MOLECULAR WEIGHTS, ATOMIC WEIGHTS, FORMULAS

Introduction. In the chapter on The Atomic Theory, it was shown that if it were true that two elements uniting to form a compound always combined in the ratio of one atom of one element to one atom of the other element, it would be a very easy matter to decide upon figures which would represent the relative weights of the different atoms. It would only be necessary to select some one element as a standard and determine the weight of every element which combines with a definite weight (say 1 g.) of the standard element. The figures so obtained would evidently represent the relative weights of the atoms.

But the law of multiple proportion at once reminds us that two elements may unite in several proportions; and there is no simple way to determine the number of atoms present in the molecule of any compound. Consequently the problem of deciding upon the relative atomic weights is not an easy one. To the solution of this problem we must now turn.

Dalton's method of determining atomic weights. When Dalton first advanced the atomic theory he attempted to solve this problem by very simple methods. He thought that when only one compound of two elements is known it is reasonable to suppose that it contains one atom of each element. He therefore gave the formula HO to water, and HN to ammonia. When more than two compounds were known he assumed that the most familiar or the most stable one had the simple formula. He then determined the atomic weight as 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:

molecular weight = density × 0.9046 × 32;

or, still more briefly,

M. = D. × 28.9.

The value found by this method for the determination of molecular weights will of course agree with those found by calculating the weight of 22.4 l. of the gas, since both methods depend on the same principles.

Fig. 69

Determination of densities of gases. The relative weights of equal volumes of two gases can be easily determined. The following is one of the methods used. A small flask, such as is shown in Fig. 69, is filled with one of the gases, and after the temperature and pressure have been noted the flask is sealed up and weighed. The tip of the sealed end is then broken off, the flask filled with the second gas, and its weight determined. If the weight of the empty flask is subtracted from these two weighings, the relative weights of the gases is readily found.

3. Deduction of atomic weights from molecular weights and equivalents. We have now seen how the equivalent of an element and the molecular weight of compounds containing the element can be obtained. Let us see how it is possible to decide which multiple of the equivalent really is the true atomic weight. As an example, let us suppose that the equivalent of nitrogen has been found to be 7.02 and that it is desired to obtain its atomic weight. The next step is to obtain the molecular weights of a large number of compounds containing nitrogen. The following will serve:

	DENSITY BY EXPERIMENT	APPROXIMATE MOLECULAR WEIGHT (D. × 28.9)	PERCENTAGE OF NITROGEN BY EXPERIMENT	PART OF MOLECULAR WEIGHT DUE TO NITROGEN
Nitrogen gas	0.9671	27.95	100.00	27.95
Nitrous oxide	1.527	44.13	63.70	27.11
Nitric oxide	1.0384	30.00	46.74	14.02
Nitrogen peroxide	1.580	45.66	30.49	13.90
Ammonia	0.591	17.05	82.28	14.03
Nitric acid	2.180	63.06	22.27	14.03
Hydrocyanic acid	0.930	26.87	51.90	13.94

Method of calculation. The densities of the various gases in the first column of this table are determined by experiment, and are fairly accurate but not entirely so. By multiplying these densities by 28.9 the molecular weights of the compounds as given in the second column are obtained. By chemical analysis it is possible to determine the percentage composition of these substances, and the percentages of nitrogen in them as determined by analysis are given in the third column. If each of these molecular weights is multiplied in turn by the percentage of nitrogen in the compound, the product will be the weight of the nitrogen in the molecular weight of the compound. This will be the sum of the weights of the nitrogen atoms in the molecule. These values are given in the fourth column in the table.

If a large number of compounds containing nitrogen are studied in this way, it is probable that there will be included in the list at least one substance whose molecule contains a single nitrogen atom. In this case the number in the fourth column will be the approximate atomic weight of nitrogen. On comparing the values for nitrogen in the table it will be seen that a number which is approximately 14 is the smallest, and that the others are multiples of this. These compounds of higher value, therefore, contain more than one nitrogen atom in the molecule.

Accurate determination of atomic weights. Molecular weights cannot be determined very accurately, and consequently the part in them due to nitrogen is a little uncertain, as will be seen in the table. All we can tell by this method is that the true weight is very near 14. The equivalent can however be determined very accurately, and we have seen that it is some multiple or submultiple of the true atomic weight. Since molecular-weight determinations have shown that in the case of nitrogen the atomic weight is near 14, and we have found the equivalent to be 7.02, it is evident that the true atomic weight is twice the equivalent, or 7.02 × 2 = 14.04.

Summary. These, then, are the steps necessary to establish the atomic weight of an element.

1. Determine the equivalent accurately by analysis.

2. Determine the molecular weight of a large number of compounds of the element, and by analysis the part of the molecular weight due to the element. The smallest number so obtained will be approximately the atomic weight.

3. Multiply the equivalent by the small whole number (usually 1, 2, or 3), which will make a number very close to the approximate atomic weight. The figure so obtained will be the true atomic weight.

Molecular weights of the elements. It will be noticed that the molecular weight of nitrogen obtained by multiplying its density by 28.9 is 28.08. Yet the atomic weight of nitrogen as deduced from a study of its gaseous compounds is 14.04. The simplest explanation that can be given for this is that the gaseous nitrogen is made up of molecules, each of which contains two atoms. In this respect it resembles oxygen; for we have seen that an entirely different line of reasoning leads us to believe that the molecule of oxygen contains two atoms. When we wish to indicate molecules of these gases the symbols N₂ and O₂ should be used. When we desire to merely show the weights taking part in a reaction this is not necessary.

