenting the coefficient of expan- sion in the above formulte by Jc, we can obtain, by transposing and re- ducing the equation, V'-V , , , V" — V '' = -lir' (60 and ^ = v' (£/-«)• CO From these we can calculate the coefficient of expansion when we know the volume of a gas at two diflfereat temperatures. 2. A volume of gas measured 560 c. c. at 15°. What would it measure at 95° ? 3. A glass globe holding 450 c. c. of air at 0° was heated to 300°. At this temperature the neck was her- metically sealed, and the globe cooled again to 0°. The neck was then opened under mercury, and the air remain- ing in the globe passed up into a graduated jar, and meas- ured. How much was it found to measure ? Solution. — By substituting the values for V and t in the equation V = V (1 + < X 0.00003), we obtain the increased capacity of the globe, and of course the number of cubic centimetres of expanded air which it contains at 300^. It is then only necessary to substitute this value for V and 300 for t, in equation (3), in order to find what will be the volume of this expanded air when cooled again to 0°. 4. "What is the weight of air contained in an open glass globe of 250 c. c. capacity, at the temperature of 20°, and when the barometer stands at 74 centimetres ? Solution. — In order to make the solution general, we will represent the capacity of the globe, the temperature, and the height of the barometer, by V, t, and H, respectively. One cubic centimetre of air at 0°, and when the barometer stands at 76 centimetres, weighs 0.00129 grammes. To find what one cubic centimetre would weigh when the barometer stands at H centimetres, we make use of proportion (3), page 51: H : H' = W : W; or 76 : H = 0.00129 : W; whence W = 0.00129 . ^, the weight of one cubic centimetre at 0^, and under a pressure of H centimetres. To find what one cubic centimetre NON-METALLIC ELEMENTS, OR METALLOIDS. 55 would weigh at P, it must be remembered that one cubic centimetre at 0= becomes (1 ■+• t 0.00366) c. c. at t^; therefore, at P and at H centimetres of the barometer, (1 + i 0.00366) c. c. weigh 0.00129 . ^ grammes. By equating these two terms we obtain (1 + 1 0.00366) = 0.00129 . ^, whence 1 = 0.00129 • ]-^ x7o"oo366 ' 76' *^® weight of one cubic centimetre at P and under a pressure of H centimetres. The weight of V cubic centimetres, w, is evidently zi) = 0.00129 V . H 1 + ( 0.00366 76 * (8.) 5. Wliat is the weight of air contained in an open glass globe of 560 cubic centimetres' capacity at 0°, at the temperature of 300°, and under a pressure of 77 centi- metres ? Solution. — In the soli;tion of the last example we neglected the change of capacity of the glass globe due to the change of temperature. This causes no sensible error when the change of temperature is small, but when, as in the present problem, the change of temperature is quite large, the change of capacity of the globe must be considered. If the capacity is V c. c. at 0^, it becomes atPY{l + t 0.00003). Introducing this value for V into the equations of the last section, we obtain w = 0.00129 V (1 + < 0.00003) 1 + t 0.0U366 76 ■ (9.) 99. Problems on Specific Gravity of Vapors. General Solution. — The specific gravity of a vapor is its weight com- pared with the weight of the same volume of air under the same con- ditions of temperature and pres- sure. To find, then, the specific gravity of a vapor, we must as- certain the weight of a known volume, V, at a known tempera- ture, t, and under a known pres- sure, H, and divide this by the weight of the same volume of air at the same temperature, and under the same pressure. The method may best be explained by an example. Suppose, then, that we wish to ascertain the specific gravity of alcohol vapor. We take a light glass globe having a ca- pacity of from 400 to 500 c. c, and Fig. 6. 56 NON-METALLIC ELEMENTS, OR METALLOIDS. draw the neck out in the flame of a blast lamp, so as to leave only a fine opening, as shown in the figure at a. The first step is now to ascertain the weight of the glass globe when completely exhausted of air. As this cannot readily be done directly, we weigh the globe full of air, and then subtract the weight of the air, ascertained by calculation from the capacity of the globe, and from the temperature and pressure of the air, by means of equation (8). Call the weight of the globe and air W, and the weight of the air w, then W — w is the weight of the globe exhausted of air. The second step is to ascertain the weight of the globe filled with alcohol vapor at a known temperature, and under a known pressure. For this purpose we introduce into the globe a few grammes of pure alcohol, and mount it on the support represented in Fig. 6. By loosening the screw, r, we next sink the balloon beneath the oil contained in the iron vessel, V, and secure it in this position. We noAV slowly raise the temperature of the oil to between 300° and 400°, which we observe by means of the thermometer, B. The alcohol changes to vapor and drives out the air, which, with the excess of vapor, escapes at a. When the bath has attained the requisite temperatui'e, we close the opening a, by suddenly melting the end of the tube at a with a mouth blow-pipe, and as nearly as possible at the same moment observe the temperature of the bath and the height of the barometer. We have now the globe filled with alcohol vapor at a known temperature, and under a known pressure. Since it is hermetically sealed, its weight cannot change, and we can therefore allow it to cool, clean it, and weigh it at our leisure. This will give us the weight of the globe filled with alcohol vapor at a known temperature, t', and under a known pressure, H'. Call this weight W'. The weight of the vapor is W' — W + w. The third step is to ascertain the weight of the same volume of air at the same temperature and under the same pressure. This can easily be found by calculation from equation (9). The last step is to find the capacity of the globe, which, although we have supposed it known, is not actually ascertained experimentally until the end of the process. For this purpose we break off the tip of the tube (a), under mercury, which, if the experiment has been carefully conducted, rushes in and fills the globe completely. We then empty this mercury into a carefully graduated glass cylinder, and read oS" the volume. We find then the specific gravity by dividing the weight of the vapor by the weight of the air. The formula for the calculation are then Weight of the globe and air, W. " " air, ,„ = 0.00129 V.pp-^^. 5. " " globe exhausted of air, W — w. " " " filled with vapor '\ at a temperatm-e t' and under a > W. pressure H', ) NON-METALLIC ELEMENTS, OR METALLOIDS. 57 Weight of the vapor, W — W + to. " " air a.t t' ) 1 H ail HI I ( =,0.00129 V(l + «' 0.00003)- i + ,. o.uosss ' 76 ' jsure H', ) ^ W — W + w Sp. Gr. = . and under a pressure 0.00129 V (l + C 0.00003). 1 + t 0.00366 75 1. Ascertain the Sp. Gr. of alcohol vapor from the fol- lowing data : — "Weight of glass globe, W 50.8039 grammes. Height of barometer, H 74.754 centim. Temperature, t 18° Weight of globe and vapor, W 50.8245 grammes. Height of barometer, H' 74.764 centim. Temperature, i' 167° Volume, V 351.5 cubic centim. Ans. 1.5795. 2. Ascertain the Sp. Gr. of camphor vapor from the following data : — Weight of glass globe, W 50.1342 grammes. Height of barometer, H 74.2 centim. Temperature, t 13°.5 Weight of globe and vapor, W 50.8422 grammes. Height of barometer, H' 74.2 centim. Temperature, t 244° Volume, V 295 cubic centim. Ans. 5.298. Carbon (C). 109. C + 20 = C02. 2 Hg 4- C = 2 ^r/ -}- C ©2- 1. HoAV many grammes and how many cubic centi- metres of carbonic acid gas are formed by burning 10 grammes of charcoal ? 58 ■ NON-METALLIC ELEMENTS, OR METALLOIDS. 2. Ho"W many grammes and hoTV many cubic centi- metres of oxygen are consumed in the process ? 3. Assuming that the volume of carbonic acid gas gen- erated during combustion is exactly equal to the volume of oxygen gas consumed, what is the Sp. Gr. of carbonic acid gas ? 4. How much oxide of mercury is required to bum up 5.672 grammes of charcoal ? 110. C + = €0. C + C02=2CO. 1. How many grammes and how many cubic centime- tres of oxide of carbon gas are formed by burning 10 grammes of charcoal ? 2. How many grammes and how many cubic centime- tres of oxygen are consumed in the process ? 8. Ten cubic centimetres of oxygen yield how many cubic centimetres of carbonic acid gas, and how many of oxide of carbon gas ? What expansion does oxygen un- dergo in combining with carbon to form oxide of carbon ? Spermaceti. 115. C64 H64 O4 + 188 O = 64 C O2 + 64 HO. 1. How many grammes of carbonic acid, and how many grammes of water, are formed by burning 10 grammes of spermaceti ? 2. The carbonic acid and water given off by a burning spermaceti candle were carefully collected and weighed. The water weighed 0.564 grammes, the carbonic acid weighed 1.3786. How much of the candle was burned? NON-METALLIC ELEMENTS, OR METALLOIDS. 59 SECOND GROUP OF METALLOIDS. Sulphur (S). 