FORMULÆ, EQUATIONS, ETC.
Formulæ and equations are a kind of short hand for expressing briefly and in the language of the atomic theory the facts of chemical composition and reaction. The convenience of this method of expressing the facts justifies a short description of it here.
On comparing the percentage composition of a series of compounds the proportions in which the elements combine appears to be regulated by no simple law. For example:
| Realgar. | Orpiment. | Mispickel. | Pyrites. | |
| Arsenic | 71.4 | 60.9 | 46.0 | — |
| Sulphur | 28.6 | 39.1 | 19.6 | 53.3 |
| Iron | — | — | 34.4 | 46.7 |
| ——— | ——— | ——— | ——— | |
| 100.0 | 100.0 | 100.0 | 100.0 |
But if in these examples the composition is calculated, not on 100 parts, but on 107, 246, 163, and 120 parts respectively, evidence of a simple law becomes apparent.
| Realgar. | Orpiment. | Mispickel. | Pyrites. | |
| Arsenic | 75.0 | 150.0 | 75.0 | — |
| Sulphur | 32.0 | 96.0 | 32.0 | 64.0 |
| Iron | — | — | 56.0 | 56.0 |
| ——— | ——— | ——— | ——— | |
| 107.0 | 246.0 | 163.0 | 120.0 |
It will be seen that the proportion of arsenic is 75 or twice 75, that of iron is 56, and that of sulphur 32 or some simple multiple of 32. The series of examples might be extended indefinitely, and it would still be found that the "combining proportions" held good. The number 75 is spoken of as the "combining weight," or, more frequently, as the "atomic weight" of arsenic. Similarly 56 is the atomic weight of iron, and 32 the atomic weight of sulphur. The importance of this law of chemical combination is altogether independent of the atomic theory; but this theory furnishes the simplest explanation of the facts. According to it a chemical compound is made up of exactly similar groups of particles. The particles of each elementary substance are all alike, but differ from those of other elements in weight. Ultimate particles are called atoms, and the groups of atoms are called molecules. The atomic weight of any particular element is the weight of its atom compared with the weight of an atom of hydrogen. The atom of sulphur, for instance, is 32 times as heavy as the atom of hydrogen, and the atomic weight of sulphur is 32. The molecular weight is the sum of the atomic weights of the group. The molecule of pyrites contains two atoms of sulphur and one of iron: on referring to the table of atomic weights it will be seen that the atomic weights are—sulphur 32, and iron 56. The molecular weight, therefore, is 32 + 32 + 56—that is, 120. The meaning of this is, 120 parts by weight of iron pyrites contain 64 parts of sulphur and 56 parts of iron; and this is true whether the "parts by weight" be grains or tons.
The symbol or formula of an atom is generally the initial letter or letters of the Latin or English name of the substance. The atom of hydrogen is written H, that of oxygen O, of sulphur S, of iron (ferrum) Fe, and so on. A list of these symbols is given in the table of atomic weights.
The formula of a molecule is obtained by placing together the symbols of the contained atoms. Thus, Fe represents an atom of iron, S an atom of sulphur, while FeS represents the molecule of sulphide of iron as containing one atom of each element.
When more than one atom of an element is present this is shown by writing a figure under and after the symbol; thus, FeS2 represents a molecule with one atom of iron and two atoms of sulphur, Fe2S3 similarly shows one with two atoms of iron and three of sulphur. When a group of atoms is enclosed in brackets, a figure after and under the bracket multiplies all within it; for example, Pb(NO3)2 is another way of writing PbN2O6. Sometimes it is convenient to represent the atoms of a molecule as divided into two or more groups; this may be done by writing the formulæ of the groups, and separating each simple formula by a full stop. Slaked lime, for instance, has the formula CaH2O2; or, as already explained, we may write it Ca(HO)2; or, if for purposes of explanation we wished to look on it as lime (CaO) and water (H2O), we could write it CaO.H2O. A plus sign (+) has a different meaning; CaO + H2O indicates quantities of two substances, water and lime, which are separate from each other. The sign of equality (=) is generally used to separate a statement of the reagents used from another statement of the products of the reaction; it may be translated into the word "yields" or "becomes." The two statements form an equation.
