Q. 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 $60.0$ ?
AMUAMU 2001
Solution:
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 $= CHCl _{3}$
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
$=\left( CHCl _{3}\right)_{1}= CHCl _{3}$
| Element | $\%(a)$ | Atomic weight $(b)$ | $\frac{a}{b} =x$ | Ratio |
|---|---|---|---|---|
| $C$ | $10.06$ | $12$ | $\frac{10.06}{12}=0.838$ | $1$ |
| $H$ | $0.84$ | $1$ | $\frac{0.84}{1}=0.84$ | $1$ |
| $Cl$ | $89.10$ | $35.5$ | $\frac{89.10}{35.5}=2.50$ | $3$ |