132. Fe S + HO, 80^-\-Aq^Fe0,S0^^Aq•\- H S. 1. How much sulphide of hydrogen can be made from 15 grammes of sulphide of iron ? How many cubic centi- metres ? 2. How much sulphide of iron, and how much sulphuric acid, is required to generate sufficient gas to saturate one litre of water ? HS-^Aq-\-i^=^-\-Aq. HS + 30= eO + SOij. Black. 133. a. Pb + iT^-f ^^ = Pb S + ^^ + H. Yellow. Black. l.YhO ^ H S -\- Aq = Y'o^ -\- Aq. Black. c.PlO, {_G^H;\ O3 4- H8 -\- Aq = PbS -f HO,lG^H,-\0,JrM- Black. d. Fe 0,S0^-\- GaO,HO-\- HS -{-Aq^YQ^ + Ga 0,S0^-\- Aq. Black. Gu 0, [O; H,-] 0, -{. ITS + Aq = Cu S + irO,[G,ff,-] 0, + Aq. Orang'e# Sb Gl3-}-3JIS-\-Aq = ^hS3-{-3irGl-{- Aq. Yellow. As Gls -i-SIIS-^ Aq = As S3 -\- dHGl-\-Aq. White. ZnGl-{-GaSA Aq =ZnS-\- Ga Gl-\- Aq. 60 NON-METALLIC ELEMENTS, OR METALLOIDS. Phosphorus (P). 140. P + 3 O = P O3 (by slow combustion). P-f-5O==P05(by rapid combustion). 144. Burnt bones consist chiefly of 3 Ca O, cP O5. 3 Ca O, cP O5 + 2 {HO, S0,)-\-aq = 2 (CaO, S O3) + [2 HO, Ca 01 eP O5 + aq. Heated to a red heat. \^H0, Ca 0], ^PO, ^ Aq -\-xC = CaO, PO5 + a;C + Aq + 2CO + 2H. Heated intensely. 3 (CaO, ,P Os) + a; C = 3 CaO, aPOs + a: C + 2 P + 10 C O. ^ 1. How much phosphorus can be manufactured from 20 kilogrammes of burnt bones, of which four fifths are phosphate of lime ? 145. 4P + 3(CaO, HO) = 3(CaO, P0) + PH3. Besides the above reaction, there take place simulta- neously the two following reactions, in the experiment described in the text-book. 3 P 4- 2 (Ca 0, H 0) = 2 (Ca 0, P 0) + P H^. P -f Ca O, H = Ca 0, P + M. THIRD GROUP OF METALLOIDS. Chlorine (CI). 150. Mn 0.,-^2HGl-\-aq = Mn CI -{- aq -{- CI. 1. HoAv much chlorine gas can be obtained from 2.467 NON-METALLIC ELEMENTS, OR METALLOIDS. 61 grammes of clilorohydric acid gas ? How many cubic centimetres ? 2. How much chlorine can be obtained from an unde- termined amount of muriatic acid by means of 4.567 grammes of hyperoxide of manganese ? How many cubic centimetres ? 3. The hyperoxide of manganese of commerce is more or less adulterated. What per cent of Mn O2 does an article contain, of which 10 grammes, when heated with strong muriatic acid, evolve 4.0135 grammes of chlorine ? 4. How much chlorine can be obtained from 25 cubic centimetres of muriatic acid* of Sp. Gr. = 1.16 ? How many cubic centimetres ? 5. In order to prepare one litre of chlorine gas how much hyperoxide of manganese, and how much muriatic acid, must be used ? Calculate the amounts for pure Mn O2 and H CI gas, and also when the oxide used contains only 70 per cent of pure MnOa, and when the liquid acid used has a Sp. Gr. = 1.15. 151. Mn O2 + 2 Na CI + 2 (^ 0, aS' O3) -{- aq = Mn a + 2 (JVa 0, S Os) -{- ag -{- Ch "We might use one half as much common salt, but then we should find sulphate of manganese instead of chloride of manganese in solution. Thus, Mn O2 + Na CI + 2 {HO, S 0^) -{-aq=-.Mn 0, SO, Jf-J^aO,SOs-\-aq^ CI. 1. How much chlorine gas can be obtained by the last process from 34 kilogrammes of salt ? * See Table VI., which gives the per cent of H CI in the fluid acid of dif- ferent specific gravities. 6 62 NON-METALLIC ELEMENTS, OR METALLOIDS. 2. How many cubic centimetres of chlorine can be ob- tained from one cubic centimetre of rock salt ? Sp. Gr. of salt = 2.15. 152. /. 2 {Fe 0, S 0^) -{- H 0, S 0^ -\- Gl -\- Aq = Fe^ 0^, 3 S Os + II CI -\- Aq. 6 (Fe 0,8 0^) -\- 3 CI -\-Aq = 2 {Fe.^ O3, 3 S O3) _|_ Fe^ Ck + Aq. g. An -^ 3 CI -\- Aq =^ An CI, + Aq. ACIDS FIRST GROUP : OXYGEN ACIDS. Nitrogen and Oxygen. Nitric Acid {HO, NO,). 159. KO, N Os + 2 {HO, 8 0^) = KO, SO,. HO, S O3 NaO, NO5 + 2 {HO, SOs) = NaO, SO,. HO, S 0, + H0,J¥05. 1. How much nitric acid can be made from 250 kilo- grammes of potash nitre, and how much sulphuric acid must be used in the process ? 2. How much more nitric acid will the same weight of soda nitre yield ? 3. How much nitric acid, containing 40 per cent, of NO5, can be made from 1700 kilogrammes of potash nitre ? 4. How much soda nitre, and how much sulphuric acid, and how much water, must be used to make 450 kilo- grammes of nitric acid, which shall contain 60 per cent. of pure acid ? 64 ACIDS. Ammonia, 160. c. INH,-] 0, HO + HO, NO, -^Aq^ iNH,-\ 0, N 0, + Aq. d.VhO-{-HO,NO,-\-Aq = PhO,NOs-\-Aq. 1. How much nitric acid, of Sp. Gr. 1.14, is required to dissolve 20 kilogrammes of oxide of lead ? 2. How much nitric acid, of Sp. Gr. 1.14, and how much oxide of lead, must be used to make 10 kilogrammes of nitrate of lead ? c. 3 Pb + 4 {HO, N O5) -\- aq =^ Z (Pb 0, NO5) + aq + 1V02. 3 Cu + 4 {HO, NO,)^aq= 3 {Cu 0, NO,) + aq 1. How much nitric acid, of Sp. Gr. 1.22, is required to dissolve 450 grammes of lead ? How much nitrate of lead would be formed ? 2. How much nitric acid, of Sp. Gr. 1.362, is required to dissolve 450 grammes of copper ? /. 3 P + 5 (^ (9, NO,) -{-Aq=3{3HO,,P 0,) + Aq + 5 ]^ O2. S + HO, NO, -\-Aq = HO,SOs + Aq + JVO^. Nitric Oxide (X O). 6FeCl-\-KO,N05-{- 4.H01 -{- Aq = S Fe^ CI3 + KCl + aq-\-NO!i. Colorless. Red. 1. How much IV O2 can be obtained by dissolving 10 grammes of copper in nitric acid ? How many cubic centimetres ? ACIDS. 65 2. How much IV O2 can be obtained from 10 grammes of iron by the last reaction but one ? 3. "What volume of oxygen must be mixed with one litre of TV O2 in order to change it into ]V O4 ? Nitrous Oxide (IV O). When heated. 163. [NHi]0, N05=4HO + 2]VO. Sp.Gr. of 1.14. 4.Vh^b{H0, N0,)+Aq = 4: (Pb 0, NOs) + Aq + JVO. Mi 0, S 0^ -\- Aq -{- -X 0^ = Na 0, S 0^ -{- Aq -\- IV©. 1. Ten grammes of nitrate of ammonia yield how many grammes and how many cubic centimetres of protoxide of nitrogen ? 2. How much nitrate of ammonia must be used in order to make one litre of the gas ? 3. One litre of TV O2 yields how many cubic centime- tres of IV O by the third reaction of this section ? 4. One litre of IVO gives, when decomposed, what volume of nitrogen ? Carbon and Oxygen. Carbonic Acid (C 02)' 164. CaO, CO2 4- HO, NO, -\- Aq = Ca 0, NO^ + Aq + CO2. Ca 0, C O2 + ^0, S 0^-\- Aq= C^O, S O3 + Aq + C O2. 6* •66 ACIDS. Ca 0, NOs + HO, SO3 + aq = C?.0, ^0^-{- HO,NOs-\-aq. 1. How much sulphuric acid and how much nitric acid* must be used to drive out all the carbonic acid from 25.462 grammes of chalk ? How many grammes and how many cubic centimetres of gas would be obtained ? 2. The specific gravity of Carrara marble is 2.716. How many cubic centimetres of carbonic acid gas does one cubic centimetre of the marble contain in a condensed state ? 167. 1. Animals remove oxygen from the air, and return the whole as carbonic acid. Plants remove carbonic acid, and, having decomposed it, return the oxygen it contained. How does the volume of the oxygen in either case com- pare with that of the carbonic acid ? Sulphur and Oxygen. Sulphuric Acid (S O3). 169. S02 4-Ot= SO3. 1. How much anhydrous sulphuric acid is formed by the oxidation of 10 grammes of sulphurous acid? and how much oxygen is required in the process ? 2. How much anhydrous sulphuric acid is formed by the oxidation of one litre of sulphurous acid gas ? and how many cubic centimetres of oxygen must be mixed with it in the experiment ? a. 2 S O3 + H O = iT 0, 2 ^ ^3 (the Nordhausen Acid). * When the strength of the nitric acid is not stated, monohydrated acid (H 0, N O5) is always intended. t The two gases are mixed together and led over heated platinum sponge in a glass tube. ACIDS. 67 S O3 + H O = ^ 0, aS' 6>3 (the common Acid). 170. Fe O, S O3 . 6 H is the symbol of ciystallized green vitriol. When healed. 2(FeO, SO3 . 6HO) = Fe203, S O3 + 12IIO + S©2. By further heating", Fe2 03, S03=Fe2 03 + §©3. By conducting the anhydrous acid fumes into HO, S O3 we get IIO,2S03. 1. How much anhydrous acid, and how much Nord- hausen, can be made by the above process from 20 kilo- grammes of green vitriol ? If the Nordhausen acid has the specific gravity of 1.9, how many litres can be obtained from 20 kilogrammes of green vitriol ? 171. S02 + M.0,N0s + Afl = HO, SOs + Aq-^NOi. 1. How much HO, S O3 will be formed from one gramme of S O2 ? How much from one litre ? While. Ba Ol-{- HO, S03-\-Aq = BiiO,SO3-^HCl-\- Aq. BaO,NOs + HO, S 0^ -\- Aq = BaO, SO3 + HGl-\-Aq. 1. Why must sulphuric acid or a soluble sulphate pro- duce a precipitate when added, in solution, to the solution of any salt of baryta ? 2. The precipitate produced by adding an excess of Ba CI to a solution of HO, S O3 was collected, and weighed 4.567 grammes. How much sulphuric acid* was present in solution ? * When the name sulphuric acid is used, H 0, S O3 is always meant, unless otherwise specified. 69 ACIDS. 3. The precipitate produced by adding an excess of HO, S O3 to a solution of Ba O, N O5 weighed 5.942 grammes. How much Ba 0, N O5 did the solution con- tain? 172. 1st stage. S + O2={SO2,andK0,N05 + 2(ir0, ^Og) = KO,SO,. HO, SO, + MO, IVOs. 2d stage. SO2 + HO, ]¥05 = HO, S 0, + ]\04. 3d stage. 3]^04 + ccHO=2(HO,]\05) + J\02. r l¥02 + 02 = ]^®4• 4th stage. < 2 SO, + 2 (H O, ]¥ O5) = 2 (^0, ^ 0,) ( +2WO4. The last two stages are now repeated indefinitely, so long as there is a supply of sulphurous acid, oxygen, and steam, with the same amount of N O4 . 