CHAPTER VII

THE LAW OF DEFINITE PROPORTIONS

IF Dalton's atomic theory be correct, every compound must contain its elements in certain definite proportions by weight. Thus a water molecule, made up of two atoms of hydrogen, weighing 2 (or more precisely 2.016), and one atom of oxygen weighing 16, has evidently 8 times as much of oxygen by weight as of hydrogen. And since a cupful (or a barrelful) of water is made up of individual molecules, every one of which contains 8 times as much oxygen as hydrogen, the cupful (or barrelful) of water if decomposed would yield 8 times the weight of oxygen as of hydrogen. Experiment shows that this is true no matter what the quantity of water used. The same principle applies to every compound.

A chemical compound always contains the same elements in the same proportions by weight. This is known as the Law of Definite Proportions.

This principle of Definite Proportions is not confined to reactions of combination. It governs chemical reactions of all kinds.

The molecule of calcium carbonate, CaCO3, being made up of one atom of calcium, weighing 40, one atom of carbon, weighing 12, and three atoms of oxygen, each weighing 16, has a total weight of 40+ 12 +48 100. Similarly, the molecular weight of calcium oxide, CaO, is 56 (40 + 16) and that of carbon dioxide, CO2, is 44 (12 + twice 16). According, then, to the equation

100 parts of calcium carbonate decompose into 56 parts of calcium oxide and 44 parts of carbon dioxide. And this is

true, whatever the units in which we express the weight. Thus 100 grains of calcium carbonate give 56 grains of calcium oxide and 44 grains of carbon dioxide; 100 ounces of calcium carbonate give 56 ounces of calcium oxide and 44 ounces of carbon dioxide; and so on for pounds, tons, grams, kilograms, and all other units of weight.

So also 170 parts of silver nitrate by weight react with 58.5 parts of salt (sodium chloride), no more and no less:

Every pound of silver nitrate takes 58.5 170 = 0.344 pound of salt, and every ton of silver nitrate 0.344 ton of salt. If more salt is used, the excess will be left unchanged. If, for instance, one pound of silver nitrate and one pound of salt are brought together in solution, 0.344 pound of the salt will be used up in precipitating the silver nitrate and the remaining 0.656 pound of salt will be left in the water unchanged.

This principle is of some importance in the household applications of chemistry. Baking powders, for instance, should be so made as to contain exactly the right quantity of soda to react with all the cream of tartar. If too much of either constituent is used, the excess will remain in the cake or biscuit, making it sour or bitter as the case may be.

EXERCISES

1. The reaction between baking soda and cream of tartar is represented by the equation:

Using the atomic weights of the table on the inside of the back cover of the book, calculate how many pounds of pure cream of tartar are required for every pound of baking soda.

2. How many ounces of carbon dioxide are obtainable from a pound (16 ounces) of baking soda by the reaction with cream of tartar?

3. 100 pounds of cream of made up into baking powder. soda should be used?

tartar 90 per cent pure are to be How many pounds of pure baking

4. When baking soda is heated, washing soda (Na2CO3) is formed according to the following equation:

This equation signifies that two molecules of sodium bicarbonate are converted into one each of sodium carbonate, water, and carbon dioxide.

How many pounds of washing soda would 100 pounds of baking soda make?

5. 100 grams each of baking soda and pure cream of tartar are mixed and used as a baking powder in making a batch of biscuits. How much baking soda will be left over when the cream of tartar has all been acted upon?

6. Assuming that the baking soda left over in Example 5 is all converted into washing soda in the baking, how many grams of washing soda will the batch of biscuits contain?

Experiment 23.

Materials:

Potassium bitartrate.

Sodium bicarbonate.

Litmus solution.

25 cc. graduated cylinders.

Dissolve 0.84 gram Divide each solution

Dissolve 1.88 grams potassium bitartrate in 50 cc. of hot water in a beaker. Keep the liquid hot, but not boiling, and stir with a glass rod until all the solid is dissolved. sodium bicarbonate in 20 cc. of cold water. into two exactly equal portions. To one portion of each add enough litmus solution to color it distinctly. Observe the color the litmus takes in the two liquids. Put the colored bicarbonate

solution into a beaker or dish and gradually pour the colored bitartrate solution (still hot) into it. What do you observe? Hold a film of limewater (calcium hydroxide) in the escaping gas.

What is this gas? If the color of the bicarbonate solution is not changed, add some of the second portion of bitartrate solution little by little, until the color does change. Then add the uncolored portion of bicarbonate solution, and finally the remainder of the bitartrate solution. If necessary, dissolve a little more bitartrate and add it. If the materials are perfectly pure (which they often are not) and the weighings exact, the color changes will occur when the half portion of the two solutions are mixed and again when the whole of the bitartrate solution has been added.