58 gravity to the atoms of differe pose that the different weights magnitude; e. g., that an atom volume. It follows from this t stances must depend upon the their atoms. These atoms m unalterable volume, and the e nution of external pressure, as crease of pressure, as proceedi the intervening spaces. In op of necessity from the hypothesi 92, 409) and more lately Gra which a body occupies is com vening spaces. According to body multiplied by the volume (for the greater the sp. gr. of this view of its atoms, and th be their weight). Conversel gravity are known, the volum first by the second. Thus the the specific gravity of the diar carbon 1714; and the at by its sp. gr. = 0·865 gives i numbers be compared with th in column D, it will be found the reciprocal of the atomi 45.32. 0·0221 = 1. (It is numbers will come out diffe = weight of hydrogen 1, we done). That the atomic vo atomic number is easily seen (which must be regarded as product of the atomic numbe if S specific gravity, G we have S G. Z. If no atomic weight G by the sy we have the atomic volum is the reciprocal of the at the expression atomic volu for we can understand by with the adjacent and su stices vary according t temperature, the atomi by Pol. Boullay (An N. Tr. 23, 1, 208) w' lation be adopted, question and adop quotients equival lent of 126 iod 8 cubic me 126: 4.048 = sures, whose iodine wh irabis and ambers of the mpal atering to its elements we are the ital letters of the me the the page 30, column If the rate cass, have Eikewise Water=Aq: Cyanogen art = T. Aretie acid = A; Quinine The aim was annexed a the symbols denote - tye wa montana eristag in the com- un groupe raste in the compound. Electroart as minus ant alfa, je lose precede electromsn the formule. This alties in the nomenclature, but it would - v, as 12 the firmale to give precedence to Devet frem a map and e stains proximate and sme de zitate is expressed by means of - ia takas, Capsa, which occurs so fre17, 15 points placed over the sym 2 2 2 7 2 2 ematen. the number of these Camren tout are reed to denote atoms of A paid passim sad 1 At, oxy · 5 war 5 HD = Hala 2At alamicum and 3 At. NON; = S_atric acid is NO1 = N; ammonia is NH3 — 25%) - S0' = Pb0), SO' = P6 S; bicarbonate of Maza, 242, carbonic acid, and 1 At. water) is KO + C90 = KC H; crystallized sulphate of ammoA spearic acid, and 2 At. water) is NH1 + 252 = NSH; crystallized potash alum (1 YFS 15 Auric acid, and 24 At. water) is 24 HO) = KO, SO3 + AFO, SO * sacházet of potassium (1 At. potassium asaj barve of molybdenum is MS = " KS, MoS3 = KMo. A number placed on the part of a symbol denotes merely the number of › denoted by the symbol: e. g., SO must be underus of oxygen and only 1 of sulphur; when oxygen , the number placed on the right refers to the whole the oxygen; thus in bicarbonate of potash KC H 2 At. CO, not merely 2 At. C. On the other hand, re several symbols multiplies them all as far as the ma, or if the number stands before a bracket, it mulols and numbers included within the brackets: thus bO, NO3 means a combination of 6 At. oxide of lead id; and KCl + 4 HgCl + 4HO, or KCl, 4HgCl + on of 1 At. chloride of potassium with 4 At. chloride t. water. Many chemists write the number or index instead of the right above the symbols (thus, SO, for cause, in algebraical formulæ, a number on the right Dower: but there is no risk of confusion between algeal formulæ, and the number when written above is an when placed below. Stoichiometrical Calculation. e number of atoms of any substance in a given comreater the weight of those atoms, the greater will be hat substance in the compound. Hence the quantity uents in a given quantity of the compound (in 100 parts determined by multiplying the number of atoms (Z) of by the atomic weight (G). Hence we have the three rmula comes into use in determining the quantities of the uents contained in a given quantity of any compound. onsists in multiplying the atomic weight of each constituent er of its atoms contained in the compound atom, and adding es so obtained: the sum is the atomic weight of the compound, ne quantity of each element contained in it has also been dethe quantities of those several elements in any other quantity apound may be found by the Rule of Three. t are the quantities of the several elements of sulphate of lead O3) contained in 100 parts? PbO is 103.8 + 8 = 111.8; SO3 is 2440; therefore PbO + SO3 is 1118 + 40 = 151·8. We then that the proximate elements of 1518 parts of sulphate of lead 111.8 oxide of lead and 40 sulphuric acid; the ultimate elements are 8 lead, 16 sulphur and 32 oxygen. If now 1518 parts of sulphate lead contain 1118 oxide of lead, 100 parts must contain (151·8:111.8 100: 73-65) 73.65 parts; similarly the proportion 1518: 40 = 100: x gives 26-35 per cent. of sulphuric acid; 1518: 1038 100 x gives 68.38 per cent. lead; 151·8: 16 = 100: x gives 10.54 per cent. sulphur, and 151.8:32 = 100 x gives 21.08 per cent. oxygen.