1. How much sulphuric acid may be made by the above process from 100 kilogrammes of sulphur ? How many litres of acid having the Sp. Gr. 1.842 ? How many of acid of Sp. Gr. 1.734? How many cubic metres of oxy- gen must be used in the process ? How many cubic metres of air must pass through the lead chamber, sup- posing all its oxygen to be removed ? 173. a. HO, SOs -\- Ati = H 0, S 0-\- Aq, 1. How much water must one kilogramme of the mono- hydrated acid withdraw from the air in order to reduce its Sp. Gr. to 1.398 ? /. Na 0, CO^ -\-H0,S03-{-Aq = Na0,S0, + Aq ^+C02. 1. How much HO, S O3 is required to exactly neu- tralize 5.645 grammes of anhydrous carbonate of soda? How much acid of the Sp. Gr. 1.306 ? ACIDS. 69 2. How much NaO, CO2 must be dissolved in one litre of water so as to make a solution such that one cubic centimetre will exactly neutralize 0.01 of a gramme of H 0, S O3 ? Yellow. White. ^. Pb O + HO, SO^-\-Aq = PbO, S O3 + Aq. Black. Blue Solution. h. QxxQ -^ H 0, S 0^^ Aq = 0u0,8 0^-\- Aq, which, when evaporated, gives crystals of the composition Blue Vitriol. CuO, SO3.5HO. 1. How much sulphate of lead can be made from twenty grammes of litharge ? How much sulphuric acid must be used in the process ? 2. How much crystallized Blue Vitriol can be made from one kilogramme of oxide of copper ? How much sulphui'ic acid of Sp. Gr. 1.615 must be measured out for the process ? 174. Cu + 2 {HO, SOs) = CuO, SO3 + 2 HO + SOa- 175. C-\-2(HO,SOs) = 2HO + 2S02 + C02. WaO, GO,-\-SO^-}-Aq = JVaO,SO^-\-Aq-{-€Oi. 1. How much sulphurous acid can be made from 4.562 grammes of sulphuric acid by means of copper ? How much by means of charcoal ? How much anhydrous sul- phate of copper would be formed in the first case ? How much carbonic acid in the second ? How much carbonate of soda will the sulphurous acid in the two examples neu- tralize ? What is the volume of S O2, and what the volume of C O2 evolved in the second case ? 2. How much copper and how much sulphuric acid must be used to make one litre of sulphurous acid gas ? 70 ACIDS. How much to make 500 grammes of anhydrous sulphite of soda? 3. By burning sulphur in one litre of oxygen, how much S O2 gas is obtained ? What is the Sp. Gr. of S O2 ? Phosphorus and Oxygen, Phosphoric Acid (POs). 176. P + 5 O = P O5 (white powder). Colorless Fluid. 3P + 6 {HO, NO,) + Aq = ZHO, ,PQ,-\- ^^ + 5]\02. For preparation of phosphoric acid from bones, see § 144. At the ordinary temperature. 3CaO, ePOs + 2(110, SOs)-\-aq= 2(CaO, S O3) + [2^0, Ca 0] ,P O5 + aq. Intensely heated. 2(CaO, S03)+ [2H0, CaOjePOs = SCaOePOs 4-2(HO, SO3). Before ijcnition. 3 HO, P O5 + 3 ilNH,-] 0,II0)-\-3 (Ag 0,NOs) + Aq Yellow. = 3AgO, ,P05 + 3 ([i^^4] 0, NO,) + Aq. After ignition. HO,PO,-^ INH,-] 0, HO + AgO, NO,-\-Aq = White. Ag 0, ,P O5 + [^^J 0,N0, + Aq. 3H0, ,P0, + 2{MgO, SO,) + 3 [iV^^^] 0, HO -\-Aq == [N H4] 0, 2 Mg 0, eP O5 . 12 H O + 2(lNH,^0,S0s)+Aq. ACIDS. 71 [N H4] 0, 2 Mg 0, eP Og . 12 H when ignited resolves into 2 Mg O, bP O5 + W H3 + 13 H O. 1. How much P Og and how much 3 H 0, cP Og can be obtained from 16 grammes of phosphorus ? 2. By boiling one gramme of phosphorus in nitric acid until it dissolves, diluting, neutralizing with aqua ammonia, and precipitating with a solution of sulphate of magnesia, collecting and igniting the precipitate, how much wiU it be found to weigh ? 3. How much nitrate of silver is required to precipitate the phosphoric acid made from one gramme of phos- phorus before ignition ? How much after ignition ? Cyanogen and Oxygen. 179. Cy O = Cyanic Acid, which is mo7iobasic. Cy2 O2 = Fulminic Acid, which is hihasic. Cys O3 = Cyanuric Acid, which is trihasic. Boron and Oxygen. Boracic Acid (B O3). 180. NaO, 2BO3 + HCl + Aq =. 2 {H 0, B 0^ + iVa CT + Aq. The crystallized boracic acid is H O, B O3 . 2 H O. Dried at 100° it becomes H O, 2 B O3 . 2 H O. At a red heat it loses its water and melts, and on cooling it hardens to a vitreous mass. 72 ACIDS. Silicon and Oxygen. Silicic Acid (Si O3). Na 0, Si O3 -{-HCl +Aq = HO, Si 0^ -\- Na 01 -\-Aq. If the quantity of water is large, the hydrated silicic acid remains in solution. If the amount of water is small, it separates as gelatinous precipitate. SECOND GROUP : HYDROGEN ACIDS, OR COMPOUNDS OF THE HALOGENS WITH HYDROGEN. Chlorine and Hydrogen. Chlorohydric Acid (H CI). 185. Na CI + HO, S Os = m 0, SO^-^ EL CI; or NaCl + 2 (HO, SO,) = Na 0, S 0, . HO, SO, + HC1. Only one equivalent 0^ HO, S 0, is necessary to de- compose one equivalent of salt ; but then the last half of the H CI can be driven off only at a temperature suffi- ciently high to melt glass, so that with these proportions the process cannot be conducted in glass vessels. If two equivalents of HO, S 0, are used, the whole of the H CI is expeUed at a moderate temperature, and the process can then be conducted to its end in a glass flask or retort. Hence, in the manufactories, where the acid is generally generated in iron retorts, only one equivalent of HO, S O3 is used, while in the laboratory, where glass vessels are employed in the process, two equivalents are taken to ACIDS. 73 each equivalent of salt. The last we wiU assume to be the case in the following problems. 1. How much chlorohydric acid gas can be made from 4.562 grammes; from 25 kilogrammes; from 34.567 grammes of common salt ? 2. How much sulphuric acid is required to decompose the above amounts of common salt ? and how much bisul- phate of soda is in each case formed ? 3. How many cubic centimetres of H CI can be made from 1 gramme ; from 5,643 grammes of Na CI ? 4. How much salt and how much sulphuric acid are required in order to make one kilogramme, to make 5.463 grammes, and to make one litre, of ffl CI ? 5. How much H CI is contained in one litre of the liquid acid of Sp. Gr. 1.16, of Sp. Gr. 1.17, of Sp. Gr. 1.14? 6. How many cubic centimetres of gas are dissolved in one litre of the liquid acid of the above strengths ? 7. How much Na CI, and how much HO, SO^, and how much water in the receiver, are required to make, — a. 20 kilogrammes of liquid acid of Sp. Gr. 1.13 ? h. 560.4 grammes of liquid acid of Sp. Gr. 1.18 ? c. 4 litres of liquid acid of Sp. Gr. 1.16 ? H + CI = M CI. 1. One litre of hydrogen gas combines with what vol- ume of chlorine gas ? and what is the volume of hydro- chloric acid gas formed ? 186. a.YQ^HCl-\-Aq = Fe 01 -{-Aq^ H. 1. How much liquid acid, by weight and by measure, of Sp. Gr. 1,16 is required to dissolve 250 grammes of 7 74 ACIDS. iron ? How much chloride of iron would be obtained, and how much hydrogen gas, by measure, evolved ? ^n -^ H CI -\- Aq = Sn Gl ^ Aq-^ H. 1. Solve the last problem, substituting tin for iron. h. FegOs, 3 HO H- 3 ^C; + ^^ = Fe^ Ck + Aq. c. 2Fem-\- Cl^Aq= Fe^ Ck + Aq. d.NaO,G02-\-HCl-\-Aq = NaCl-\-Aq-\-€Oi. e. AgO,NO^-^ HCl-\-Aq = AgC\-\- HO, NOs-^Aq. 1. How much liquid acid of Sp. Gr. 1.16 is required to dissolve 4 grammes of iron-rust? 2. How many cubic centimetres of chlorine are required to convert one gramme of protochloride of iron into ses- quichloride ? 3. How much liquid acid of Sp. Gr. 1.13 is required to neutralize one gramme of carbonate of soda ? 4. How much H CI do 50 c. c. of a liquid contain which is exactly neutralized by one gramme of anhydrous carbonate of soda ? 5. How much H CI do 50 c. c. of a liquid contain which gives, with an excess of nitrate of silver, a precipi- tate weighing 5.643 grammes ? Aqua Eegia. H0,N0,-\-ZHCl-\-aq = N02Ck-^Cl^ aq. Au + {N 0^ Ck -\-0l-\- aq) = Au Ck + aq + N O^. ACIDS. 75 Bromohydric Acid (HBr). lodohydric Acid (HI). PBrs + SiTO = P03 + 3HBr. PI3 _^ 3 iTO = PO3 + 3 H I. Hydrofluoric Acid (H Fl). CaFl 4- RO,SO-\- aq = CaO, SO3 + HFl-\-aq. Si03 + 3HFl = 3HO + SiFl3. Tartaric Acid (2 H 0, Cs H4 0,o) • 194. 2 UN H,-] 0, HO) + ^ H 0, O, H, 0,, -^ Aq =^ ^INH,-] 0, GsJT,0,o + Aq. 2 (K 0, CO,) + 2 ^ 0, Gs IT, O.o -}- Aq = 2K0, Gsffi 0,0 + ^? + 2 C O2. 2K0, GsJI, 0,0 + HGl + ^^ = HO, KO, Cs H4O10* -^KGl-\-Aq. 2 (CaO, 00 + 2 (irO, KO, Gs H, 0,,) -\- Aq = 2 CaO, C4 HsO.o + 2K0,GsH, 0,, + ^^ + 2 CO2. 2/f 0, Gs H^ 6>,o -^2GaGl^Aq=2 Ca 0, Cg H4 0,o ^2KGl^Aq. 2 CaO, C4 HsOio + 2 {HO, S0-^^Aq==2 (Ca 0, S O3) ^2 HO, G,HsO,,^Aq. * This salt is not absolutelj^ insoluble, but only difficultly soluble in -water, and hence is not completelj'- deposited in this reaction. 76 ACIDS. 1. How much KO, CO2 is required to exactly neu- tralize 5.462 grammes of tartaric acid ? 2. How much H CI is required to convert 4.678 grammes of 2 K O, Cs H, 0,0 into H O, K O, C3 H4 Oio ? 3. From ten kilogrammes of cream of tartar how much tartaric acid can be made ? OxaKc Acid (H 0, C2O3). The above is the sypabol of the acid dried at 100°. When crystallized, it contains two more equivalents of water, and corresponds to the formula H 0, C2 O3 . 2 H O. 196. H O, Ca O3 . 2 ^0 + ^ {HO, SOs) = x {HO, SO,) 197. h.KO,CO.,-{- HO, G^ 0, + Aq = KO, G^ 0, + Aq + C O2. KO, 0, O3 + HO, C^O, + Aq = KO,^ a 6)3 + Aq. d. Ca 0, SO, ^ HO, Oi 0, -\- Aq = Ca O, C2O3 -\-HO,SO, + Aq. Ca 0,SOs^ iNH,-] 0, G, 0, ^ Aq = Ca 0, C2 O3 + {NH,) 0,S0,-^ Aq. 1. How much C O2 and how much C O will be ob- tained by decomposing five grammes of crystallized oxalic acid by sulphuric acid ? How many cubic centimetres of each gas ? 