An Elementary Study of Chemistry by William McPherson (best beach reads .txt) 📖
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When hydrochloric acid is used zinc chloride and hydrogen are the products of reaction:
When iron is used in place of zinc the equation is
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,
and the action of heated iron upon water,
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
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
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
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
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
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
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
from which we get x = 270.
In the preparation of hydrogen by the action of sulphuric acid upon zinc, according to the equation,
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
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
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
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
will show at once that they are true. The equations
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 representing water of crystallization. An examination of substances containing water of crystallization has shown that in every case the water is present in such proportion by weight as can readily be represented by a formula. For example, copper sulphate (CuSO4) and water combine in the ratio of 1 molecule of the sulphate to 5 of water; calcium sulphate (CaSO4) and water combine in the ratio 1: 2 to form gypsum. These facts are expressed by writing the formulas for the two substances with a period between them. Thus the formula for crystallized copper sulphate is CuSO4·5H2O; that of gypsum is CaSO4·2H2O.
Heat of reaction. Attention has frequently been directed to the fact that chemical changes are usually accompanied by heat changes. In general it has been found that in every chemical action heat is either absorbed or given off. By adopting a suitable unit for the measurement of heat, the heat change during a chemical reaction can be expressed in the equation for the reaction.
Heat cannot be measured by the use of a thermometer alone, since the thermometer measures the intensity of heat, not its quantity. The easiest way to measure a quantity of heat is to note how warm it will make a definite amount of a given substance chosen as a standard. Water has been chosen as the standard, and the unit of heat is called a calorie. A calorie is defined as the amount of heat required to raise the temperature of one gram of water one degree.
By means of this unit it is easy to indicate the heat changes in a given chemical reaction. The equation
means that when 2.016 g. of hydrogen combine with 16 g. of oxygen, 18.016 g. of water are formed and 68,300 cal. are set free.
means that an expenditure of 19,000 cal. is required to cause 12 g. of carbon to unite with 64.12 g. of sulphur to form 76.12 g. of carbon disulphide. In these equations it will be noted that the symbols stand for as many grams of the substance as there are units in the weights of the atoms represented by the symbols. This is always understood to be the case in equations where the heat of reaction is given.
Conditions of a chemical action are not indicated by equations. Equations do not tell the conditions under which a reaction will take place. The equation
does not tell us that it is necessary to keep the mercuric oxide at a high temperature in order that the decomposition may go on. The equation
in no way indicates the fact that the hydrochloric acid must be dissolved in water before it will act upon the zinc. From the equation
it would not be suspected that the two gases hydrogen and chlorine will unite instantly in the sunlight, but will stand mixed in the dark a long time without change. It will therefore be necessary to pay much attention to the details of the conditions under which a given reaction occurs, as well as to the expression of the reaction in the form of an equation.
EXERCISES1. Calculate the percentage composition of the following substances: (a) mercuric oxide; (b) potassium chlorate; (c) hydrochloric acid; (d) sulphuric acid. Compare the results obtained with the compositions as given in Chapters II and III.
2. Determine the percentage of copper, sulphur, oxygen, and water in copper sulphate crystals. What weight of water can be obtained from 150 g. of this substance?
3. What weight of zinc can be dissolved in 10 g. of sulphuric acid? How much zinc sulphate will be formed?
4. How many liters of hydrogen measured under standard conditions can be obtained from the action of 8 g. of iron on 10 g. of sulphuric acid? How much
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