Question

A camphor sample melts at $176^{\circ }C.K_f$ for camphor is $40Kkgmo{l}^{-1}$. A solution of $0.02g$ of a hydrocarbon in $0.8g$ 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?

• A. $C_6{H}_6$
• B. $C_{12}{H}_{12}$
• C. $C_4{H}_4$
• D. $C_8{H}_8$

Answer 1

Answer: (C)

Melting point of camphor $=176^{\circ }C$

$\left(K_f\right)$ for camphor $=40Kkgmo{l}^{-1}$

Mass of hydrocarbon $\left(w_B\right)=0.02g$

Mass of camphor $\left(w_A\right)=0.8g$

Temperature of camphor (at which it melts) $=156.77^{\circ }C$

Thus, $\Delta T_f$ for camphor $=176-156.77$

$=19.23^{\circ }C=19.23K$

$\therefore$ 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 $=(CH{)}_4=C_4{H}_4$