Q. $0.30\, g$ of an organic compound containing $C, H$ and $O$ on combustion yielded $0.44\, g \,CO_2$ and $0.18\, g \,H_2O$. If $1$ mole of compound weighs $60\, g$, then molecular formula of the compound is
Solution:
$\%$ of $C$
$=\frac{12}{44} \times \frac{\text { Weight of } CO _{2} \text { formed }}{\text { Weight of organic compound }} \times 100$
$=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 $
$\%$ of $ H =\frac{2}{18} \times \frac{\text { Weight of } H _{2} O \text { formed }}{\text { Weight of organic compound }} \times 100 $
$=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.66$
$\%$ of $O =100-(\%$ of $C +\%$ of $ H )$
$=100-(40+6.66)=53.34$
For empirical formula
Element
Atomic mass
Relative number of moles
Simplest ratio
C
12
$\frac{40}{12}=3.33$
1
H
1
$\frac{6.66}{1}=6.66$
2
O
16
$\frac{53.34}{16}=3.33$
1
So, empirical formula (EF) is $CH _{2} O$.
Molecular formula (MF) $=n \times EF$
$n=\frac{\text { MF mass }}{\text { EF mass }}=\frac{60}{30}=2$
So, MF is $\left( CH _{2} O \right)_{2}= C _{2} H _{4} O _{2}$
| Element | Atomic mass | Relative number of moles | Simplest ratio |
|---|---|---|---|
| C | 12 | $\frac{40}{12}=3.33$ | 1 |
| H | 1 | $\frac{6.66}{1}=6.66$ | 2 |
| O | 16 | $\frac{53.34}{16}=3.33$ | 1 |