-What are the constituents of 85 parts of morphia (CNH2 O)? 35 At. carbon weigh 356 210: 1 At. nitrogen 14; 20 At. hydrogen 20.1 = 20; and 6 At. Oxygen 6.8 = 48; and 210 + 14 + 20 + 48 292; now 292: 210 =85x gives 61.13 carbon in 85 parts of morphia; 292: 14 = 85: x that when the proximate elements of such compounds of the second order, sulphate of lead for example, contain a common ultimate element, as oxygen in this case, the quantities of this ultimate element contained in the two proximate elements must bear a simple relation to each other: e. g., the quantity of oxygen in sulphuric acid is exactly 3 times as great as that in the oxide of lead combined with it. + Chemical Formula. A chemical formula is an expression by symbols and numbers of the composition of a definite chemical compound according to its elements and their relative quantities. The symbols are the initial letters of the names of the elementary substances given in the table, page 50, column B. Certain compounds, particularly of the organic class, have likewise particular symbols appropriated to them: e. g., Water Aq; Cyanogen = Cy; Tartaric acid = T; Citric acid = C; Acetic acid = A; Quinine = Ch; Morphia =M, &c. The numbers annexed to the symbols denote the numbers of atoms of the several constituents existing in the compound; a symbol with no number annexed to it implies that one atom only of the corresponding substance exists in the compound. Electropositive substances, such as metals and salifiable bases, precede electronegative substances, such as oxygen and acids in the formulæ. This order is the reverse of that adopted in the nomenclature, but it would perhaps be better in this as well as in the formula to give precedence to the electro-positive element. When a compound contains proximate and ultimate elements, the mode of combination is expressed by means of points, commas, + signs, and brackets. Oxygen, which occurs so frequently in compounds, is often expressed by points placed over the symbol of the body with which it is in combination, the number of these points being equal to the number of atoms of oxygen present. In a similar manner, strokes leaning from right to left are used to denote atoms of sulphur, and points under the symbol of the other body, atoms of hydrogen. Oxide of lead is PbO = Pb; potash (1 At. potassium and 1 At. oxygen) is KO = K; water is HO = Í; alumina (2 At. aluminum and 3 At. oxygen) is Al2 03 Äl; carbonic acid is COC; silica is Si O2 = sulphuric acid is SO = = N; sulphate of lead is S; nitric acid is NO3 = Si; S; nitric acid is NON; ammonia is NH3 — PbO SO PbO, SO3 = Pb S; bicarbonate of + = potash (1 At. potash, 2 At. carbonic acid, and 1 At. water) is KO + 2CO2 + HOKO, CO, HO= KC 11; crystallized sulphate of ammonia (1 At. ammonia, 1 At. sulphurie neid, and 2 At. water) is NH3 + SO3 + 2HO NH3, SO, 2HO, NSH'; crystallized potash alum (1) At. potash, 1 At. alumina, 4 At. sulphuric acid, and 24 At. water) is (KO+ SO3) + (Al 03 + 3SO) + (24 HO) = KO, SO+ AF 03, 3SO + 24 HO = K S, Al S, H; sulphuret of potassium (1 At. potassium and 1 At. sulphur) is KS K; tersulphuret of molybdenum is MoS = Mo; the combination of these two metallic sulphurets in equal numbers = of atoms is KS + MoS = KS, MOS3 = KMo. A number placed on the right and at the upper part of a symbol denotes merely the number of atoms of the substance denoted by the symbol: e. g., SO must be understood to denote 3 atoms of oxygen and only 1 of sulphur; when oxygen is expressed by points, the number placed on the right refers to the whole compound containing the oxygen; thus in bicarbonate of potash KC H we must understand 2 At. CO, not merely 2 At. C. On the other hand, a number placed before several symbols multiplies them all as far as the next sign or comma, or if the number stands before a bracket, it multiplies all the symbols and numbers included within the brackets: thus 6PbO + NO3 or 6PbO, NO3 means a combination of 6 At. oxide of lead with 1 At. nitric acid; and KCl + 4 HgCl + 4HO, or KCl, 4HgCl + 4HO, the combination of 1 At. chloride of potassium with 4 At. chloride of mercury and 4 At. water. Many chemists write the number or index on the right below, instead of the right above the symbols (thus, SO, for sulphuric acid), because, in algebraical formula, a number on the right above expresses a power: but there is no risk of confusion between algebraical and chemical formulæ, and the number when written above is more easily read than when placed below. Stoichiometrical Calculation. The greater the number of atoms of any substance in a given compound, and the greater the weight of those atoms, the greater will be the quantity of that substance in the compound. Hence the quantity (M) of