Question

Percentage composition of a organic compound is as follows $C=10:06,H=0.84,Cl=89.10$ Which of the following correspond to its molecular formula if the vapour density is $60.0$ ?

• A. $C{H}_{3}Cl$
• B. $CHC{l}_3$
• C. $C{H}_{2}C{l}_2$
• D. None of these

Answer 1

Answer: (B)

Molecular formula
=(empirical formula)${}_n$ where,
$n=\frac{\text{ molecular formula weight }}{\text{ empirical formula weight }}$
Molecular weight $=2\times$ vapour density
Vapour density $=60$

Element	$\%(a)$	Atomic weight $(b)$	$\frac{a}{b}=x$	Ratio
$C$	$10.06$	$12$	$\frac{10.06}{12}=0.838$	$1$
$H$	$0.84$	$1$	$\frac{0.84}{1}=0.84$	$1$
$Cl$	$89.10$	$35.5$	$\frac{89.10}{35.5}=2.50$	$3$

Empirical formula $=CHC{l}_3$
Empirical formula mass $=119.5$
Molecular weight $=2\times 60=120$
$n=\frac{\text{ molar mass }}{\text{ empirical formula mass }}=\frac{120}{119.5}$ $n=1$
$\therefore$ Molecular formula
$={\left(CHC{l}_3\right)}_1=CHC{l}_3$