0.1 Atomic masses and molecular formulas  (Page 5/10)

This is a wonderful result because we have now determined the molecular formula of hydrogen chloride, HCl. We have found a way to “count” the numbers of atoms in the reaction, at least in ratio, by measuring the volumes of the gases that react and that are produced. This gives us a chemical reaction showing the atoms and molecules that participate in the reaction in the correct ratio:

1 H ₂ molecule + 1 Cl ₂ molecule → 2 HCl molecules

This chemical equation is consistent with all of our known observations and the postulates of the Atomic Molecular Theory.

Since the Law of Combining Volumes is a general result, we can look at many chemical reactions with the same analysis. Let’s apply this to the hydrogen and oxygen reaction discussed earlier. Remember that 2 L of hydrogen react with 1 L of oxygen to form 2 L of water vapor. This means that two particles of hydrogen (which we know to be H ₂ ) react with one particle of oxygen to form two particles of water. Once again, we have the problem that one atom of oxygen cannot make two molecules of water. Therefore, an oxygen gas particle cannot be an oxygen atom, so oxygen gas exists as oxygen molecules, O ₂ . Since two H ₂ molecules react with one O ₂ molecule to form two water molecules, each water molecule must be H ₂ O. We can write the chemical equation:

2 H ₂ molecules + 1 O ₂ molecule → 2 H ₂ O molecules

We can use these observations to finally solve the riddle which is posed in [link] . We need to observe the volumes of oxygen and nitrogen which react to form Oxides A, B, and C. In separate experiments, we find:

1 L N ₂ + 2 L O ₂ → 2 L Oxide A

1 L N ₂ + 1 L O ₂ → 2 L Oxide B

2 L N ₂ + 1 L O ₂ → 2 L Oxide C

(At this point, it is pretty clear from the data and using our previous reasoning that nitrogen gas must consist of nitrogen molecules, N ₂ , since 1 L of nitrogen gas can make 2 L of Oxide B.) From these data, we can conclude that Oxide B has molecular formula NO, since 1 L of oxygen plus 1 L of nitrogen produces 2 L of Oxide B with nothing left over. Similarly and with the use of [link] , we can say that Oxide A is NO ₂ and Oxide C is N ₂ O.

Observation 2: relative atomic masses

In the Introduction, we presented a dilemma in developing the Atomic Molecular Theory. To find the molecular formula of a compound, we needed to find the relative atomic masses. And to find the relative atomic masses, we needed to find the molecular formula of a compound. Using Avogadro’s Law, we have found a way to break out of this dilemma. By measuring gas volumes during reactions, we can essentially count the numbers of atoms in a molecule, giving us the molecular formula. Our task now is to use this information to find atomic masses.

We can begin by looking at the data in [link] and focusing on Oxide B at first. We know now that Oxide B has molecular formula NO. As such, it is given the name Nitric Monoxide, or more commonly Nitric Oxide. We also know from [link] that the mass ratio of oxygen to nitrogen in NO is 1.14 to 1.00. Since there are equal numbers of nitrogen atoms and oxygen atoms in any sample of NO, then the mass ratio of an oxygen atom to a nitrogen atom is also 1.14 to 1.00. Stated differently, an oxygen atom has mass 1.14 times greater than a nitrogen atom.