An Elementary Study of Chemistry, William McPherson [icecream ebook reader txt] 📗

224]"> explained above. The results he obtained were contradictory and very far from satisfactory, and it was soon seen that some other method, resting on much more scientific grounds, must be found to decide what compounds, if any, have a single atom of each element present.

Determination of atomic weights. Three distinct steps are involved in the determination of the atomic weight of an element: (1) determination of the equivalent, (2) determination of molecular weights of its compounds, and (3) deduction of the exact atomic weight from the equivalent and molecular weights.

1. Determination of the equivalent. By the equivalent of an element is meant the weight of the element which will combine with a fixed weight of some other element chosen as a standard. It has already been explained that oxygen has been selected as the standard element for atomic weights, with a weight of 16. This same standard will serve very well as a standard for equivalents. The equivalent of an element is the weight of the element which will combine with 16 g. of oxygen. Thus 16 g. of oxygen combines with 16.03 g. of sulphur, 65.4 g. of zinc, 215.86 g. of silver, 70.9 g. of chlorine. These figures, therefore, represent the equivalent weights of these elements.

Relation of atomic weights to equivalents. According to the atomic theory combination always takes place between whole numbers of atoms. Thus one atom unites with one other, or with two or three; or two atoms may unite with three, or three with five, and so on.

When oxygen combines with zinc the combination must be between definite numbers of the two kinds of atoms. Experiment shows that these two elements combine in the ratio of 16 g. of oxygen to 65.4 g. of zinc. If one atom of oxygen combines with one atom of zinc, then this ratio must be the ratio between the weights of the two atoms. If one atom of oxygen combines with two atoms of zinc, then the ratio between the weights of the two atoms will be 16: 32.7. If two atoms of oxygen combine with one atom of zinc, the ratio by weight between the two atoms will be 8: 65.4. It is evident, therefore, that the real atomic weight of an element must be some multiple or submultiple of the equivalent; in other words, the equivalent multiplied by 1/2, 1, 2, or 3 will give the atomic weight.

Combining weights. A very interesting relation holds good between the equivalents of the various elements. We have just seen that the figures 16.03, 65.4, 215.86, and 70.9 are the equivalents respectively of sulphur, zinc, silver, and chlorine. These same figures represent the ratios by weight in which these elements combine among themselves. Thus 215.86 g. of silver combine with 70.9 g. of chlorine and with 2 × 16.03 g. of sulphur. 65.4 g. of zinc combine with 70.9 g. of chlorine and 2 × 16.03 g. of sulphur.

By taking the equivalent or some multiple of it a value can be obtained for each element which will represent its combining value, and for this reason is called its combining weight. It is important to notice that the fact that a combining weight can be obtained for each element is not a part of a theory, but is the direct result of experiment.

Elements with more than one equivalent. It will be remembered that oxygen combines with hydrogen in two ratios. In one case 16 g. of oxygen combine with 2.016 g. of hydrogen to form water; in the other 16 g. of oxygen combine with 1.008 g. of hydrogen to form hydrogen dioxide. The equivalents of hydrogen are therefore 2.016 and 1.008. Barium combines with oxygen in two proportions: in barium oxide the proportion is 16 g. of oxygen to 137.4 g. of barium; in barium dioxide the proportion is 16 g. of oxygen to 68.7 g. of barium.

In each case one equivalent is a simple multiple of the other, so the fact that there may be two equivalents does not add to the uncertainty. All we knew before was that the true atomic weight is some multiple of the equivalent.

2. The determination of molecular weights. To decide the question as to which multiple of the equivalent correctly represents the atomic weight of an element, it has been found necessary to devise a method of determining the molecular weights of compounds containing the element in question. Since the molecular weight of a compound is merely the sum of the weights of all the atoms present in it, it would seem to be impossible to determine the molecular weight of a compound without first knowing the atomic weights of the constituent atoms, and how many atoms of each element are present in the molecule. But certain facts have been discovered which suggest a way in which this can be done.

Avogadro's hypothesis. We have seen that the laws of Boyle, Charles, and Gay-Lussac apply to all gases irrespective of their chemical character. This would lead to the inference that the structure of gases must be quite simple, and that it is much the same in all gases.