2. How much crystallized oxalic acid must be used to yield one litre of C O2 and one litre of C O ? 3. How much crystallized oxalic acid will exactly neu- tralize 1.456 grammes of carbonate of potassa ? ACIDS. 77 Acetic Acid {HO, [C4 ^3] O3). 198. PbO + HO, \_G^H,-] 0^ -\- Aq = Pb 0, [O.Hs] 0, ■^Aq. CrystaUized acetate of lead = Pb O, [Q H3] O3 . 3 H 0. PhO, (GiHs) Os + HO, SOs + Aq = PbO, SO3 -\-HO,lC,H,-] 0,-\-Aq. 7* LIGHT METALS. FIKST GROUP : ALKALI METALS. Potassium (K). 202. KO, aO^-\-HO,iG,H,-] 0,-]- Aq = KO,\_G,H,-\ 0, + Jl^ 4- C 02. KO, CO2+ HO, S03 + ^^=XC>, ^^(93 + ^g+COj. 1. How much HO, S O3 must be dissolved in water in order to make a litre of test acid such that one cubic centimetre will exactly neutralize one decigramme of KO, CO2? 2. How much crystallized oxalic acid must be dissolved in water in order to make a litre of test acid such that one hundred cubic centimetres will exactly neutralize 6.92 grammes of K O, C O2 ? 203. K 0, G O2 -\- Ga 0, H -\- Aq = Ca O, C O2 -^KO, HO -\- Aq. Melted together. 204. d.SiOs-\-KO,I£0= K 0, Si O3 + H ©. Light Blue. e. Gic 0, S O3 -\- K 0, If -\- Aq === Cu O, H + KO,SO, + Aq. LIGHT METALS, 79 Intensely heated. 205. KO, C O2 -(- 2 C = K + 3 C O. 206. KO, 00^ + 2 {HO, SO^) -{-aq = KO, SO3. HO, S O3 + a^ + C02. 207. KO, G0,-\- HO, N 0^ -\- aq = KO, N O5 -{- aq + CO2. When heated. a. KO, NO5 = KO, NO3 + O2. J. K 0, NOg + 3 C + S = K S + 3 € O2 + TV. 1. How many cubic centimetres of mixed gases are formed by the burning of one kilogramme of gunpowder, when measured at the standard temperature and pressure ? Assuming that the temperature at the time is 1000°, what would be the volume of the gases the moment after the explosion. 2. Assuming that gunpowder occupies the same bulk as an equal weight of water, into how many times its own volume does it expand on burning ? Calculate both for 0° and for 1000°. 3. Assuming that the temperature, the moment after explosion, is 1000°, what would be the pressure on the in- terior surface of a bomb of 20 centimetres internal diame- ter when exploded filled with gunpowder ? 208. a. K O, CI O5 = K CI + 6 O. 1. How much does the oxygen contained in chlorate of potassa expand when the salt is decomposed ? Sp, Gr. of chlorate of potassa is 2 nearly. 2. How much mechanical force would be required to reduce oxygen gas to the same degree of condensation in which it exists in the salt ? c. 3 (KO, CI 0,) + 4 {HO, ^ O3) = 2 (K 0, SO3 . HO, SO3) + K0, C10; + 2ClO4. 80 LIGHT METALS. f. -KO, C\Os+ ^ H Gl -^r Aq == KGl -{- Aq -\- 2 CI. 6 {KO, HO) -\- , ^ 0^) -\- Aq ^ 4 Si O3 + 9 [j%, Fe-] 0,S0,-{- Aq. Since sulphate of magnesia and sulphate of protoxide of iron have the same crystalline form, we shall obtain, on evaporating the solution, crystals containing both salts. We can prevent the sulphate of protoxide of iron from 88 LIGHT METALS. crystallizing, by converting it into sulphate of sesquioxide of iron by means of nitric acid. Thus, 6 {Fe 0, aS' O3) + 3 {HO, S 0,) + HO, NO, -^ Aq ^ 3 {Fe^ O3, 3 ^ O3) + ^^ + W ©2. Talc = 6MgO, 5 SiOs, 2 H 0. Meerschaum = Mg O, Si O3, H 0. Hornblende = 4 [Mg, Ca, Fe] 0, 3 Si O3. Augite = 3 [Ca, Fe, Mg] 0, 2 Si O3. 250. 5 {Mg 0, SO,) + 5 {KO, 00^) -{-Aq=d (MgO, CO2 . aq) + MgO, HO + % 0, 2 (76>2 + 5 {K 0, S 0^) -^ Aq. The relative proportions of carbonate of magnesia and of hydrate of magnesia vary with the temperature and other circumstances attending the precipitation. 251. M.g0,Q0^-\-HGl^Aq = Mga^Aq-\-i^O2. 2 {Mg 0, S0s)-\-H0,2 Na 0,P0,-\- {N H,-\ 0,H0 -{-Aq= [NHJ 0, 2MgO, PO5 + 2 {Na 0, S 0,) JrAq. When heated. (NH4)0,2MgO,P05=2MgO,P054-WM3 + HO. 1. What is the percentage composition of talc ? What that of hornblende and augite, assuming that the whole of the base in either case is Mg O ? 2. From a solution of sulphate of magnesia the whole of the magnesia was precipitated by phosphate of soda and ammonia. This precipitate, after ignition, was found to weigh 2.456 grammes. How much sulphate of mag- nesia was contained in the solution ? LIGHT METALS. 89 Aluminum (Al). 258. AI2O3, Si O3 . 2 H O + 3 {HO, S O3) -\- aq = SiOs + AI2O3, 3S03 + a(?. AU 0^,^SO^ + Z{N'aO, G 0^)^Aq = A\0^,ZB.O + 3 {Na 0, S03)+Aq-\-C ©2. AI2O3, 3 HO + KO, HO-\-Aq = KO, Ak O3 + Aq. KO,SO^. Ak O3, 3 aS C>3 + 3 {Fa 0, O 0,) -{- Aq = AI2 O3, 3 HO + 3 (iVa O^S O3) +KO,SOs-{- Aq + C®2. . Al, Os, 3S0, -\- S (Pb 0, 10, H,-\ O3) + Aq = 3(PbO, SO3) +^4 0„ 3 lO,H,-\ 0, + Aq. Symbols of Isomorphous Alums. Potassa, Alumina, Alum. K 0, S O3 . AI2 O3, 3 S O3 . 24 H O. Soda, Alumina, Alum. NaO, S O3 . AI2O3, 3 S O3 . 24 HO. Ammonia, Alumina, Alum. [N H4] O, SO3 . AI3 O3, 3 SO3 . 24HO. Potassa, Chrome, Alum. KO, S O3 . Cr2 O3, 3 S O3 . 24 HO. Soda, Chrome, Alum. Na 0, S O3 . Crg O3, 3 S O3 . 24 H O. Ammonia, Chrome, Alum. [N HJ 0, S O3 . Cr2 O3, 3 S O3 . 24HO. 90 LIGHT METALS. Potassa, Iron, Alum. K O, S O3 . Fea O3, 3 S O3 . 24 HO. Soda, Iron, Alum. NaO, S O3 . Fes O3, 3 S O3 . 24:HO. , Ammonia, Iron, Alum. [NH4] 0, S O3 . FeaOs, 3 S O3 . 24HO. Symbols of the most important Silicates of Alumina. 2AI2O3, SiOg, Staurotide. 3 AI2O3, 2 SiOs, Andalusite. 3 AI2 O3 , 2 Si O3 , Ky anite. 3 AI2 O3, 2 Si [O, Fl]3, Topaz, K O, Si O3 . AI2 O3, 3 Si O3, Common Felspar. Na O, Si O3 . AI2 O3, 3 Si O3, Albite. [Ca, Na] 0, Si O3 . AI2 O3, Si O3, Labradorite. K 0, Si O3 . 4 (AI2 O3, Si O3), Common Mica. 3 R O,* Si O3 . R. 03,t Si O3, Garnet. 1. What is the percentage composition of staurotide ? What is that of kyanite ? 2. An analysis of one of the above silicates would give the foUowing percentage composition. Silica, 64.76 Potassa, 16.87 Alumina, 18.37 100.00 What is the symbol of the mineral ? * K = Fe 0, Mn 0, Mg 0, or Ca 0. t Ks Os = AI2 O3, Fe2, O3, or Cra O3. LIGHT METALS. 91 Solution. — This problem is evidently the reverse of deducing the per- centage composition from the symbol ; but it does not admit, like that, of a definite solution, for while there is but one percentage composition corresponding to a given symbol, there may be an infinite number of symbols corresponding to a given percentage composition. This can easily be made clear by an example. The commonly received symbol of alcohol is [C4 H5] 0, H = C4 Ho O2 . The percentage composition is easily ascertained. Thus, C4 Hg O2 24 + 6 + 16 = 46. 46 : 24 = 100 : a; = 52.18 per cent of carbon. 46 : 6 = 100 : x = 13.04 per cent of hydrogen. 46 : 16 = 100 : x = 34.78 per cent of oxygen. Percent. Carbon, 52.18 = Ca = 12 or C4 = 24 or Cs = 36 Hydrogen, 13.04 = H3 = 3 " Hs = 6 " H9 = 9 Oxygen, 34.78 = =_8 " O2 =_16 " O3 = 2i 100.00 23 46 69 This percentage composition evidently corresponds not only to C4 Hs O2 , but also to C2 H3 0, to Cg H9 O3 , and to any other symbol -which is a multiple of the first; for, taking the per cent of carbon as an example, ■we have 100 : 52.18 = 23 : 12 = 46 : 24 = 69 : 36 = 92 : 48, &c. If, then, we had given the percentage composition of alcohol, it would be impossible to determine, without other data, whether the symbol was C2 Ha 0, or some multiple of it. If, however, we had also given that the sum of the equivalents of the elements of alcohol equalled 46, then we could easily reverse the above process. Thus, 100 : 52.18 = 46 : a; = 24, the sum of the equivalents of carbon. 100 : 13.04 = 46 : a; = 6, " " " " hydrogen. 100 : 34.78 = 46 : a; = 16, " " " « oxygen. — = 4, number of equivalents of carbon. hydrogen. — == 6, " " " 16 — = 2, " " " " oxygen. Assuming, however, that we had no means of ascertaining the sum of the equivalents in alcohol, then, although we could not definitely fix its sjmibol, yet nevertheless we could easily find which of all the possible symbols expressed its composition in the simplest terms ; in other words, 92 LIGHT METALS. ■with the fewest number of whole equivalents. For this purpose, assume for a moment that the sum of the equivalents is equal to 100, then 52.18 = the sum of the equivalents of carbon. 13.04= " " " " hydrogen. 34.78 = " " " " oxygen. — ^ = 8.697, number of equivalents of carbon. 13^ = 13.04, « " " " hydrogen. 84 78 — ^ = 4.348, " " " " oxygen. 