0.1 Atomic masses and molecular formulas  (Page 8/10)

To convince ourselves, we should repeat this process for many gaseous compounds of carbon, including ones that contain hydrogen and oxygen both. In every experiment, we find that the relative mass of carbon in each molecule is either 11.9 or a simple multiple of 11.9. It is never less than 11.9. We can conclude that the relative mass of a carbon atom on the same scale we have been using is 11.9. We can also conclude that Oxide A and Oxide B each contain one carbon atom, so the molecular formulas of Oxide A and Oxide B are CO and CO ₂ .

This procedure can be used to find molecular formulas of compounds containing other non-gaseous elements. This is the actual procedure that was used around 1850 to provide the first set of relative atomic masses and the first definite molecular formulas for common compounds. This is a major stride forward from the postulates of the Atomic Molecular Theory.

As one last step, we note that the standard agreed on by chemists for the relative atomic masses does not take hydrogen to have mass 1.00. Rather, on the agreed upon scale, the relative atomic mass of hydrogen is 1.008, that of carbon is 12.01, that of nitrogen is 14.01, and that of oxygen is 15.999. These ratios are the same as the ones we observed in our calculations.

Chemical algebra: stoichiometry

In the Introduction , we decided that we could determine molecular formulas if we knew the relative atomic masses. At that point, we didn’t have the relative atomic masses, but now we do. Once we know all the relative atomic masses, we no longer need the Law of Combining Volumes and Avogadro’s Law to determine molecular formulas.

Let’s show this by an example, with a compound which contains only carbon, oxygen, and hydrogen. An analysis from the Law of Definite Proportions gives us that the compound is 40.0% carbon, 53.3% oxygen, and 6.7% hydrogen by mass. In other words, if we have a 100.0 g sample of the compound, it consists of 40.0 g of carbon, 53.3 g of oxygen and 6.7 g of hydrogen. But we also know that the relative masses of carbon, oxygen, and hydrogen are 12.01:15.99:1.008. This will allow us to determine the relative numbers of atoms of each type in the compound.

To do this, we create a method of chemical algebra. Let’s start by assuming that we have exactly N atoms of carbon, N atoms of hydrogen, and N atoms of oxygen. N is some very large number, and it doesn’t matter what it is, as long as we have taken the same N for all three elements. The relative mass of 1 carbon atom to 1 hydrogen atom is 12.01 to 1.008. Therefore the relative mass of N carbon atoms to N hydrogen atoms is also 12.01 to 1.008. Let’s pick a very specific N: let’s make N be whatever number it is such that that a sample of N carbon atoms has mass 12.01g. Interestingly, we don’t need to know what N is – we just need to find a sample of carbon which has a mass of 12.01 g.

What is the mass of N hydrogen atoms (for the exact same N)? It must be 1.008 g, since each hydrogen atom has mass ratio to each carbon atom 1.008 to 12.01. Therefore, if we weigh out a sample of carbon with mass 12.01 g and a sample of hydrogen with mass 1.008 g, we know that we have exactly the same number of atoms of each type.