the constituents in a given quantity of the compound (in 100 parts for example) is determined by multiplying the number of atoms (Ż) of each constituent by the atomic weight (G). Hence we have the three following formula: 1). MZ. G; 2) G = Z; 3) z = M M Z G The first formula comes into use in determining the quantities of the several constituents contained in a given quantity of any compound. The process consists in multiplying the atomic weight of each constituent by the number of its atoms contained in the compound atom, and adding the quantities so obtained: the sum is the atomic weight of the compound, and since the quantity of each element contained in it has also been determined, the quantities of those several elements in any other quantity of the compound may be found by the Rule of Three. What are the quantities of the several elements of sulphate of lead (PbO, SO3) contained in 100 parts? PbO is 103-8 +8=111·8; SO3 is 16+ 24 = 40; therefore PbO + SO3 is 1118 + 40 151·8. We know then that the proximate elements of 151.8 parts of sulphate of lead are 111.8 oxide of lead and 40 sulphuric acid; the ultimate elements are 103-8 lead, 16 sulphur and 32 oxygen. If now 1518 parts of sulphate of lead contain 111.8 oxide of lead, 100 parts must contain (1518:111·8 100: 73.65) 73.65 parts; similarly the proportion 1518: 40 = 100 x gives 26-35 per cent. of sulphuric acid; 1518:1038 = 100: a gives 68.38 per cent. lead; 1518:16 = 100x gives 10.54 per cent. sulphur, and 1518:32 = 100 x gives 21.08 per cent. oxygen.-What are the constituents of 85 parts of morphia (C NH O)? 35 At. carbon weigh 356 210: 1 At. nitrogen 14; 20 At. hydrogen 20.1 = : 20; and 6 At. oxygen 6.8 48; and 210 +14 +20 + 48 292; now 292 : 210 =85 gives 61.13 carbon in 85 parts of morphia; 292: 14 = 85; x gives 4.08 nitrogen; 292: 2085 : x gives 5·82 hydrogen; and 292 : 48 85 x gives 13.97 oxygen. The first formula also serves to find how much of any substance, simple or compound, is required to convert a given quantity of another substance into a given compound, or to decompose a given quantity of any compound. How much sulphur is required to convert 135 parts of copper into disulphuret of copper? In this compound, Cu S, there are 2.31.8 636 copper combined with 16 sulphur: since then 63.6 copper require 16 sulphur, 135 parts of copper will require 33.96 parts of sulphur, for 636 16 135: 33·96.-How much oil of vitriol is required for the decomposition of 79 pts. of nitrate of potash, so that bisulphate of potash may be formed while the nitric acid escapes? In nitrate of potash, 1 At. potash, (KO) 39.2 + 8 = 47.2 is combined with 1 At. nitric acid (NO) = 14 + 40 = 54: the atomic weight of nitrate of potash is therefore 472 + 54 = 101·2. Oil of vitriol (SO3 HO) contains 1 At. sulphuric acid = 16 + 3.8 40, and 1 At. water = = 1 + 89: therefore the atomic weight of oil of vitriol is 40+ 9 = 49. Since then 2 At. sulphuric acid are to combine with 1 At. potash, 1 At. or 101.2 parts of nitrate of potash (containing 47-2 potash) will require 2 At. or 2. 49 = 98 parts of oil of vitriol (containing 80 sulphuric acid); now 1012: 98 = 79 765 the quantity of oil of vitriol required to decompose 79 parts of nitrate of potash. By means of the second formula we may find the atomic weight of a substance when we know the relative quantity of it in a given compound, and also the number of its atoms which probably enter into the constitution of that compound. Suppose that in 100 parts of selenious acid we have found 71.43 parts of selenium and 28.57 parts of oxygen, and assume as probable that it contains 1 At. selenium and 2 At. oxygen; we have then, in order to perform the operation indicated by the second formula, to divide the quantity of each element by the number of its atoms: thus M :Z= G Se 71.43 1= 71.43 O 28.57: 2 = 14.285 According to this, the atomic weight of selenium would be 71.43 if that of oxygen were 14-285; but taking 8 for the atomic weight of oxygen, we have 14.285 71.43 = 8: 40: if the atomic weight of oxygen be 100 we have 14.2857143 = 100: 500. Hence the atomic weight of selenium is 40 when that of oxygen is 8, and 500 when that of oxygen is 100. By the third formula we find in what numbers the atoms of the different constituents are united, when their relative quantities and atomic weights are known. If 100 parts of nitric acid contain 25.926 nitrogen, and 74.074 oxygen, and the atomic weights of nitrogen and oxygen are 14 and 8 respectively, what are the numbers of atoms of these two elements contained in nitric acid? According to the formula, the relative quantity divided by the atomic weight gives the number of atoms; therefore M : G = Ꮓ N 25.926: 14 = 1.852 O 74074 8 = 9.260 According to this, 1852 At. nitrogen are united with 9260 At. oxyThis complicated ratio may however (as in most other cases) be |