8 These are the number of equivalents of each element on the supposition that the sum of the equivalents in alcohol is equal to 100. Any other possible number of equivalents must be either a multiple or a submul- tiple of these, and we can easily find the fewest number of whole equivalents possible, by seeking for the three smallest whole numbers ■which stand to each other in the relation of 8.697 : 13.08 : 4.348, ■which will be found to be 2:3:1. Hence the simplest possible symbol is C2 H3 0, but, from anything we are assumed to know, the symbol may be any multiple of this ; and for con- siderations which cannot be discussed in this connection, chemists usually assign to alcohol the symbol C4 He O2 , which is double the above. The symbol thus obtained expresses merely the relative number of equivalents of each element present in the compound, and gives no in- formation in regard to the grouping of the elements. Such symbols are called empirical symbols, to distinguish them from the rational symbols, which indicate the manner in which the elements are supposed to be arranged. The rational symbol of alcohol is [C4 H5] 0, H 0. This in- dicates not only that alcohol consists of four equivalents of carbon, six of hydrogen, and two of oxygen, but also that it is the hydrated oxide of a compound radical called ethyle. It must be carefully noticed, how- ever, that the empirical symbols fully represent all our positive knowl- edge. They alone are not hable to be changed. The gi-ouping of elements in a compound is a matter of theory, and the rational symbols are liable to constant changes, as the opinions of chemists on this sub- ject vary. From the example just discussed we can easily deduce the following rule for finding the empirical symbol of a compound from its percentage composition. Divide the per cent of each element entering into the com- poundhy its chemical equivalent^ and find the simplest series ofichole num- bers to ivMch these results correspond. To apply this rule to the problem under consideration. LIGHT METALS. 93 — 1-— = 1.43, number of equivalents of silica. 45.3 —1— = 0.3575, " " " " potassa. 47.2 i-— - = 0.3575, " " " " alumina. 51.4 143 : 0.3575 : 0.3575 = 4:1:1. Empirical symbol, AI2 O3 , K 0, 4 Si O3 . Rational symbol, K 0, Si O3 , AI2 O3 , 8 Si Oa . 3. An analysis of one of the silicates of alumina would give the following percentage composition. Per cent. Silica, 53.29 Lime, 16.47 Alumina, 30.24 100.00 What is the symbol of the minei'al ? Ans. Ca 0, Si O3 . A\, O3 Si O3. Second Method of Solution. — By inspecting the formula obtained by solving the problem according to the method just described, the student will see that the amount of oxygen in the acid stands in a very simple relation to that in the bases. This relation is 1 : 3 : 6, corresponding to Ca 0, AI2 O3 , and 2 Si O3 . It has been shown in the text-book, § 200, that a similar simple ratio exists between the amount of oxygen in the acid and that in the bases of all oxygen salts. The I'atio can easUy be found from the percentage composition. For this purpose we have merely to calculate the amount of oxygen in the per cent of the acid and bases indicated by analysis, and find the simplest ratio in which these amounts stand to each other. In our example, 53.29 per cent of silica contains 28.24 parts of oxygen. 16.47 " " lime " 4.71 " " " 30.24 " " alumina " 14.12 " " " According to the principle just stated, these numbers ought to stand to each other in some simple ratio, and it can easily be seen that 4.71 : 14.12 : 28.24 = 1:3:6. From this ratio we can easily deduce the symbol, for one equivalent of oxygen corresponds to one equivalent of Ca 0, three equivalents of oxy- gen correspond to one equivalent of AI2 O3 , and sis of oxygen to two 94 LIGHT METALS. equivalents of Si O3 . Hence the symbol is Ca 0, AI2 O3 , 2 Si O3 , which we may write as above, Ca 0, Si O3 . AI2 O3 , Si O3. For convenience in calculating the amount of oxygen from the per cent of acids or bases indicated by analysis, Table VII. has been added at the end of the book, which gives the per cent of oxygen, together with its logarithm con- tained in the bases and acids of most common occurrence. In deducing empirical symbols from the results of actual analysis, it must be remembered that our processes are not absolutely accurate, and that therefore we must not expect to find more than a close approxima- tion to a simple ratio between the oxj-gen in the base and that in the acid. Again, in mineral compounds, it is very frequently the case that isomorphous bases replace each other to a greater or less extent. This is the case in common garnet, the symbol of which may be written thus : 8 [Fe, Jin, Mg, Ca] 0, Si O3 . [Al2Fe2] O3, Si O3. We generally, however, write the symbol as on page 90 : 3 R 0, Si O3 . E2 O3 , Si O3 . Here E stands for the sum of all the protoxide bases, which make together but one equivalent of base, and E2 O3 for the sum of all the sesquioxide bases, which also make together but one equivalent of base. Such general symbols as these give all the information in regard to the constitution of the mineral which is required. In deducing such sym- bols, it is evident that the oxygen of all the protoxide bases must be added together to obtain the amount of oxygen in the assumed base E 0, and aU the oxygen of the sesquioxide bases must be added together in the same way in order to obtain the amount of oxygen in the assumed base E2 O3 . From these sums we can easily obtain the required oxygen ratio. 4. An analysis of andesine (a mineral allied to felspar) yielded the following result. Proportion of Oxyg'en. Silicic Acid, 59.60 30.90 in Si 0. Alumina, Sesquioxide of Iron, 24.28 1.58 ":S}="-™""^o Lime, 5.77 1.61- Magnesia, 1.08 0.37 - = 3.79 in RO. Soda, 6.53 1.65 Potassa, 1.08 0.16. 99.92 What is the symbol of the mineral ? LIGHT METALS. 95 Solution. — The ratio of the oxygen in E 0, R2 O3, and Si O3 is 3.79 : 11.70 : 30.90 = 1 : 3.08 : 8.1, for which we may substitute, for reasons stated above, 1:3:8. One equivalent of oxygen corresponds to R 0. • Three equivalents of oxygen correspond to Eg O3 . Eight equivalents of oxygen correspond to ^ Si O3 • But as we do not admit fractional equivalents, we may multiply the whole by three, when we obtain the empirical symbol 3E0, 3E2 03,8Si03; from which we may deduce the rational symbol 3 E 0, 2 Si O3 . 3 (E2 03, 2 Si O3). 5. Deduce the symbols of the silicious minerals of which the following are analyses. 1. Silicic Acid, 65.72 Sesquioxide of Iron, Alumina, 18.57 Lime, 0.34 Magnesia, 0.10 Potassa, 14.02 Soda, 1.25 100.00 100.0 100.04 100.16 2. 68.4 3, 44.12 63.70 0.1 0.70 0.50 20.8 35.12 23.95 0.2 19.02 2.05 0.66 0.65 0.25 1.20 10.5 0.27 8.11 HEAVY METALS. FIRST GROUP OF THE HEAVY METALS. Iron (Fe). 286. a. 4 Fe + O = Fe4 O. b. 3 Fe4 O + 13 O = 4 (Fe 0, Fe^ O3). c. 2 (Fe 0, Fe.2 O3) + O = 3 Fe^ O3. d. 2 (Fe O, S O3. 6 H O) when heated resolves into Fg O3, S O3 + S O2 + 12 H O ; by further heat- ing, Fe^ O3, S O3 = Fea O3 + S O3. e. 3 Fe + 4 O = Fe 0,'Fe2 O3. /. 2 (Fe 0, Fea O3) H- 9 ^ + © = 3 (Fe^ oTs H 0). g. Fe O, Fes O3 -\- 2 G 0^ + Aq = Fe^ O3 ^Fe 0,2 O 0,-\-Aq. Red. 2 (Fe 0, 2 O2) + ^^ + -O = Fe^ O3, 3 H ^4,0 0,-^ Aq. 285. Fe + IlO,S03-{-Aq = FeO,SO,-j-Aq-{-n^ HEAVY METALS. 97 285. Fe + Cii 0, ^ O3 + ^^ = Cu + Fe 0, S 0, Red. a. 6 (Fe 0, S 0,) -\- Aq + 3 O = ¥e, O3, 3 H Yellow-Brown. -{-2(Fe^0s, 3 S 0,)+Aq. h. 6 {Fe O, S 0,) + 3{H0, S 0,) -{- H 0, N 0^ Yellow-Brown. ■\-Aq=3 {Fe^ Os, 3 S O3) + .4^ + ]¥ O2 . Light Green. c. FeO,SOs-\-lIfJI,2 0,ffO-^Aq = FeO,B.O + [iV^^4] 0, SOs-\-Aq. Fe^ Os, 3SOs + 3 ([JSTIf,] 0, H 0) -\- Aq = Red. Fe2 O3, 3 H + 3 ([J^IIi'] 0, S 0^) + Aq. d. '2^YQ-}-A:HO,NO,-^rAq = Fe^Os,3NO,-\-Aq 288. \_H0, 2 Na 0] P Os + 3 {Fe 0, S 0^) -\- Aq = White. 3 Fe O, V 0,^2Na 0, SO^^HO, SOs ~\-Aq. [HO, 2NaO']POs-\- Fe^ Os, 3 S Os + Aq = While. Fe^Os ,P05.4HO+2 {M 0, SOs)-\-HO, SO^+Aq. 290. 3 Fe O + 2 Fea O3 + ^ H Cy -^ Aq = Blue. 3 Fe Cy, 2 Fca €73 + 9 HO + Aq. Blue. 291. 3 Fe Cy, 2 Fcg Cyg + % {K 0, HO) + Aq = , Red. 2 (Fe2 O3, 3 H 0) + 3 (2 ir (7y, Fe Cy) + Aq. 292. a. 3 (2 ^ (7y, Fe Cy) + 2 (^eg Os, 3 SOs)+Aq = 3 Fe Cy^^'Fea Cyg + 6 (.ff 0, >«? O3) + Aq,. 9 98 HEAVY METALS. 292. 5. 2 ^ Gy, Fe Cy ^ 2 {Fe 0, S 0,) -^ Aq = White. 3 Fe Cy 4- 2 Z" 0, 5 O3 + Aq. White. Blue. 9 Fe Cy + 3 O = Fe, O3 + 3 Fe Cy, 2 Fez Cyg. c. 2 K Cy, Fe Cy -\- 2 {Cu 0, S O3) -\- Aq = Purple. 2 Cu Cy, Fe Cy. Aq + 2 {KO, S 0,) + Aq. 2 Z- (7y, Fe Cy -i- 2 (Pb 0, ]^ 0,) -\- Aq = White. 2 Pb Cy, Fe Cy. Aq + 2 (JT 0, iV^Os) + Aq. A large number of similar compounds having the general symbol 2 R Cy, Fe Cy. Aq, may be formed thus : — White. Pale Yellow. 2 H Cy, Fe Cy. Aq. 2 [N H/] Cy, Fe Cy. Aq. Yellow. Yellow. 2 Na Cy, Fe Cy. Aq. 2 Ba Cy, Fe Cy. Aq. Pale Yellow. White. 2 Mg Cy, Fe Cy. Aq. 2 Zn Cy, Fe Cy. Aq. There are also compounds in which the two equiva- lents of R in the general symbol are replaced by different metals thus : — White. Yellow. Yellow Salt. Red Salt. 293. 2 (2 X Cy, Fe Cy)^ Cl-{- Aq =Z K Cy, Fe^ Cy, -^KCl-\- Aq. 3 K Cy, Fe^ Cy, + 3 {Fe 0, S 0,) -\- Aq = 3 Fe Cy,'"Fe2 Cys + 3 {K 0, S 0,) + Aq. HEAVY METALS. 99 293. There also may be formed a large nmnber of similar compounds having the general symbol 3 E. Cy, Fcg Cys, such as : — Brown. Lijht-Red. 8 H Cy, Fe, Cy,. 8 Ca Oy, Fe^ Cyj. {^B^Cyl-'^'^Cys. Aq. Light-Gtecn Solution. Black. 294 Fe 0, S 0, + \_N H,-\ ^ + ^^ = Fe S + IN-H,-] 0,SO,-\-Aq. Black. Lig-ht-Green. 295. Fe S + a^ + 4 O = J^e 0, ^ O3 + a^. Manganese (Mn). 299. Mn O2 + iT 0, S 0^ -\- aq = Mn 0, S03-\-aq + 0. Black. Lijht-Pink Solution. MnO^-^^ 2 H Gl -]- Aq = Mn 01 -^ Aq -\- CI. Light-Pink Solution. Brown. 300. a. 6 {Mn 0, /S O3) + .ig + 3 O = Muj O3, 3 H O + 2 {Mn^ Os, S S Os)+ Aq. Light-Pink Solution. 'White. b. MnO,S03-{-l]^iri']0,H:0-\-Aq = MnO,'H.O -\-[i^ir,-]o,so, + Aq. White. Brown. 2 (Mn 0, H 0) + «? + O = Muj O3, 3 H O -\-aq. Lig-ht-Pink Solution. Flesh-colored. Mn 0, S Os -\- [iV^ ff,-] S -\- Aq = Mn S -{-[JSTH,-] 0,SO,-i-Aq. Melted together. 301. MnO^-f K 0,11 0-^0= K0,Mn03 + UO. 100 HEAVY METALS. Green Solution, 301. 