Question

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 $B{r}_2=73.4\%$. The structural formula of the compound is

• A. $C{H}_{3}C{H}_{2}C{H}_{2}OH$
• B. $C{H}_{3}CH=CHOH$
• C. $C{H}_2=CH-C{H}_{2}OH$
• D.

Answer 1

Answer: (C)

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 $C_3{H}_{6}O$.