A Text-book of Assaying: For the Use of Those Connected with Mines Part 10

[7] NaC_{2}H_{3}O_{2} + HCl = H_{4}C_{2}O_{2} + NaCl.

CHAPTER VII.

FORMULae, EQUATIONS, ETC.

Formulae 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.

a.r.s.enic 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.

a.r.s.enic 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 a.r.s.enic 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 a.r.s.enic. 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, FeS_{2} represents a molecule with one atom of iron and two atoms of sulphur, Fe_{2}S_{3} 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(NO_{3})_{2} is another way of writing PbN_{2}O_{6}. 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 formulae of the groups, and separating each simple formula by a full stop. Slaked lime, for instance, has the formula CaH_{2}O_{2}; 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 (H_{2}O), we could write it CaO.H_{2}O. A plus sign (+) has a different meaning; CaO + H_{2}O indicates quant.i.ties 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 quant.i.tative relation, the meaning of the equation CaO + H_{2}O = CaO.H_{2}O is: "lime and water yield slaked lime." By referring to a table of atomic weights we can elicit the quant.i.tative relations thus:--

CaO + H_{2}O = CaH_{2}O_{2} | | | V V V Ca = 40 H_{2} = 2 = 12 Ca = 40 O = 16 O = 16 H_{2} = 2 = 12 -- -- O_{2} = 32 = 162 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 (56112)/74, or about 84-3/4 lbs.

As another example having a closer bearing on a.s.saying take the following question:--"In order to a.s.say 5 grams of 'black tin' (SnO_{2}) by the cyanide process, how much pota.s.sic cyanide (KCN) will be required?" The reaction is

SnO_{2} + 2KCN = Sn + 2KCNO | | V V Sn = 118 K = 39 O_{2} = 32 C = 12 --- N = 14 150 -- 652 = 130

What is sought for here is the relation between the quant.i.ties of SnO_{2} 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 2Fe_{2}O_{3}.3H_{2}O?" Bringing this formula together we have Fe_{4}H_{6}O_{9}. Find the molecular weight.

Fe_{4} = 224 = 564 H_{6} = 6 = 16 O_{9} = 144 = 169 --- 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 (CuSO_{4}.5H_{2}O) 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 O_{4} = 64.0 = 164 5H_{2}O = 90.0 = 185 ----- 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 cla.s.s 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 t.i.tration of iron by permanganate solution there are two reactions. First in dissolving the iron

Fe + H_{2}SO_{4} = FeSO_{4} + H_{2} | V 56

and second, in the actual t.i.tration,

10FeSO_{4} + 2KMnO_{4} + 9H_{2}SO_{4}= 2MnSO_{4} | + 5Fe_{2}(SO_{4})_{3} + 2KHSO_{4} + 8H_{2}O V K = 39 Mn = 55 O_{4}= 64 --- 158 2 = 316

As before, attention is confined to the two substances under consideration--viz., Fe and KMnO_{4}. In the second equation, we find 316 parts of the permanganate are required for 10 molecules of FeSO_{4}; and in the first equation 56 parts of iron are equivalent to one molecule of FeSO_{4}, 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 t.i.tration" (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

2FeSO_{4} + MnO_{2} + 2H_{2}SO_{4} | = Fe_{2}(SO_{4})_{2} + MnSO_{4} + 2H_{2}O | V Mn = 55 O_{2} = 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 a.n.a.lysis. The method of calculating is shown by the following example. Required the formula of a mineral which gave the following figures on a.n.a.lysis:--

Cupric oxide (CuO) 10.58 Ferrous oxide (FeO) 15.69 Zinc oxide (ZnO) 0.35 Sulphuric oxide (SO_{2}) 28.82 Water (H_{2}O) 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.

Cla.s.sify the results as follows:--

Bases. Acids. Water.