3 {K 0, Mn O3) + ^y + 2 C O2 = Mn O2 Crimson Solution. -^KO, Mn^ Ot-\-2{KO,G 0^) + Aq. Green Solution. 3 {K 0, Mn Os) -}- 2 (JIO, S O3) -{- Aq = Crimson Solution. Mn Oi+KO, Mn^ Ot^2{KO,S 0,)J^Aq. Cobalt (Co) + Nickel (Ni). Pink Solution. Blaclc. 303. Co 0, S Os -{- [N- H,-\ S -[- Aq = Qo ^ \_NH,-] 0,SO, + Aq. Green Solution. + [i\rir,] o,sOs + Aq. Green Solution. Black. M 0, S Os -^ [JSr H,-] S -{- Aq = m ^ Zinc (Zn). 311. Zn -{- H 0, S 0^ -{- Aq = Za 0, S Os -\- Aq + H. White. 312. a. Zn 0, S Os -\- KO, H 0-{- Aq =^ZnO, no ■\-KO,SOs-\- Aq. This precipitate dissolves in an excess of K 0, HO + Aq. White. b. 2k 0, S Os -i- [li H,-] S -\- Aq = Zn ^ ^iNH,-\0,SOs + Aq. HEAVY METALS. ^01, 312. c. 5 {Zn 0, S 0,) -^ 6 {Na 0, G 0,) -{- Aq = While. White, 2 (Zn 0, C Oa) + 3 (Zn O, H 0) + 5 (iVa 0, S 0^) -\- Aq -\- 3 € O,. This, precipitate is a mixture of Zn O, C O2, and Zn O, H O, but in variable proportions. Yellow. 315. Cd 0,S Os + JIS-\-Aq=CdS-\-irO,S03 Dark-Brown. Gray-White. White. 317. The oxides of tin are Sn O ; Sng O3 ; and Sn Oj. Cadmium (Cd). a + Aq. Tin (Sn). Dark-Bi tin are Sn 319. Sn -^ H Gl ^ aq = Sn 01 -\- aq -{- n. White. 320. Sn Gl + \_NH^ 0, H -{- Aq = SnO, HO + [iV^^4] Gl^Aq. 321. Sn Gl -\- Gl -\- Aq = Sn Gl, + Aq. Sn -\- 2 I£ Gl -i- IT 0, W Os -\- Aq = Sn CI, -\-Aq + NOs. White. SnGk-\-2 ilJ^Ifi'] 0,irO)^Aq = lIO, Sn 0^ + 2 [i^^,] Gl-\-Aq. SnO+KO,HO-^Aq = KO,Sn -\- Aq.. HO, Sn O2 + ^ 0, HO -\-Aq = KO, Sn Q, + Aq. 9* 102 HEAVY METALS. By evaporation of the last solution we can obtain crystals of K O, Sn Oj. 4 H O. We may also prepare Na O, Sn Og . 4 H O. White. S24. 5 Sn + 10 {H 0, N 0,) -\- Aq = Sng 0,o. 10 H O ^Aq -\-10NOi. Sng Oio. 10 H O + K 0, H -\- Aq = KOySn, 0,,^Aq. By evaporation we can obtain crystals of White. K 0, Sng Oio. 4 H O. Brown. 325. Sn CI -{- ITS -{- Aq =^ Sxi S -}- H Gl -{- Aq. Yellow. Sn Ck + 2 HS-\-Aq = Sn ^^ -\- ^ H Gl -\- Aq. SECOND GROUP OF THE HEAVY METALS. Lead (Pb). Black. Red or Yellow. Red-Yellow. 331. The oxides of lead are Pbj O ; Pb O ; Pba O3 ; Dark-BrowD, and Pb Oj. 334, B Vh -\- 4. {H O, N 0,) -\- aq = B (Pb 0, W 0^) + a^ + ]¥ O2. 3 Pb 0-1- 3 (^0, iV^ O5) + a^ = 3 (Pb 0, N 0,) -\- aq. HEAVY METALS. 103 White. 335. Ph 0, N Os + HO, S 0^-\-Aq = Pb O, S O3 + HO,NOs + Aq. Ak O3, 3 >Sf (93 + 3 {Pb 0, [C; ^3] O3) +Aq = White. 3 (Pb O, S O3) + Ak O3, 3 [Ci H,-] O3 + Aq. White. 336. Pb O + ^ (7? + a^ = Pb CI + aq. Vh O + H Gl -\- Aq = Pb CI + Aq. 337. Acetate of oxide of lead = Pb O, [C^ H3] O3 . 3 H O. Basic acetate of oxide of lead = 3 PbO, [QHsjOa. HO. 338. 2 (Pb 0, [ C4 ^3] 6>3) + 2 IT 0, 0^ B, 0,, -\-Aq^ White. 2 Pb O, Cs H, Oio + 2 (IT (9, [ Oi H,-\ 0,).+ Aq. White. PbO,N'0,-\-lNH,-\ 0,HO-\-Aq = VhO,B.O JrWH,-]0, NO,-^Aq. 339. Pb 0, [C, Bs] Os + Aq 4- 2 Pb O = 3Pb 0,lG,ff,-] 0, + Aq. 3 Pb 0, IC ffs] O3 + ^? + 2 C O2 = While. 2 (Pb O, C O2) + P& 0, [a ^3] O3 + Aq. 340. Zn -f P5 (9, [C; Hs] O3 -{- Aq == Pb + ^« 6>, [en,-] Os-i-Aq. Blaclc. 341. P^ 0, [O; 1^3] O3 + ^ .S' + ^^ = Pb s + ^0, [(7,^3] 0,-\-Aq. 342. Pb S + 3 O = Pb O + S O2, also Pb S White. + 4O = Pb0, SO3. 104 HEAYT METALS. 342. By roasting galena we obtain a mixture of Pb O with a small amount of Pb O, S O3. By then melting together Pb O or Pb 0, S O3 and an excess of Pb S, we obtain metallic lead and sul- phurous acid, thus : — 2 Pb O + Pb S = 3 Pb + S O2, also Pb 0, S O3 + PbS = 2Pb + 2S02. Bismuth (Bi). 847. 2 Bi + 4 (^ (9, II O5) + «^ = Bi^ O3, 3 J^ 0^ -\-aq-\-N O2. White. 4 (Bi, Oe, SI^Os) -\- Ag = 3 (Bi^ O3, N O5) + Bh O3, 9 i\r O5 + Aq. Brownish-Black. Bh O3, d If Os -\- 3 II S -{- Aq = Bk S3 -\- 9 (HO, If 0,) + Aq. Copper (Cu). Green. 349. Malachite = Ou 0, HO. Cu O, CO2. Blue Carbonate of Copper = Blue. CuO, HO. 2 CuO, CO2. Red. Black. 350. The oxides of copper are Cu2 and Cu O. Blue Solution. Blue. 352. Cu 0, S 0, + K 0, H -\- Aq = Cvi O, n O ■ -\-K0, SO,-\-Aq. HEAVY METALS. l05 Blue. Black. 352. By boiling, Cu 0, B. -{- Aq ^ Cn O -}- Aq. Light-Blue Solution. 353. GuO, ^03 + 2 ilNH,-] 0, H 0) -^ Aq =^ Very deep Blue Solution. Oit 0, S Os- 2 IT Bs. H -i- Aq = [^^l] 0, S 0,. [_NH,-\ 0,H0^Aq,2. solution which, when treated with alcohol, yields crystals having the composition Deep Blue. JN^^I 0, SO3. [NHJO, HO. These, when heated, are resolved into Green Powder. |n H^3 I 0, S O3 + []¥ H4] O, H O. 854. 2((7mO, >S03) + 2(^0, ir(9)+^g — 0* = Yellow-Red. Cu2 O + 2 {K 0, S O3) + Aq. Heated together. 855. 2 (Cu 0, S O3) + 2 (Na O, C O2) -f C = 2 Cu + 2 (Na O, S O3) + 3 C O2* 856. Zn + Gu 0, S O3 -\- Aq = Cu ^ Zn 0, S O3 + Aq. 357. Cu 0, H O + H = Cu + 2 H O. Black. Green Solution. 359. CnO-^HOl-\-Aq= Cu Cl + Aq. Blue Solution. 360. 3 Cu + 4 (^ (9, i\r Og) + «y = 3 (Cic 0, JSf 0,) + a^ + TV O2. * The oxygen is removed by adding grape sugar to the solution. 106 HEAVY METALS. Blue Salt. Black. White. 360. 2(CuO, NO5. 4H0) -f Sn=2CuO + Sn02 2 ((7m 0, S O3) -{- 2 {Fa 0, G 0,) -\- Aq = Blue or Green. Cu O, H O. Cu 0, C O2 + 2 {Na 0, S'O,) + ^? + co,. 3 (Cu 0, S 0,) -^IHO, 2 Na 0], POs + Aq== Greenish-Blue. 3 Cu O, P O5 + 2 {Na 0, S 0,) + H 0, S 0, JrAq. Emerald-Green. Dioptase = 3 Cu 0, 2 Si O3. 3 H 0. 361. Basic Acetate of Copper = Green. Cu 0, [C4 H3] O3. Cu O, H O. 5 H O. Neutral Acetate of Copper = Green. CuOCaHsiOg. 5 HO. Blue Solution. Black. 362. GuO,SO^-^rHS-\-Aq=Cvi^-\-HO,SO^ -\-Aq. Black. Green Solution. CxxS^HGl-{-ciq= Ou Gl-i-aq-^JS.^. Mercury (Hg). 367. Q Bff -{- A (ff 0, N Os) -i- aq = S (ffff^ 0, N Os) + «^ + ]¥ O2. Black. 368. Hff^ 0, NO, -{- K 0, H -\- Aq == Hg^ J^KO, N0, + Aq. HEAVY METALS. 107 368. Hahnemann's Suboxide of Mercury has not a constant composition. Some chemists, however, assign to it the symbol 2 Hga 0, N O5. N H3. White. 370. Hg^ 0, N 0^ -\- Na CI ^ Aq = Hg^ CI -\-Na 0, N 0,-[- Aq. 371. d Hg-^4.{H0,N 0,)-\-aq = d{Hg 0,N 0,) + aq-\-N O,. Yellow. ITg 0, NO, -{- KO, IT -{- Aq = Hg + K O, N O5 + Aq. When heated. Red. 372. Hg O, N O5 = Hg O + ]¥ O4 + O. 373. ffg 0, N Os -{- Na Gl -\- Aq, gives no precipitate, because Hg CI is soluble in water. White. B.g O -\- IT CI -\- aq = Ug CI -{- IT -{- aq. Heated together. Residue. Sublimate. Hg O, S O3 + Na CI = Na O, S O3 + Hg CI. Black. Hg2 CI + ^ 0, ITO -\- Aq== Hg^O-^-K CI + Aq. Yellow. Hg CI + .AT 0, i^ -i- ^^ = Hg O + ^ (7/ + Aq. White. 374. 2 Bg Gl-\- [iV^iT,] 0, HO-\-Aq = {n^^ X ci -\-Ha-\-Aq. Black Powder. 375. Hg Cl-^ Sn Gl-\-Aq = Hg + Sn Ck + Aq. Black. 376. HgCl + HS-^-Aq^-S-g^^ HCl-[-Aq. 108 HEAVY METALS. Silver (Ag). 380. 3 Ag + 4 {HO, NO,)-{-Aq = Z {Ag 0, NO,) + ^^ + W O2. Heated together. 381. a. Ag O, N O5 + 2 C = Ag + 2 C Oa + W O2. Brown, h. AgO,NO,-\-KO,irO+Aq = AgO,E.O -i-KO, NOs + Aq. White. Ago, nO-\-lNH,-\ 0, HO + Aq== {^Ag} + Aq. White. c. Ag 0, N 0, -^ Na Gl -\- Aq ^ Ag CI + iVa 6>, A^ O5 + Aq. Black e. AgO, NO,-\-HS-\-Aq^Ag^-\-HO, N 0, -{-Aq, Gold (Au). 385. Axi-\-ZHGl + HO, NO^ ■{- Aq=Au Ok + ^g + W O2. Dark-Brown, 387. Au Ok + 6 {Fe 0, S O3) + .4^ = Au -f Fe^ Ok + 2{Fe, O3, 3^03)+^^. 388. Aurate of Potassa (K O, Au O3 + 4 H 0) is a compound of oxide of potassium and teroxide of gold, in which the last plays the part of an acid. HEAVY METALS. 109 Platinum (Ft). 391, BVi -^ ^HGl -\- 2{H0, N 0^) -{■ Aq = 3 P< C4 + ^g + 2 W O2. Yellow. 392. lNH,-\ CI -\-PtCk^-Aq= [N HJ CI, Pt Cl^ + Aq. Yellow. 394, K 01 + Pt Ok -\-Aq=K CI, Pt CI2 + Aq. Black. The Oxides of Platinum are Pt 0* and Pt Oj. Greenish-Brown. Reddish-Brown. The Chlorides of Platinum are Pt CI and Pt CI2. Black. Black. The Sulphides of Platinum are Pt S and Pt Sj. Black. Pt 01 + H S -^ Aq = -2tS -{- H 01 ^ Aq. Black, Pt (74 + 2 ir5+ ^^ = Pt Sa + 2 iTCZ + Aq. THIRD GROUP OF THE HEAVY METALS. Chromium (Cr). 397. The symbol of chrome iron ore is Fe O, Crg O3 ; but almost invariably a portion of the Fe O is replaced by Mg 0, and a portion of the Crj O3 Heated togrether in contact with air. 2 (Fe 0, Cra O3) + 2 (K O, C 0^) + 7 O = by AI2 O3. Heated t (Fe 0, Cra Fe^ O3 + 2 (K O, 2 Cr O3). + 2 C O2 * Only known in combination with water. 10 110 HEAVY METALS. 397. The above process is hastened by mixing with the pulverized mineral a portion of nitre, which yields when heated a large supply of oxygen. Red. Yellow. 898. K0,2Cr 0^-^K0, O0^^Aq=2 {KO, Cr O3) + 4? + C O2. Yellow. Red. 2 (KO, Cr 0,)-\-HO,NOs + Aq=KO, 2 Cr O3 J^KO, N0,-\- Aq. 399. K 0, Cr O3 + Pb 0, IC, H,'] 0^ -\- Aq =. Yellow. Pb O, Cr O3 + ^ 0, [C; ^3] O.^Aq. Yellow. Red. - 2 (PbO, Cr03) -^KO, HO-\-Aq = 2 Pb 0, Cr O3 + K0, Cr 6>3 + Aq. Yellow. White. 400. 2 (Pb O, CvO,) -^ S H CI -\- Aq = 2 Pb CI Green. + Cr^ Ck + .4^ + 3 CI. Cr^ Ck + 3 {IN H,-\ 0, H 0) + Aq = Cra O3, 10 H O + 3 [Nil,-] CI + Aq. KO, 2 Cr Os-\- H 0, S 0,-\- ^ S C,-\- Aq = KO, S 0,. Cr^ 6>3, 3 ^ O3 + ^^. The symbol of crystallized chrome alum is KO, SO3. Cr^Os, 3SO3. 24 HO. 401. KO, 2 Cr Os -\- X* {JI 0, S0,)-\-aq=2 CrOg -\- KO, SO^. H 0, S Os + X {IT 0, S 0,) -\-aq. * X is here used to express an indefinite amount. HEAVY METALS. Ill Red. Green. 401. a. 2 Cr O3 — 3 O = Cv^ O3. The oxygen in the last reaction may be removed by alcohol or any other reducing agent. Antimony (Sb). The oxides of antimony are as follows : — White. 