A compound contains 69.94 percent iron and 30.06 percent oxygen. What is its molecular formula if the molar mass of the compound is 199.55 per mole?

Aug 23, 2016

$F {e}_{2} {O}_{3}$

Explanation:

As with all these problems, we ASSUME, that there is a $100 \cdot g$ mass of unknown compound.

We break this quantity down into atoms.

And, there is $\frac{30.06 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.88 \cdot m o l$ $O$.

And $\frac{69.94 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.25 \cdot m o l$ $F e$.

If we divide thru by the lowest molar amount, we get an empirical formula of $F e {O}_{1.5}$, but because the empirical formula is the simplest whole number ratio defining constituent atoms in a species, we double this result to get WHOLE numbers, i.e. $F {e}_{2} {O}_{3} , \text{ ferric oxide.}$

Note that an examiner would be quite justified at A level to say that $\text{an oxide of iron has 70% iron content}$ without quoting the oxygen percentage, and expect you to realize that the balance is oxygen.