Ignoring the quantitative relation, the meaning of the equation CaO + H2O = CaO.H2O is: "lime and water yield slaked lime." By referring to a table of atomic weights we can elicit the quantitative relations thus:—
| CaO | + | H2O | = | CaH2O2 | |
| ↓ | ↓ | ↓ | |||
| Ca = 40 | H2 = 2 | = 1×2 | Ca = 40 | ||
| O = 16 | O = 16 | H2 = 2 | = 1×2 | ||
| —— | —— | O2 = 32 | = 16×2 | ||
| 56 | 18 | —— | |||
| 74 |
Or, putting it in words, 56 parts of lime combine with 18 parts of water to form 74 parts of slaked lime. This equation enables one to answer such a question as this:—How much lime must be used to produce 1 cwt. of slaked lime? for, if 74 lbs. of slaked lime require 56 lbs. of lime, 112 lbs. will require (56 × 112)/74, or about 84-3/4 lbs.
As another example having a closer bearing on assaying take the following question:—"In order to assay 5 grams of 'black tin' (SnO2) by the cyanide process, how much potassic cyanide (KCN) will be required?" The reaction is
| SnO2 | + | 2KCN | = Sn + 2KCNO |
| ↓ | ↓ | ||
| Sn = 118 | K = 39 | ||
| O2 = 32 | C = 12 | ||
| —— | N = 14 | ||
| 150 | —— | ||
| 65 | ×2 = 130 |
What is sought for here is the relation between the quantities of SnO2 and KCN. Note that a figure before a formula multiplies all that follows up to the next stop or plus or equality sign. The question is now resolved to this: if 150 grams of oxide of tin require 130 grams of cyanide, how much will 5 grams require?
150 : 130 :: 5 : x
x = 4.33 grams.
A problem of frequent occurrence is to find the percentage composition of a substance when its formula has been given. For example: "What percentage of iron is contained in a mineral having the formula 2Fe2O3.3H2O?" Bringing this formula together we have Fe4H6O9. Find the molecular weight.
| Fe4 | = 224 | = 56×4 |
| H6 | = 6 | = 1×6 |
| O9 | = 144 | = 16×9 |
| —— | ||
| 374 |
Then we get: 374 parts of the mineral contain 224 of iron. How much will 100 contain?
374 : 224 :: 100 : x
x = 59.89.
And the answer to the question is 59.89 per cent.
Again, suppose the question is of this kind:—"How much crystallised copper sulphate (CuSO4.5H2O) will be required to make 2 litres of a solution, 1 c.c. of which shall contain 0.0010 gram of copper?"
A litre is 1000 c.c., so, therefore, 2 litres of the solution must contain 0.001 gram × 2000, or 2 grams. How much crystallised copper sulphate will contain this amount of metal?
| Cu | = 63.3 | |
| S | = 32.0 | |
| O4 | = 64.0 | = 16×4 |
| 5H2O | = 90.0 | = 18×5 |
| ———— | ||
| 249.3 |
If 63.3 grams of copper are contained in 249.3 grams of sulphate, in how much is 2 grams contained.
63.3 : 249.3 :: 2 grams : x
x = 7.8769 grams.
The answer is, 7.8769 grams must be taken.
As a sample of another class of problem similar in nature to the last (but a little more complicated) take the following:—"What weight of permanganate of potash must be taken to make 2 litres of a solution, 100 c.c. of which shall be equivalent to 1 gram of iron?" In the first place the 2 litres must be equivalent to 20 grams of iron, for there are 20 × 100 c.c. in two litres. In the titration of iron by permanganate solution there are two reactions. First in dissolving the iron
Fe + H2SO4 = FeSO4 + H2
↓
56
and second, in the actual titration,
10FeSO4 + 2KMnO4 + 9H2SO4= 2MnSO4 + 5Fe2(SO4)3 + 2KHSO4 + 8H2O
↓
K = 39
Mn = 55
O4= 64
——
158 × 2 = 316
As before, attention is confined to the two substances under consideration—viz., Fe and KMnO4. In the second equation, we find 316 parts of the permanganate are required for 10 molecules of FeSO4; and in the first equation 56 parts of iron are equivalent to one molecule of FeSO4, therefore 560 of iron are equivalent to 316 of permanganate; and the question is, How much of the permanganate will be equivalent to 20 grams of iron?
560 : 316 :: 20 grams : x.
x = 11.286 grams.
The answer is 11.286 grams.
Very similar to this last problem is the question suggested under the head "Indirect Titration" (p. 43). "If 100 c.c. of the standard permanganate solution are equivalent to 1 gram of iron, how much peroxide of manganese will they be equivalent to?" The equation for dissolving the iron is already given; the second equation is
2FeSO4 + MnO2 + 2H2SO4 = Fe2(SO4)2 + MnSO4 + 2H2O
↓
Mn = 55
O2 = 32
——
87
It will be seen that 87 grams of peroxide of manganese are equivalent to 112 grams of iron. How much then is equivalent to 1 gram of iron?
