Q. Two organic compounds (A) and (B) both containing only carbon and hydrogen, on quantitative analysis gave the same percentage composition by weight $C = \left( \frac{12}{12} \right) \times 100\%,\ \ H = \left( \frac{1}{13} \right) \times 100\%$ A decolourises bromine water but B does not. A and B respectively are
A) $C_{2}H_{2}\ and\ C_{6}H_{6}$
B) $C_{6}H_{6}\ and\ C_{2}H_{2}$
C) $C_{2}H_{4}\ and\ C_{2}H_{6}$
D) $C_{2}H_{2}\ and\ C_{2}H_{6}$
Explanation
(a)
Given, $C = \left( \frac{12}{13} \right) \times 100\%,\ \ H = \left( \frac{1}{13} \right) \times 100\%$
$\therefore$ $C = 92.3\%\ H = 7.69\%$
$C = \frac{92.3}{12} = 7.69 = \frac{7.69}{7.69} = 1$
$H = \frac{7.69}{1} = 7.69 = \frac{7.69}{7.69} = 1$
$\therefore$ Empirical formula of hydrocarbon is $C_{1}H_{1} = CH$
$\therefore$ A has empirical formula $CH$ and decolourises bromine water.
$\therefore$ It is alkyne which is $C_{2}H_{2}$.
$\therefore$ B has empirical formula $CH$ and does not decolourise bromine water.
$\therefore$ It is benzene $C_{6}H_{6}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.