To master the syınbolical language of chemistry, so as to understand fully what it expresses, is a great step toward mastering the science; and so important is this part of my subject that I propose to occupy the hour this evening with a number of illustrations of the use of symbols for expressing chemical changes.

First, I will recur to the experiment of the last lecture, for we have not yet learned all that it is calculated to teach.

Let us again write on the black-board the symbols which represent the chemical process:

(Na2CO3 + 2HCl + Aq.) = (2NaCl + H2O + Aq.) + CO2.

Sodic

Carbonate.

Hydrochloric
Acid.

Common
Salt.

Water.

Carbonic Dioxide Gas.

We bring together a solution of sodic carbonate and hydrochloric acid; and there are formed as products a solution of common salt, water, and carbonic dioxide gas. I need not refer again to the circumstance that the state of solution is an essential condition of the change, for this point was fully discussed at the time; but, before we pass on to another experiment, I wish to call your attention to the fact that the several terms in this equation stand for absolutely defi

nite weights of the quantities they represent. Each symbol stands for the known weights of the atoms which are tabulated in this diagram (table, page 128), and the weights of the molecules, which the several terms represent, are found by simply adding up the weights of the several atoms of which they consist. When the substance is capable of existing in the aëriform condition, its molecular weight can be found, as I have shown, from its specific gravity; but these symbols assume that either by this or by some other method the constitution of the molecule has been determined; and, now that the result is expressed in symbols, nothing is easier than to interpret what they have to tell us. To begin with the sodic carbonate, Na2CO3. The weight of this molecule is 2× 23 +12+3 × 16 = 46+ 12+48 106 m.c. The weight of the molecule HCl is 1+35.5 36.5, and two such molecules would weigh 73 m.c. Next, for the products, we have NaCl = 23+ 35.5 58.5, and 2NaCl = 117.0, also CO2 = 12+32 44, and H2O2+16= 18. Hence the terms of our equation stand for the weights written over them below:

=

=

106

=

(Na2CO3 + 2HCl + Aq.) = (2NaCl + H2O + Aq.) + CO2.

We leave out of the account the water represented by Aq., for this, being merely the medium of the reaction, is not changed. Now we can prove our work; because, if we have added correctly, the sum of the weights of the factors must exactly equal the sum of the weights of the products and so it is 106+73= 179, and 117+18+44 = 179. Besides the information which the equation gives us in regard to the manner in which the chemical change takes place, the symbols also inform us that 106 parts by weight of sodic carbonate are acted upon by 73 parts by weight of hydro

CHEMICAL ARITHMETIC.

167

chloric acid, and that the yield is 117 parts of common salt, 18 parts of water, and 44 parts of carbonic-dioxide gas.

We learn from this, in the first place, the exact proportion in which the sodic carbonate and hydrochloric acid can be most economically used; for, if the least excess of one or the other substance over the proportions indicated is taken, that excess will be wasted. It will not enter into the chemical change, but will be left behind with the salt and water.

Assume, then, that we have 500 grammes of sodic carbonate, and we wish to know what amount of hydrochloric acid to use, we simply make the proportion as 106 73 500 x 344,3%. Again, suppose we wish to know how much common salt would be produced from these amounts of sodic carbonate and acid, we write a similar proportion

106 117 500 x 552, nearly.

So, then, in any process, after we have written the reaction as above, if the weight of any factor or product is given, we can calculate the weight of any other factor or product by this simple rule:

As the total molecular weight of the substance given is to the total molecular weight of the substance required, so is the given weight to the required weight. By total molecular weight we mean, evidently, not the weight of a single molecule, but the weight of the number of molecules which the equation indicates.

This may be called the golden rule of chemistry.

In the laboratory we never mix our materials at random, but always weigh out the exact proportions found by this rule. When one of the products is a gas, as in the present case, a simple modification of the

rule enables us to calculate the volume of the resulting gas. Suppose, for example, we wished to calculate what volume of carbonic-dioxide gas could be obtained from 500 grammes of sodic carbonate. We should first find the weight by the above rule:

106 44 500: x = 207, nearly.

The answer is 207 grammes of carbonic dioxide. To find the corresponding volume in litres, we have merely to divide this value by the weight of one litre of the gas. Now, there are tables, in which the weight of one litre of each of the common gases is given; but such tables, although convenient, are not necessary, when, as in a written reaction, we know the molecular weights of the substances with which we are dealing. You remember that the molecular weight is always twice the specific gravity with reference to hydrogen. Half the molecular weight is, then, the specific gravity with reference to hydrogen. For example, the molecular weight of carbonic dioxide (CO2) is 44, and its specific gravity with reference to hydrogen 22-in other words, a litre of carbonic dioxide weighs 22 times as much as a litre of hydrogen. Now, a litre of hydrogen, under the normal pressure of the atmosphere, and at the freezing-point of water, weighs one crith, or 0.0896 gramme, or, near enough for common purposes, 0.09 gramme. If, then, a litre of carbonic dioxide. is 22 times as heavy, its weight is 22 criths, or 22 × 0.09 1.98 gramme. Our total product, above, being 2074 grammes, the number of litres will be 207÷ 1.98, or very nearly 104 litres. A litre, as I have said, is very nearly 1 pint, but we always use these French weights and measures in the laboratory, so that the values are as significant to the chemist as are pounds

DECOMPOSITION OF CARBONIC DIOXIDE.

169

and pints to the trader. The general rule, then, is this: We first find the weight of one litre of the gas in grammes, by simply multiplying one-half of its molecular weight by 18, and then we reduce the weight of the gas in grammes to litres by dividing the weight by this product.

Let us pass, now, to another case of chemical change, and the example which I have selected is closely related to the last. One of the products of that reaction was carbonic-dioxide gas, and here we have a jar of that aëriform substance. On the other hand, I have in this bottle an elementary substance, called sodium. It belongs to the class of metals, and is one of the constituents of sodic carbonate, which we used in the former experiment. I now propose to cause these two substances to act chemically upon each other; but, as before, no chemical action will result unless the molecules have sufficient freedom of motion. Those of the carbonic-dioxide gas are already as free as the wind, moving with immense velocity through this jar. But not so with those of the sodium. In the usual solid condition of this metal, the motion of its molecules is restricted within very narrow limits. Before, we gave freedom to the molecules of sodic carbonate and hydrochloric acid by dissolving the substances in water. That method is not applicable here, for sodium acts chemically on water, and with great violence; but we can reach a similar result by melting the sodium, and heating the molten metal until it begins to volatilize. Then, on introducing the crucible containing the seething metal into the gas, the molecules of the sodium, as they are forced up by the heat, will come into contact with those of the carbonic dioxide, and a violent chemical action will be the result.