Question

In a compound $C,H$ and $N$ are present in $9:1:3.5$ by weight. If the 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. $C_9{H}_{12}{N}_3$

Answer 1

Answer: (C)

$C_6{H}_8{N}_2$

Find ratio of atoms, empirical formula, n and molecular formula.
Given, $C:H:N=9:1:3.5$ (ratio of weights)
$Mol.wt.=108$

Element	Ratio of wt.	Ratio of atoms	Simplest ratio
C	$9$	$9/12=0.75$	$0.75/0.25=3$
H	$1$	$1/1=1$	$1/0.25=4$
N	$3.5$	$3.5/14=0.25$	$0.25/0.25=1$

$\therefore$ Empirical formula $=C_3{H}_{4}N$
Empirical formula wt.
$=12\times 3+1\times 4+14=54$
$n=\frac{mol.wt.}{\text{empirical formula wt.}}=\frac{108}{54}=2$
Molecular formula $=$ empirical formula $\times n$
$=2\left(C_3{H}_{4}N\right)$
$=C_6{H}_8{N}_2$