Q. What is empirical formula of an oxide of iron which has $69.9 \%$ iron and $30 \%$ oxygen by mass?
Solution:
$\%$ of
mass
Atomic
mass
Relative no.of
moles
Simple
ratio
Simple whole no
Iron
$Fe$
$69.9$
$55.85$
$69.9 / 55.85=1.25$
$\frac{1.25}{1.25}=1$
$2$
Oxygen
$O$
$30.1$
$16.0$
$30.1 / 16=1.88$
$\frac{1.88}{1.25}=1.5$
$3$
Formula $= Fe _{2} O _{3}$
| $\%$ of mass | Atomic mass | Relative no.of moles | Simple ratio | Simple whole no | ||
|---|---|---|---|---|---|---|
| Iron | $Fe$ | $69.9$ | $55.85$ | $69.9 / 55.85=1.25$ | $\frac{1.25}{1.25}=1$ | $2$ |
| Oxygen | $O$ | $30.1$ | $16.0$ | $30.1 / 16=1.88$ | $\frac{1.88}{1.25}=1.5$ | $3$ |