# Do I have to consider the molecular mass of the oxygen atom or the diatomic oxygen molecule when determining the empirical formula of an iron oxide?

In determining the empirical formula of a compound which constitutes a diatomic molecule, should we calculate the molar mass of the divalent molecule by multiplying the molar mass of the element by 2, or should we leave it as it is. Because, I happened to come across a problem, where I had to calculate the empirical formula for an oxide of iron. A similar question has been answered before, but it doesn't seem to clear my doubt - Why not to consider hydrogen and oxygen moles to determine an empirical formula. Hence, I restate it:

If an oxide of Iron has $69.9\%$ iron and $30.1\%$ dioxygen by mass, what is the empirical formula?

Now, if I assume the mass of compound to be $100~\mathrm{g}$, then it contains $69.9~\mathrm{g}$ of iron and $30.1~\mathrm{g}$ of oxygen. If I calculate the no. of moles,

$$\text{Moles of }\ce{Fe} = \frac{\text{given mass}}{\text{molar mass}} = 69.9/58.5 = 1.25~\mathrm{mol}$$

Moles of oxygen - this is where the problem comes. If I take molar mass of oxygen as $32~\mathrm{g}$, as in $\ce{O2}$, I get $0.94~\mathrm{mol}$. It will not make any sense, as the ratio of iron to oxygen will not be in any reasonable quantity. However, if I take the molar mass to be $16~\mathrm{g}$, I get $1.88~\mathrm{mol}$ of oxygen, and I can divide both the numbers by $1.25$ to get the ratio as $1.88/1.25: 1.88/1.88 = 1:1.5 = 2:3$, hence the compound will be $\ce{Fe2O3}$, which makes perfect sense. But the problem is, why should I calculate the no. of moles from the molar mass of $\ce{O}$ rather than $\ce{O2}$? After all, its is the oxygen molecule that will exist, and no the oxygen atom right?

Also, the solution is given as $\ce{Fe2O3}$, so I need to take the molar mass of oxygen to calculate the number of moles as $16~\mathrm{g}$. The question is, Why? Why not $32~\mathrm{g}$?

Does it have something to with the fact that the question states "dioxygen" rather than "oxygen"? If yes, what is the difference between dioxygen being present in $30.1\%$ or oxygen present in $30.1\%$? I am in a real mess.

• The question is badly worded. Iron oxide does not contain "dioxygen" in its structure and–even if it did–this would be irrelevant for calculating the mass percent of oxygen. May 28 '21 at 13:58