Q. In a compound C, H and N are present is 9 : 1 : 3.5 by weight. If molecular weight of the compound is 108, then the molecular formula of the compound is
A) $C_{2}H_{6}N_{2}$
B) $C_{3}H_{4}N$
C) $C_{6}H_{8}N_{2}$
D) $C_{9}H_{12}N_{3}$
Explanation
(c)
[table] C | H | N ; 9
9/12=0.75
$\frac{0.75}{0.25} = 3$ | 1
1/1=1
$\frac{1}{0.25} = 4$ | 3.5
3.5/14=0.25
$\frac{0.25}{0.25} = 1$ [/table]
So, empirical formula=$C_{3}H_{4}N$
$n = \frac{108}{54} = 2$
Molecular formula=$\left( C_{3}H_{4}N \right)_{2} = C_{6}H_{8}N_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.