2.13 Relative atomic masses and empirical formulae (ticket #3497,  (Page 4/5)

What if nitrogen were actually $N_4$ ? Then the first oxide would be $N_{2}O$ , the second would be $N_2{O}_2$ , and the third would be $N_{4}O$ . Furthermore, the mass of a nitrogen atom would be 7. Why don't we assume this? Simply because in doing so, we willalways find that the minimum relative mass of nitrogen in any molecule is 14. Although this might be two nitrogen atoms, there isno reason to believe that it is. Therefore, a single nitrogen atom weighs 14, and nitrogen gas particles are $N_2$ .

Determination of atomic weights for non-gaseous elements

We can proceed with this type of measurement, deduction, and prediction for any compound which is a gas and whichis made up of elements which are gases. But this will not help us with the atomic masses of non-gaseous elements, nor will it permitus to determine the molecular formulae for compounds which contain these elements.

Consider carbon, an important example. There are two oxides of carbon. Oxide A has oxygen to carbon mass ratio1.33 : 1 and oxide B has mass ratio 2.66 : 1. Measurement of reacting volumes shows that we find that 1 liter of oxide A isproduced from 0.5 liters of oxygen. Hence, each molecule of oxide A contains only half as many oxygen atoms as does an oxygen molecule.Oxide A thus contains one oxygen atom. But how many carbon atoms does it contain? We can't determine this yet because theelemental carbon is solid, not gas. This means that we also cannot determine what the mass of a carbon atom is.

But we can try a different approach: we weigh 1 liter of oxide A and 1 liter of oxygen gas. The result we find isthat oxide A weighs 0.875 times per liter as much as oxygen gas. Since we have assumed that a fixed volume of gas contains a fixednumber of particles, then 1 liter of oxide A contains just as many particles as 1 liter of oxygen gas. Therefore, each particle of oxide A weighs 0.875 times as much as a particle of oxygen gas(that is, an $O_2$ molecule). Since an $O_2$ molecule weighs 32 on our atomic mass scale, then a particle of oxide A weighs $0.875\times 32=28$ . Now we know the molecular weight of oxide A.

Furthermore, we have already determined from the combining volumes that oxide A contains a single oxygen atom,of mass 16. Therefore, the mass of carbon in oxide A is 12. However, at this point, we do not know whether this is one carbonatom of mass 12, two atoms of mass 6, eight atoms of mass 1.5, or one of many other possibilities.

To make further progress, we make additional measurements on other carbon containing gas compounds. 1 liter ofoxide B of carbon is formed from 1 liter of oxygen. Therefore, each oxide B molecule contains two oxygen atoms. 1 liter of oxide Bweighs 1.375 times as much as 1 liter of oxygen. Therefore, one oxide B molecule has mass $1.375\times 32=44$ . Since there are two oxygen atoms in a molecule of oxide B, the mass of oxygen inoxide B is 32. Therefore, the mass of carbon in oxide B is 12, the same as in oxide A.

We can repeat this process for many such gaseous compounds containing carbon atoms. In each case, we findthat the mass of carbon in each molecule is either 12 or a multiple of 12. We never find, for examples, 6 or 18, which would bepossible if each carbon atom had mass 6. The simplest conclusion is that a carbon atom has mass 12. Once we know the atomic mass ofcarbon, we can conclude that the molecular formula of oxide A is $CO$ , and that of oxide B is $C{O}_2$ .