112 : 87 :: 1 gram : x
x = 0.7767 gram.
It is sometimes convenient to calculate the formula of a substance from its analysis. The method of calculating is shown by the following example. Required the formula of a mineral which gave the following figures on analysis:—
| Cupric oxide (CuO) | 10.58 |
| Ferrous oxide (FeO) | 15.69 |
| Zinc oxide (ZnO) | 0.35 |
| Sulphuric oxide (SO2) | 28.82 |
| Water (H2O) | 44.71 |
| —————— | |
| 100.15 |
First find the molecular weights of CuO, FeO, &c., and divide the corresponding percentages by these figures. Thus, CuO = 63.3+16 = 79.3 and 10.58 divided by 79.3 gives 0.1334. Similarly FeO = 56+16 = 72 and 15.69 divided by 72 gives 0.2179. Treated in the same way the oxide of zinc, sulphuric oxide and water give as results 0.0043, 0.3602 and 2.484.
Classify the results as follows:—
| Bases. | Acids. | Water. |
| CuO 0.1334 | SO3 0.3602 | H2O 2.484 |
| FeO 0.2179 | ||
| ZnO 0.0043 | ||
| —————————— | —————————— | —————————— |
| RO 0.3556 | RO3 0.3602 | R2O 2.484 |
The figures 0.3556, 0.3602 and 2.484 should be then divided by the lowest of them—i.e., 0.3556; or where, as in this case, two of the figures are very near each other the mean of these may be taken—i.e., 0.3579. Whichever is taken the figures got will be approximately 1, 1 and 7. The formula is then RO.SO3.7H2O in which R is nearly 2/5ths copper, 3/5ths iron and a little zinc.
This formula requires the following percentage composition, which for the sake of comparison is placed side by side with the actual results.
| Calculated. | Found. | |
| Cupric oxide | 11.29 | 10.58 |
| Ferrous oxide | 15.37 | 15.69 |
| Zinc oxide | nil | 0.35 |
| Sulphuric oxide | 28.47 | 28.82 |
| Water | 44.84 | 44.71 |
| ——— | ——— | |
| 99.97 | 100.15 |
Trimming the results of an analysis to make them fit in more closely with the calculations from the formula would be foolish as well as dishonest. There can be no doubt that the actual analytical results represent the composition of the specimen much more closely than the formula does; although perhaps other specimens of the same mineral would yield results which would group themselves better around the calculated results than around those of the first specimen analysed. It must be remembered that substances are rarely found pure either in nature or in the arts; so that in most cases the formula only gives an approximation to the truth. In the case of hydrated salts there is generally a difficulty in getting the salt with exactly the right proportion of water.
PRACTICAL EXERCISES.
The following calculations may be made:—
1. Calculate standards in the following cases—
(a) Silver taken, 1.003 gram. Standard salt used, 100.15 c.c.
(b) Iron taken, 0.7 gram. Bichromate used, 69.6 c.c.
2. Calculate percentages:—
(a) Ore taken, 1 gram. Solution used, 65.2 c.c. Standard, 0.987 gram.
(b) Ore taken, 1 gram. Barium sulphate got, 1.432 gram. Barium sulphate contains 13.73 per cent. of sulphur, and the percentage of sulphur in the ore is wanted.
(c) Barium sulphate is BaSO4. Calculate the percentage of sulphur it contains, for use in the preceding question.
3. A method of estimating the quantity of peroxide in a manganese ore is based on the following reactions:—
(1) MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O.
(2) Cl + KI = KCl + I.
To how much MnO2 is 1 gram of Iodine (I) equivalent?
4. A mineral has the following composition:—
Carbonic acid (CO2) 19.09
Copper oxide (CuO) 71.46
Water (H2O) 9.02
What is its formula?
5. How much copper is contained in 1.5 gram of crystallized copper sulphate (CuSO4.5H2O)? How much of these crystals must be taken to give 0.4 gram of copper?
6. How much ferrous sulphate crystals (FeSO4.7H2O) must be taken to yield 2 litres of a solution, 100 c.c. of which shall contain 0.56 gram of iron?
7. Galena is PbS, and hæmatite Fe2O3. What percentages of metal do these minerals contain?