Question

Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Answer 1

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,
= $1.25 : 1.88$
= $\frac {1.25}{1.25} : \frac {1.88}{1.25}$
= $1 : 1.5$
= $2 : 3$
The empirical formula of the oxide is $Fe_2O_3$.
Empirical formula mass of $Fe_2O_3 = [2(55.85) + 3(16.00)]\ g$
Molar mass of $Fe_2O_3 = 159.69\ g$
∴ $n = \frac {\text {Molar\ mass}}{\text {Emperical\ formula\ mass }}$

$n = \frac {159.69\ g}{159.7\ g}$
$n = 0.999$
$n = 1$ (approx)
Molecular formula of a compound is obtained by multiplying the empirical formula with $n$.
Thus, the empirical formula of the given oxide is $Fe_2O_3$ and $n$ is $1$.

Hence, the molecular formula of the oxide is $Fe_2O_3$.