Question

One per cent composition of an organic compound $A$ is, carbon : $85.71\%$ and hydrogen $14.29\%$. Its vapour density is $14$ . Consider the following reaction sequence

• A.
• B. $HO-C{H}_2-C{H}_2-C{O}_{2}H$
• C. $HO-C{H}_2-C{O}_{2}H$
• D. $C{H}_3-C{H}_2-C{O}_{2}H$

Answer 1

Answer: (B)

$C=85.71\%=\frac{85.71}{12}=7.14;\frac{7.14}{7.14}=1$
$H=14.29\%=\frac{14.29}{1}=14.29;\frac{14.29}{7.14}=2$
$\therefore$ Empirical formula $=C{H}_2$
and, empirical formula weight $=12+2=14$
Again, molecular formula weight
$=2\times$ vapour density
$=2\times 14=28$
$\therefore n=\frac{28}{14}=2$
$\therefore$ Molecular formula $={\left(C{H}_2\right)}_2=C_2{H}_4$