An Elementary Study of Chemistry. William Edwards Henderson

An Elementary Study of Chemistry - William Edwards Henderson


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parted into its constituents. The decomposition of mercuric oxide is an example: HgO = Hg + O.

      3. Substitution. It is sometimes possible for an element in the free state to act upon a compound in such a way that it takes the place of one of the elements of the compound, liberating it in turn. In the study of the element hydrogen it was pointed out that hydrogen is most conveniently prepared by the action of sulphuric or hydrochloric acid upon zinc. When sulphuric acid is used a substance called zinc sulphate, having the composition represented by the formula ZnSO4, is formed together with hydrogen. The equation is

      Zn + H2SO4 = ZnSO4 + 2H.

      When hydrochloric acid is used zinc chloride and hydrogen are the products of reaction:

      Zn + 2HCl = ZnCl2 + 2H.

      When iron is used in place of zinc the equation is

      Fe + H2SO4 = FeSO4 + 2H.

      These reactions are quite similar, as is apparent from an examination of the equations. In each case 1 atom of the metal replaces 2 atoms of hydrogen in the acid, and the hydrogen escapes as a gas. When an element in the free state, such as the zinc in the equations just given, takes the place of some one element in a compound, setting it free from chemical combination, the act is called substitution.

      Other reactions illustrating substitution are the action of sodium on water,

      Na + H2O = NaOH + H;

      and the action of heated iron upon water,

      3Fe + 4H2O = Fe3O4 + 8H.

      4. Double decomposition. When barium dioxide (BaO2) is treated with sulphuric acid two compounds are formed, namely, hydrogen dioxide (H2O2) and barium sulphate (BaSO4). The equation is

      BaO2 + H2SO4 = BaSO4 + H2O2.

      

      In this reaction it will be seen that the two elements barium and hydrogen simply exchange places. Such a reaction is called a double decomposition. We shall meet with many examples of this kind of chemical reactions.

      Chemical equations are quantitative. The use of symbols and formulas in expressing chemical changes has another great advantage. Thus, according to the equation

      H2O = 2H + O,

      1 molecule of water is decomposed into 2 atoms of hydrogen and 1 atom of oxygen. But, as we have seen, the relative weights of the atoms are known, that of hydrogen being 1.008, while that of oxygen is 16. The molecule of water, being composed of 2 atoms of hydrogen and 1 atom of oxygen, must therefore weigh relatively 2.016 + 16, or 18.016. The amount of hydrogen in this molecule must be 2.016/18.016, or 11.18% of the whole, while the amount of oxygen must be 16/18.018, or 88.82% of the whole. Now, since any definite quantity of water is simply the sum of a great many molecules of water, it is plain that the fractions representing the relative amounts of hydrogen and oxygen present in a molecule must likewise express the relative amounts of hydrogen and oxygen present in any quantity of water. Thus, for example, in 20 g. of water there are 2.016/18.016 × 20, or 2.238 g. of hydrogen, and 16/18.016 × 20, or 17.762 g. of oxygen. These results in reference to the composition of water of course agree exactly with the facts obtained by the experiments described in the chapter on water, for it is because of those experiments that the values 1.008 and 16 are given to hydrogen and oxygen respectively.

      It is often easier to make calculations of this kind in the form of a proportion rather than by fractions. Since the molecule of water and the two atoms of hydrogen which it contains have the ratio by weight of 18.016: 2.016, any mass of water has the same ratio between its total weight and the weight of the hydrogen in it. Hence, to find the number of grams (x) of hydrogen in 20 g. of water, we have the proportion

      18.016 : 2.016 :: 20 g. : x (grams of hydrogen).

      Solving for x, we get 2.238 for the number of grams of hydrogen. Similarly, to find the amount (x) of oxygen present in the 20 g. of water, we have the proportion

      18.016 : 16 :: 20 : x

      from which we find that x = 17.762 g.

      Again, suppose we wish to find what weight of oxygen can be obtained from 15 g. of mercuric oxide. The equation representing the decomposition of mercuric oxide is

      HgO = Hg + O.

      The relative weights of the mercury and oxygen atoms are respectively 200 and 16. The relative weight of the mercuric oxide molecule must therefore be the sum of these, or 216. The molecule of mercuric oxide and the atom of oxygen which it contains have the ratio 216: 16. This same ratio must therefore hold between the weight of any given quantity of mercuric oxide and that of the oxygen which it contains. Hence, to find the weight of oxygen in 15 g. of mercuric oxide, we have the proportion

      216 : 16 :: 15 : x (grams of oxygen).

      On the other hand, suppose we wish to prepare, say, 20 g. of oxygen. The problem is to find out what weight of mercuric oxide will yield 20 g. of oxygen. The following proportion evidently holds

      216 : 16 :: x (grams of mercuric oxide) : 20;

      from which we get x = 270.

      In the preparation of hydrogen by the action of sulphuric acid upon zinc, according to the equation,

      Zn + H2SO4 = ZnSO4 + 2 H,

      

      suppose that 50 g. of zinc are available; let it be required to calculate the weight of hydrogen which can be obtained. It will be seen that 1 atom of zinc will liberate 2 atoms of hydrogen. The ratio by weight of a zinc to an hydrogen atom is 65.4: 1.008; of 1 zinc atom to 2 hydrogen atoms, 65.4: 2.016. Zinc and hydrogen will be related in this reaction in this same ratio, however many atoms of zinc are concerned. Consequently in the proportion

      65.4 : 2.016 :: 50 : x,

      x will be the weight of hydrogen set free by 50 g. of zinc. The weight of zinc sulphate produced at the same time can be found from the proportion

      65.4 : 161.46 :: 50 : x;

      where 161.46 is the molecular weight of the zinc sulphate, and x the weight of zinc sulphate formed. In like manner, the weight of sulphuric acid used up can be calculated from the proportion

      65.4 : 98.076 :: 50 : x.

      These simple calculations are possible because the symbols and formulas in the equations represent the relative weights of the substances concerned in a chemical reaction. When once the relative weights of the atoms have been determined, and it has been agreed to allow the symbols to stand for these relative weights, an equation or formula making use of the symbols becomes a statement of a definite numerical fact, and calculations can be based on it.

      Chemical equations not algebraic. Although chemical equations are quantitative, it must be clearly understood that they are not algebraic. A glance at the equations

      7 + 4 = 11, 8 + 5 = 9 + 4

      will show at once that they are true. The equations

      HgO = Hg + O, FeO = Fe + O

      are equally true in an algebraic sense, but experiment shows that only the first is true chemically, for iron oxide (FeO) cannot be directly decomposed into iron and oxygen. Only such equations as have been found by careful experiment to express a real chemical transformation, true both for the kinds of substances as well as for the weights, have any value.

      Chemical formulas and equations, therefore, are a concise way of representing qualitatively and quantitatively facts which have been found by experiment to be true in reference to the composition of substances and the changes which they undergo.

      Formulas


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