0.2 Relative atomic masses and empirical formulae  (Page 3/5)

More explicitly, we observe that

Assuming that each liter volume contains an equal number of particles, then we can interpret this observationas

(Alternatively, there could be any fixed even number of atoms in each hydrogen molecule and in each chlorinemolecule. We will assume the simplest possibility and see if that produces any contradictions.)

This is a wonderful result, for it correctly accounts for theLaw of Combining Volumes and eliminates our concerns about creating new atoms. Most importantly, we now know themolecular formula of hydrogen chloride. We have, in effect, found a way of "counting" the atoms in the reaction bymeasuring the volume of gases which react.

This method works to tell us the molecular formula of many compounds. For example,

This requires that oxygen particles contain an even number of oxygen atoms. Now we can interpret this equation assaying that

Now that we know the molecular formula of water, we can draw a definite conclusion about the relative massesof the hydrogen and oxygen atoms. Recall from the Table that the mass ratio inwater is 8:1 oxygen to hydrogen. Since there are two hydrogen atoms for every oxygen atom in water, then the mass ratiorequires that a single oxygen atom weigh 16 times the mass of a hydrogen atom.

To determine a mass scale for atoms, we simply need to choose a standard. For example, for our purposes here, wewill say that a hydrogen atom has a mass of 1 on the atomic mass scale. Then an oxygen atom has a mass of 16 on this scale.

Our conclusions account for the apparent problems with the masses of reacting gases, specifically, thatoxygen gas weighs more than water vapor. This seemed to be nonsensical: given that water contains oxygen, it would seem thatwater should weigh more than oxygen. However, this is now simply understood: a water molecule, containing only a single oxygen atom,has a mass of 18, whereas an oxygen molecule, containing two oxygen atoms, has a mass of 32.

Determination of atomic weights for gaseous elements

Now that we can count atoms and molecules to determine molecular formulae, we need to determine relative atomicweights for all atoms. We can then use these to determine molecular formulae for any compound from the mass ratios of the elements inthe compound.

We begin by examining data on reactions involving theLaw of Combining Volumes. Going back to the nitrogen oxide data given here , we recall that there are three compounds formed from nitrogen and oxygen. Now wemeasure the volumes which combine in forming each. We find that 2 liters of oxide B can be decomposed into 1 liter of nitrogen and 1liter of oxygen. From the reasoning above, then a nitrogen particle must contain an even number of nitrogen atoms. We assume for nowthat nitrogen is $N_2$ . We have already concluded that oxygen is $O_2$ . Therefore, the molecular formula for oxide B is $NO$ , and we call it nitric oxide. Since we have already determined that the oxygen tonitrogen mass ratio is 1.14 : 1, then, if we assign oxygen a mass of 16, as above, nitrogen has a mass of 14. (That is $\frac{16}{1.14}=14$ .) 2 liters of oxide A is formed from 2 liters of oxygen and 1 literof nitrogen. Therefore, oxide A is $N{O}_2$ , which we call nitrogen dioxide. Note that we predict an oxygen to nitrogen mass ratioof $\frac{32}{14}=2.28:1$ , in agreement with the data. Oxide C is $N_{2}O$ , called nitrous oxide, and predicted to have a mass ratio of $\frac{16}{28}=0.57:1$ , again in agreement with the data. We have now resolved the ambiguity in the molecular formulae.