403. Oxide of Antimony, Sb O3. White. Antimonious Acid, Sb O4 = ^ (Sb O3, Sb O5). Pale-Yellow. Antimonic Acid, Sb O5. 404. 3Sb+4(jy(9,i\^C»5)=3Sb04+4HO+4]\O2. When antimony is treated with an excess of con- centrated nitric acid, only Sb O4 appears to be formed. If, however, the acid is dilute, the anti- monious acid is mixed with more or less of basic nitrate of antimony (2 Sb O3, N O5) according to the degree of dilution. By heating together one part of metallic antimony and four parts of nitre in a crucible, there is formed a white mass, which is a mixture of anti- moniate of potassa (K 0, Sb O5) ; nitrite of potassa (K O, N O3) ; and undecomposed nitre (K O, N O5). Warm water will dissolve the two last, but not the antimoniate of potassa. If, 112 HEAVY METALS. however, this anhydrous salt is boiled with water for one or two hours, it combines with five equiv- alents of water, forming a soluble compound (K O, Sb O5 . 5 H 0). The white mass, which seemed at first insoluble, dissolves in great meas- ure, leaving in suspension only a small amount of binantimoniate of potassa. 405. ^hS3-{-3irCl-^aq=SbCls-{-aq-}-Sn^. Sb + 3 ^ CZ + (II 0, N 0,) -^aq^Sb Ck + ag + ]V O2. Sb + X CI = ^5 (74 + X CI. Sh Ck^- Aq = ^h O^^ B H CI -\- Aq. The precipitate which is first formed on diluting a concentrated solution of Sb CI3 with water, always contains some chloride with the oxide ; but by con- tinued washing with water, or still better, with a weak solution of Na 0, C Og, the whole will be converted into oxide. Sh Gk + ^^ = Sb O5 + 5 IT CZ + Aq. 406. Sb O3 4- [^ 0, K 0] a H, 0,, + Aq = IKO, ShO,-] G,H, 0,,-\-Aq. The symbol of crystallized tartar emetic is [K 0, Sb O3] Cs H4 Oio. 2 H O. HEAVY METALS. 113 407. IK 0, Sh Os] OsH, 0,0 + S JiS-\-Aq = Sb'Sj [IT 0, K 0] G, H, 0:o + Aq. Kermes mineral is an amorphous modification of Sb Sg. Bri^ht-Ycllow. Sb Ol,-{-5 ffS-\-Aq= Sh^,-{-5IICl-{-Aq. Golden Sulphuret is a mixture of Sb S3 and Sb S5 . Melted together. 408. Sb S3 -j- 3 Fe = Sb + 3 Fe S. Arsenic (As). 412. As -j- 3 O = As O3 (arsenious acid). 413. 2 As O3 + 3 C = 2 As + 3 C O2. 414. As O3 -\- K 0, IT -\- Aq = K 0, As O3 -}- Aq. a. K 0, As 0, -\- 2 (Ou 0, S O3) -{- Aq = 2 Cu OrAs O3 + ^ 0, SO,. BO, S 0,-{- Aq. ■ h. The symbol of Schweinfurth green is Cu O, [Q H3] O3. 2 Cu O, As O3. 415. As O3 + 2 {H 0, N 0,) -[- aq = As 0, -\- aq + 2 W O4. The symbol of crystallized binarseniate of potassa is [2 H O, K O] As O5. Yellow. 416. As 03-\-d H S-\-Aq = As^s+M- Yellow. As 0,-\-?> HS^ Aq= A&^,-\- Aq. Yellow. Red. 2 As S3 + S = 2 As Sj. 10* 114 HEAVY METALS. 416. The symbol of Mispickel (arsenical pyrites) is Fe [As, SJ. 2 (Fe [As, S^]) + 13 O = Fea O3 + 2 As O3 + 2SO2. 417. As Zng + 3 {HO, SO,)-\-Aq^Z {Zn 0, S 0,) + ^9' + As H3. 418. Sb Zng + 3 (^0, aS O3) + ^^ = 3 {Zk 0, S O3) ^Aq-\-Sh Ha. The compounds of arsenic and antimony with hy- drogen are always mixed with more or less free hydrogen. When prepared as described in sections 417, 418 of Stockhardt's Elements, the gas consists almost entirely of hydrogen, con- taining only a very minute amount of either me- tallic compound. TABLES. EXPLANATION OF TABLES. Table I. — This table, Tvhich has been reprinted from the " Elementary Instructions in Chemical Analysis" by Fresenius, indicates by means of figures the solubility or insolubility in water and acids of some of the more fre- quently occurring compounds; thus, 1 means a substance soluble in water; 2, a substance insoluble in water, but soluble in chlorohydric or nitric acid; 8, a substance insoluble either in water or acids. For those substances stand- ing on the limits between these three classes, the figures are jointly expressed; thus 1-2 signifies a substance difficultly soluble in water, but soluble in chlorohydric or nitric acid ; 1 - 3, a body difficultly soluble in water, and the solubility of which is not increased on the addition of acids ; and 2 - 3, a sub- stance insoluble in water, and difficultly soluble in the acids. When the rela- tion of a substance to hydrochloric acid is different from that to nitric acid, this is stated in the notes. The figure indicating the solubility of a given salt will be found opposite to the symbol of its acid, in the column headed by the symbol of its base; that of a given binary, under the symbol of the correspond- ing oxide, and opposite to the symbol of its electro-negative element. Table II. — The values of the French measures and weights, in terms of the corresponding English units, given in this table, were taken from the second volume of the Cavendish Edition of " Gmelin's Hand-Book of Chemistry." The logarithms of these values and their arithmetical complements have been added to facilitate the reduction from one system to the other. The use of the table can be illustrated best by a few examples. 1. It is required to reduce 560.367 metres to English feet. Solution. — No. of feet = No. of metres X No. of feet in one metre. log. No. of feet =: log. No. of metres + log. No. of feet in one metre, log. 560.367 2.7484726 log. 3.2809 (value in feet of one metre from Table II.) 0.5159930 " 3.2644656 Ans. = 1838.51 feet. 118 EXPLANATION OF TABLES. 2. It is required to reduce 30.964 inches to centimetres. Solution. — No. of centimetres = No. of inches -^ No. of inches in one cen- timetre, log. No. of centimetres = log. No. of inches + log. (ar. co.) No. of inches in one centimetre, log. 30.964 1.4908571 log. (ar. CO.) 0.3937 (value of one centimetre in inches) 0.4048^58 1.8956829 Ans. 78.6471 centimetres. 3. It is required to reduce 23.576 kilometres to feet. Solution. — 23.576 kilometres = 23576 metres. No. of feet = No. of metres X No. of feet in one metre. ' log. 2.3576 4.3724701. log. 3.2809 0.5159930 4.8884631 Ans. 77350.5 feet. In the above examples logarithms of seven places have been used ; but where great accuracy is not required, logarithms of four places are sufficient. In such cases the last three figures of the logarithm given in the table may be neglected, and the problems solved with great expedition by means of the table of four-place logarithms which accompanies this book. Table III. — This table has been taken, with some few alterations, from Weber's " Atomgewichts-Tabellen." The atomic volumes assigned to the ele- ments are the same as those generally given in English and American text- books on Chemistrj', with the exception of those of Carbon, Boron, and Silicon, which are assumed to yield a one-volume gas like oxygen for convenience in calculation. The calculated specific gravities are deduced from the observed specific gravity of ox3'gen and the chemical equivalent of the given substance by means of the proportion, Equiv. of Oxygen : Equiv. of given substance = Sp. Gr. of Oxygen : Sp. Gr. of given substance. This proportion yields the specific gravity directly when one equivalent of the substance occupies the same volume as one equivalent of oxygen. If it occupies twice, three times, or four times this volume, the results must be divided by two, three, or four, as the case may be. The method of calculating may best be illustrated by a few examples. 1. It is required to calculate the Specific Gravity of Nitrogen. ' Equiv. of 0. Equiv. ofN. Sp. Gr. of O. Solution. 8 : 14 = 1.10563 : 1.93485. This would be the specific gravity if 14 parts of nitrogen occupied the same volume as 8 parts of oxygen ; or, in other words, if the equivalent volume of EXPLANATION OF TABLES. 119 nitrogen was 1, the same as that of oxygen. The fact is that it is 2, so that the true specific gravity of nitrogen = t (1.93485) = 0.967428. 2. It is required to calculate the specific gravity of ammonia gas. EquiT. ofO. Equiv. ofNHj. Sp. Gr. ofO. Solution. 8 : 17 = 1.10563 : 2.34946. Hence the specific gravity of ammonia gas •would be 2.34946 if one equiva- lent (or 17 parts) occupied the same volume as one equivalent of oxygen; but on referring to the table, it will be found that the equivalent volume of this gas is 4, or, in other words, that one equivalent occupies a volume four times as large as that occupied by one equivalent of oxygen, so that to find the true specific gravity of ammonia gas we must divide 2.34946 by 4, which will give a quotient 0.58736. The slight difference between this result and that given in the table arises from the fact that in Weber's table the equivalent of nitro- gen used is 14.005, and not 14, as in the solutions of the above examples. Were the law of equivalent volumes absolutely rigorous, (that is, did one equivalent of every gas precisely occupy either the same volume, or else a vol- ume two, three, or four times as great as the volume of oxygen,) then the cal- culated specific gravities ought to agree exactly with those obtained by experi- ment. On comparing together the two columns of observed and calculated specific gravities in the table, it will be found that the numbers, although approximatively equal, do not absolutely coincide. Part of these differences are unquestionably owing to errors of observation ; but after making the great- est possible allowance for all errors of that sort, there still remains (especially in the case of the easily condensed gases, such as alcohol vapor, sulphurous acid, and carbonic acid) large differences to be accounted for. The most prob- able explanation of these differences seems to be found in the assumption that the law of equivalent volumes holds rigorously only when the gases are in the state of extreme expansion. As we experiment upon them, they are more or less condensed by the pressure of the atmosphere, and it is supposed that they are not all condensed equally, or, in other words, that even under this pressure they do not obey absolutely the law of Marriotte. The more easily a gas may be reduced to a fluid, the greater is it condensed by the atmospheric pressure, and hence the greater is its specific gravity. This view is confirmed by the fact that the observed specific gravity of carbonic acid gas at 0°, and under a more feeble pressure than that of the atmosphere, approaches more nearly to that obtained by experiment. The Specific gravity of Carbonic Acid Gas, at 0° (air = 1), was. Under the pressure of 76.000 centimetres, 1.52910 " " " 37.413 " 1.52366 " " " 22.417* " 1.52145 Theoretical specific gravity, 1.52024 * It must be remembered that the specific gravity of a gas is equal to its weight, di- vided by the weight of an equal volume of air under the same conditions of temperature and pressure. 