Determine the molecular formula of an oxide of iron in which the mass per cent of iron

Solution:

Mass percent of iron (Fe) = 69.9% (Given)

Mass percent of oxygen (O) = 30.1% (Given)

Number of moles of iron present in the oxide $=\frac{69.90}{55.85}$

= 1.25

Number of moles of oxygen present in the oxide $=\frac{30.1}{16.0}$

= 1.88

Ratio of iron to oxygen in the oxide

$=1.25: 1.88$ $=\frac{1.25}{1.25}: \frac{1.88}{1.25}$

= 1 : 1.5

= 2 : 3

$\therefore$ The empirical formula of the oxide is $\mathrm{Fe}_{2} \mathrm{O}_{3} .$

Empirical formula mass of $\mathrm{Fe}_{2} \mathrm{O}_{3}=[2(55.85)+3(16.00)]$ g Molar

mass of $\mathrm{Fe}_{2} \mathrm{O}_{3}=159.69 \mathrm{~g}$



Molecular formula of a compound is obtained by multiplying the empirical formula with $n$.

Thus, the empirical formula of the given oxide is $\mathrm{Fe}_{2} \mathrm{O}_{3}$ and $n$ is 1 .

Hence, the molecular formula of the oxide is $\mathrm{Fe}_{2} \mathrm{O}_{3} .$