Q. A camphor sample melts at $176^{\circ} C . K_{f}$ for camphor is $40\, K\, kg\, mol ^{-1}$. A solution of $0.02 \,g$ of a hydrocarbon in $0.8 \,g$ of camphor melts at $156.77^{\circ} C$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?
TS EAMCET 2019
Solution:
Melting point of camphor $=176^{\circ} C$
$\left(K_{f}\right)$ for camphor $=40 \,K \,kg \, mol ^{-1}$
Mass of hydrocarbon $\left(w_{B}\right)=0.02 \,g$
Mass of camphor $\left(w_{A}\right)=0.8 \, g$
Temperature of camphor (at which it melts) $=156.77^{\circ} C$
Thus, $\Delta T_{f}$ for camphor $=176-156.77$
$=19.23^{\circ} C =19.23 \, K$
$\therefore $ Depression in freezing point,
$ \Delta T_{f} =\frac{K_{f} \times w_{B}}{M_{B}} \times \frac{1000}{W_{A}} $
$19.23 =\frac{40 \times 0.02 \times 1000}{M_{B} \times 0.8} $
Molar mass of solute $\left(M_{B}\right)$
$=\frac{40 \times 0.02 \times 1000}{19.23 \times 0.8}=52$
Element $\%$
$\frac{\%}{\text { atomic mass }}=$ mol
Mol ratio
C
$92.3 \% 7.7$
$7.7 / 7.7=1$
H
$7.7 \% 7.7$
$7.7 / 7.7=1$
$C : H = 1 : 1 $
$\therefore $ Empirical formula $= CH$
Empirical formula weight $=12+1=13$
Multiple $(n)=\frac{\text { Molecular weight }}{\text { Empirical formula weight }}=\frac{52}{13}=4$
Thus, molecular formula $=( CH )_{4}= C _{4} H _{4}$
| Element $\%$ | $\frac{\%}{\text { atomic mass }}=$ mol | Mol ratio |
|---|---|---|
| C | $92.3 \% 7.7$ | $7.7 / 7.7=1$ |
| H | $7.7 \% 7.7$ | $7.7 / 7.7=1$ |
| $C : H = 1 : 1 $ | ||