The vapor densities of many of the elements show that, like oxygen and nitrogen, their molecules consist of two atoms. In other cases, particularly among the metals, the molecule and the atom are identical. Still other elements have four atoms in their molecules.

While oxygen contains two atoms in its molecules, a study of ozone has led to the conclusion that it has three. The formation of ozone from oxygen can therefore be represented by the equation

3O₂ = 2O₃.

Other methods of determining molecular weights. It will be noticed that Avogadro's law gives us a method by which we can determine the relative weights of the molecules of two gases because it enables us to tell when we are dealing with an equal number of the two kinds of molecules. If by any other means we can get this information, we can make use of the knowledge so gained to determine the molecular weights of the two substances.

Raoult's laws. Two laws have been discovered which give us just such information. They are known as Raoult's laws, and can be stated as follows:

1. When weights of substances which are proportional to their molecular weights are dissolved in the same weight of solvent, the rise of the boiling point is the same in each case.

2. When weights of substances which are proportional to their molecular weights are dissolved in the same weight of solvent, the lowering of the freezing point is the same in each case.

By taking advantage of these laws it is possible to determine when two solutions contain the same number of molecules of two dissolved substances, and consequently the relative molecular weights of the two substances.

Law of Dulong and Petit. In 1819 Dulong and Petit discovered a very interesting relation between the atomic weight of an element and its specific heat, which holds true for elements in the solid state. If equal weights of two solids, say, lead and silver, are heated through the same range of temperature, as from 10° to 20°, it is found that very different amounts of heat are required. The amount of heat required to change the temperature of a solid or a liquid by a definite amount compared with the amount required to change the temperature of an equal weight of water by the same amount is called its specific heat. Dulong and Petit discovered the following law: The specific heat of an element in the solid form multiplied by its atomic weight is approximately equal to the constant 6.25. That is,

at. wt. × sp. ht. = 6.25.

Consequently,

		6.25
at. wt.	=	————
		sp. ht.

This law is not very accurate, but it is often possible by means of it to decide upon what multiple of the equivalent is the real atomic weight. Thus the specific heat of iron is found by experiment to be 0.112, and its equivalent is 27.95. 6.25 ÷ 0.112 = 55.8. We see, therefore, that the atomic weight is twice the equivalent, or 55.9.

How formulas are determined. It will be well in connection with molecular weights to consider how the formula of a compound is decided upon, for the two subjects are very closely associated. Some examples will make clear the method followed.

The molecular weight of a substance containing hydrogen and chlorine was 36.4. By analysis 36.4 parts of the substance was found to contain 1 part of hydrogen and 35.4 parts of chlorine. As these are the simple atomic weights of the two elements, the formula of the compound must be HCl.

A substance consisting of oxygen and hydrogen was found to have a molecular weight of 34. Analysis showed that in 34 parts of the substance there were 2 parts of hydrogen and 32 parts of oxygen. Dividing these figures by the atomic weights of the two elements, we get 2 ÷ 1 = 2 for H; 32 ÷ 16 = 2 for O. The formula is therefore H₂O₂.

A substance containing 2.04% H, 32.6% S, and 65.3% O was found to have a molecular weight of 98. In these 98 parts of the substance there are 98 × 2.04% = 2 parts of H, 98 × 32.6% = 32 parts of S, and 98 × 65.3% = 64 parts of O. If the molecule weighs 98, the hydrogen atoms present must together weigh 2, the sulphur atoms 32, and the oxygen atoms 64. Dividing these figures by the respective atomic weights of the three elements, we have, for H, 2 ÷ 1 = 2 atoms; for S, 32 ÷ 32 = 1 atom; for O, 64 ÷ 16 = 4 atoms. Hence the formula is H₂SO₄.

We have, then, this general procedure: Find the percentage composition of the substance and also its molecular weight. Multiply the molecular weight successively by the percentage of each element present, to find the amount of the element in the molecular weight of the compound. The figures so obtained will be the respective parts of the molecular weight due to the several atoms. Divide by the atomic weights of the respective elements, and the quotient will be the number of atoms present.

Avogadro's hypothesis and chemical calculations. This law simplifies many chemical calculations.

1. Application to volume relations in gaseous reactions. Since equal volumes of gases contain an equal number of molecules, it follows that when an equal number of gaseous molecules of two or more gases take part in a reaction, the reaction will involve equal volumes of the gases. In the equation

C₂H₂O₄ = H₂O + CO₂ + CO,

since 1 molecule of each of the gases CO₂ and CO is set free from each molecule of oxalic acid, the two substances must always be set free in equal volumes.

Acetylene burns in accordance with the equation

2C₂H₂ + 5O₂ = 4CO₂ + 2H₂O.

Hence 2 volumes of acetylene will react with 5 volumes of oxygen to form 4 volumes of carbon dioxide and 2 volumes of steam. That the volume relations may be correct a gaseous element must be given its molecular formula. Thus oxygen must be written O₂ and not 2O.

2. Application to weights of gases. It will be recalled that the molecular weight of a gas is determined by ascertaining the weight of 22.4 l. of the gas. This weight in grams is called the gram-molecular weight of a gas. If the molecular weight of any gas is known, the weight of a liter of the gas under standard conditions may be determined by dividing its gram-molecular weight by 22.4. Thus the gram-molecular weight of a hydrochloric acid gas is 36.458. A liter of the gas will therefore weigh 36.458 ÷ 22.4 = 1.627 g.