In 1811 Avogadro, an Italian physicist, suggested that if we assume all gases under the same conditions of temperature and pressure to have the same number of molecules in a given volume, we shall have a probable explanation of the simplicity of the gas laws. It is difficult to prove the truth of this hypothesis by a simple experiment, but there are so many facts known which are in complete harmony with this suggestion that there is little doubt that it expresses the truth. Avogadro's hypothesis may be stated thus: Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

Avogadro's hypothesis and molecular weights. Assuming that Avogadro's hypothesis is correct, we have a very simple means for deciding upon the relative weights of molecules; for if equal volumes of two gases contain the same number of molecules, the weights of the two volumes must be in the same ratio as the weights of the individual molecules which they contain. If we adopt some one gas as a standard, we can express the weights of all other gases as compared with this one, and the same figures will express the relative weights of the molecules of which the gases are composed.

Oxygen as the standard. It is important that the same standard should be adopted for the determination of molecular weights as has been decided upon for atomic weights and equivalents, so that the three values may be in harmony with each other. Accordingly it is best to adopt oxygen as the standard element with which to compare the molecular weights of other gases, being careful to keep the oxygen atom equal to 16.

The oxygen molecule contains two atoms. One point must not be overlooked, however. We desire to have our unit, the oxygen atom, equal to 16. The method of comparing the weights of gases just suggested compares the molecules of the gases with the molecule of oxygen. Is the molecule and the atom of oxygen the same thing? This question is answered by the following considerations.

We have seen that when steam is formed by the union of oxygen and hydrogen, two volumes of hydrogen combine with one volume of oxygen to form two volumes of steam. Let us suppose that the one volume of oxygen contains 100 molecules; then the two volumes of steam must, according to Avogadro's hypothesis, contain 200 molecules. But each of these 200 molecules must contain at least one atom of oxygen, or 200 in all, and these 200 atoms came from 100 molecules of oxygen. It follows that each molecule of oxygen must contain at least two atoms of oxygen.

Evidently this reasoning merely shows that there are at least two atoms in the oxygen molecule. There may be more than that, but as there is no evidence to this effect, we assume that the molecule contains two atoms only.

It is evident that if we wish to retain the value 16 for the atom of oxygen we must take twice this value, or 32, for the value of the oxygen molecule, when using it as a standard for molecular weights.

Determination of the molecular weights of gases from their weights compared with oxygen. Assuming the molecular weight of oxygen to be 32, Avogadro's hypothesis gives us a ready means for determining the molecular weight of any other gas, for all that is required is to know its weight compared with that of an equal volume of oxygen. For example, 1 l. of chlorine is found by experiment to weigh 2.216 times as much as 1 l. of oxygen. The molecular weight of chlorine must therefore be 2.216 ×32, or 70.91.

If, instead of comparing the relative weights of 1 l. of the two gases, we select such a volume of oxygen as will weigh 32 g., or the weight in grams corresponding to the molecular weight of the gas, the calculation is much simplified. It has been found that 32 g. of oxygen, under standard conditions, measure 22.4 l. This same volume of hydrogen weighs 2.019 g.; of chlorine 70.9 g.; of hydrochloric acid 36.458 g. The weights of these equal volumes must be proportional to their molecular weights, and since the weight of the oxygen is the same as the value of its molecular weight, so too will the weights of the 22.4 l. of the other gases be equal to the value of their molecular weights.

As a summary we can then make the following statement: The molecular weight of any gas may be determined by calculating the weight of 22.4 l. of the gas, measured under standard conditions.

Determination of molecular weights from density of gases. In an actual experiment it is easier to determine the density of a gas than the weight of a definite volume of it. The density of a gas is usually defined as its weight compared with that of an equal volume of air. Having determined the density of a gas, its weight compared with oxygen may be determined by multiplying its density by the ratio between the weights of air and oxygen. This ratio is 0.9046. To compare it with our standard for atomic weights we must further multiply it by 32, since the standard is 1/32 the weight of oxygen molecules. The steps then are these:

1. Determine the density of the gas (its weight compared with air).

2. Multiply by 0.9046 to make the comparison with oxygen molecules.

3. Multiply by 32 to make the comparison with the unit for atomic weights.

We have, then, the formula: