Q. A primary alcohol with a vapour density of $29$ contained $C =62.1 \%,\,\, H =10.3 \%$ and reacted with bromine to give a derivative which contains $C =16.5 \%,\,\,H =2.7 \%$ and $Br _{2}=73.4 \%$. The structural formula of the compound is
AIIMSAIIMS 2017
Solution:
Element
% of element
Atomic
mass element
Moles= %atomic mass
Simplest molar
Simplest whole molar ratio
$C$
$62.1 \%$
$12$
$\frac{621}{12}=5.17$
$\frac{5.17}{1.72}=300$
$3$
$H$
$10.3 \%$
$1$
$\frac{10.3}{1}=10.3$
$\frac{10.3}{1.72}=5.98$
$6$
$O$
$27.6 \%$
$16$
$\frac{27}{16}=1.72$
$\frac{1.72}{1.72}=1$
$1$
The emperical formula is $C _{3} H _{6} O$.
We know that $2 \times V \cdot D=$ molecular mass
$2 \times 29=$ molecular mass of organic primary alcohol
molecular mass $=58$
We know, $n=\frac{\text { molecular mass }}{\text { emperical formula mass }}$
$=\frac{58}{58}$
$n=1$
So, molecular formula of compound is $C _{3} H _{6} O$.
| Element | % of element | Atomic mass element | Moles= %atomic mass | Simplest molar | Simplest whole molar ratio |
|---|---|---|---|---|---|
| $C$ | $62.1 \%$ | $12$ | $\frac{621}{12}=5.17$ | $\frac{5.17}{1.72}=300$ | $3$ |
| $H$ | $10.3 \%$ | $1$ | $\frac{10.3}{1}=10.3$ | $\frac{10.3}{1.72}=5.98$ | $6$ |
| $O$ | $27.6 \%$ | $16$ | $\frac{27}{16}=1.72$ | $\frac{1.72}{1.72}=1$ | $1$ |