Question

Find the empirical formula of iron oxide which has $69.9\%$ of Fe and $30\%$ of oxygen by mass.

Answer 1

Consider 100 gm of the oxide,

Mass of $Fe=69.9gm$

Moles of $Fe=\frac{69.6}{56}=1.248$

Mass of oxygen $=30gm$

Moles of oxygen $=\frac{30}{16}=2.5$

$\therefore$ Empirical formula $=F{e}_{1.298}{O}_{2.5}$

$=Fe{O}_2$

[dividing by 1.298]