Question

Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Answer 1

% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]
Relative moles of iron in iron oxide:

$=\frac{ \% \text{of iron by mass} }{ \text{Atomic mass of iron}}$

$= \frac{69.9 }{ 55.85}$= 1.25

Relative moles of oxygen in iron oxide:

$= \frac{\% \text{of oxygen by mass} }{ \text{Atomic mass or oxygen}}$

$= \frac{30.1 }{ 16.00}$

= 1.88
Simplest molar ratio of iron to oxygen:
$= 1.25\ratio 1.88$
$= 1\ratio 1.5$
$≈2\ratio 3$
∴ The empirical formula of the iron oxide is $\text{Fe}_2\text{O}_3$.