120 TABLES. ^ I O I— I iJ O o i-Ih CO G?THffq»(MTHG?iH (M (N(?^C1 tH <>? tH -, ,n :S -7^ y. o S 03 pi t>i rO O ro .9 ^ ^ f 5 O. a d rS a> P 43 '« !> [>,H rC o 03 d "i^ O ^ CJ ^rO a ^ n 'S ^ g o ■73 • l-H 1^ >% r" ^ o "a-Q C) n o n-f H s s t> 4^ rQrQ 13 o ^i 'C itJ" o -H •ri •p! ^ '« — 1 ~ (=-.o o <■> a n1 .£3 s *r^ h 'S 'S 13 d o d fl o d r2 ^-^ o n -> ccco Q!z;c» 11 122 TABLES. TABLE II. FKENCH MEASUEES AND WEIGHTS. Measures of Length. 1 Kilometre 1 Hectometre 1 Decimetre 1 Metre 1 Kilometre 1 Metre 1 Centimetre 1000 Metres. 100 « 10 " 1 " 0.6214 Miles. 3.2809 Feet. 0.3937 Inches 1 Meti-e = 1.000 Metre. 1 Decimetre = 0.100 " 1 Centimetre = 0.010 " 1 MilUmetre = 0.001 " Logarithms. 9.7933712 0.5159930 9.5951742 Ar. Co. Log. 0.2066188 9.4840070 0.4048258 Measures of Volume. 1 Cubic Metre = 1000.000 Litres. 1 Cubic Decimetre = 1.000 " 1 Cubic Centimetre = 0.001 " Logarithms. Ar. Co. Log, 1 Cubic Metre = 35.31660 Cubic Feet. 1.5479790 8.4520210 1 Cubic Decimetre = 61.02709 Cubic Inches. 1.7855226 8.2144774 1 Cubic Centimetre = 0.06103 " 8.7855226 1.2144774 1 Litre = 0.22017 Gallons. 9,3427581 0.6572419 1 Litre = 0.88066 Quarts. 9.9448083 0.0551917 1 Litre = 1.76133 Pints. Weights. 0.2458407 9.7541593 1 Kilogramme = 1000 Grammes. 1 Hectogramme = 100 " 1 Decagramme =10 " 1 Gramme = 1 " 1 Gramme = 1,000 Gramme. 1 Decigramme = 0.100 " 1 Centigramme = 0.010 " 1 Milligramme = 0.001 " 1 Kilogramme = 2.67951 Pounds (Troy). 1 Gramme = 15.44242 Grains. Logarithms. Ar, Co. Log, 0.4280554 9.5719446 1.1887154 8.8112846 TABLES. 123 TABLE III. SPECIFIC GRAVITY AND ABSOLUTE WEIGHT OF ONE LITRE OF SOME OF THE MOST IMPORTANT GASES AND VAPORS. Names of Gases. > '3 Sp. Gr. Observed. Sp. Gr. Calculated. Weight of 1 Litre = 1000 c. c. Loga- rithms. Ar. Co. Log- arithms. Air, 1. 1.00000 1.29363 0.1118101 9.8881899 Alcohol, 4 1.613 1.58934 2.05602 0,3130273 9.6869727 Ammonia Gas, 4 0.5967 0.58753 0.76005 9.8808422 0.1191578 Antimony, 1 17.83274 23.06897 1.3630282 8.6369718 Antimonide of Hydr. 4 4.56239 5.90204 0.7710022 9.2289978 Arsenic, 1 10.65 10.36528 13.40884 1.1273912 8.8726088 Arsenide of Hydr. 4 2.695 2.69553 3.48702 0.5424519 9.4575481 Boron, 1 1.50591 1.94809 0.2896090 9.7103910 Bromine, 2 5.54 5.52605 7.14866 0.8542247 9.1457753 Bromoliydric Acid, 4 2.79758 3.61903 0.5585922 9.4414078 Carbon, 1 0.8469* 0.82922 1.07270 0.0304783 9.9695217 Carbonic Acid, 2 1.52908 1.52024 1.96663 0.2937226 9.7062774 Carbonic Oxide, 2 0.96779 0.96743 1.25150 0.0974309 9.9025691 Chlorine, 2 2.47 2.45052 3.17007 0.5010689 9.4989311 Chloride of Boron, 4 3.942 4.05226 5.24213 0.7195078 9.2804922 Chloride of Silicon, 3 5.939 5.92477 7.66446 0.8844816 9.1155184 Chlorohydric Acid, 4 1.2474 1.25981 1.62973 0.2121157 9.7878843 Cyanogen, 2 1.8064 1.79698 2.32463 0.3663538 9.6336462 Cyanhydric Acid, 4 0.9476 0.93304 1.20701 0.0817109 9.9182891 Ether, 2 2.586 2.55677 3.30751 0.5195012 9.4804988 Fluorine, 2 1.30151 1.68367 0.2262570 9.7737430 Fluoride of Boron, 4 2.3124 2.32875 3.01254 0.4789329 9.5210671 Fluoride of Silicon, 3 3.600 3.62677 4.69170 0.6713302 9.3286698 Fluohydric Acid, 4 0.68531 0.88654 9.9476983 0.0523017 Hydrogen, 2 0,06927 0.06910 0.08939 8.9512889 1.0487111 Iodine, 2 8.716 8.76760 11.34203 1.0547011 8.9452989 lodohydric Acid, 4 4.443 4.41835 5.71571 0.7570702 9.2429298 Marsh Gas, 4 0.5576 0.55282 0.71514 9.8543911 0.1456089 Mercuiy, 2 6.976 6.91732 8.94845 0.9517478 9.0482522 Nitrogen, 2 0.97136 0.96776 1.25192 0.0975765 9.9024235 Nitrous Oxide, 2 1.5269 1.58951 2.05624 0.3130738 9.6869262 Nitric Oxide, 4 1.0388 1.03669 1.34109 0.1274580 9.8725420 defiant Gas, 4 0.9852 0.96743 1.25150 0.0974309 9.9025691 Oxygen, 1 l.li)§63 1.43028 0.1554210 9.8445790 Phosphoms, 1 4.42 4.33452 5.60727 0.7487515 9.2512485 Phosphide of Hydr. 4 1.178 1.18728 1.53590 0.1863629 9.8136371 Selenium, 1 2.73801 3.54197 0.5492448 9.4507552 Silicon, 1 3.07120 3.97300 0.5991186 9.4008814 Sulphur, k 6.5635 6.65866 8.61384 0.9351968 9.064S032 Sulphide of Hydr. 2 1.1912 1.17888 1.52.503 0.1832784 9.8167216 Sulphurous Acid, 2 2.247 2.21541 2.86592 0.4572640 9.5427360 * Calculated from the Sp. Gr. of Carbouic Acid Gas, observed by RegnauU. 124 TABLES. TABLE IV. MEAN COEFFICIENTS OF LINEAE EXPANSION OF SOLIDS FOR ONE DEGEEE BETWEEN 0° AND 100°. Name of Substance. Coefficients. Name of Observer. Glass (flint of Clioisy le Roi), 0.00000760 Regnault. Platinum, 0.00000884 Dulong and Petit. Glass (commoa of Paris), 0.00000920 Regnault. Palladium, 0.00001000 WollastOn. Antimony, 0.00001083 Smeaton. Iron (soft forged), 0.00001220 Lavoisier and Laplace. Bismuth, 0.00001392 Smeaton. Gold, 0.00001466 Lavoisier and Laplace. Brass (English, in rods), 0.00001893 Roy. Copper, 0.00001919 Troughton. Silver, 0.00002083 « Tin (fine), 0.00002283 5 1 2 3 4 5 6 7 8 9 Proportional Parts. I 2 3 4 s 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 2 2 3 4 5 5 6 7 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 2 2 3 4 5 5 6 7 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 2 2 3 4 5 5 6 7 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 2 3 4 4 5 6 7 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 2 3 4 4 5 6 7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 2 3 4 4 5 6 6 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 2 3 4 4 5 6 6 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 2 3 3 4 5 6 6 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 2 3 3 4 5 5 6 64 8062 8069 8075 8032 8089 8096 8102 8109 8116 8122 2 3 3 4 5 5 6 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 2 3 3 4 5 5 6 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 2 3 3 4 5 5 6 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 2 3 3 4 5 5 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 2 3 3 4 4 5 6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 2 2 3 4 4 5 6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 2 2 3 4 4 5 6 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 2 2 3 4 4 5 5 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 2 2 3 4 4 5 5 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 2 2 3 4 4 5 5 74 8692 8698 8704 8710 8716 S722 8727 8733 8739 8745 2 2 3 4 4 5 5 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 2 2 3 3 4 5 5 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 2 2 3 3 4 5 5 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 2 2 3 3 4 4 5 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 2 2 3 3 4 4 5 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 2 2 3 3 4 4 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 2 2 3 3 4 4 5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 2 2 3 3 4 4 5 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 2 2 3 3 4 5 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 2 2 3 3 4 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 2 2 3 3 4 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 2 2 3 3 4 5 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 2 2 3 3 4 5 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 2 2 3 3 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 2 2 3 3 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 2 2 3 3 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 2 2 3 3 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 2 2 3 3 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 2 2 3 3 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 2 2 3 3 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 2 2 3 3 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 2 2 3 3 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 2 2 3 3 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 2 2 3 3 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 2 2 3 3 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 _i 2 2 3 3 3 niH isdVHO IV ON do AiisyaAiNn