Question

A sample of 0.25 g of organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 25 ml of 0.25 M $H_{2}C{O}_4$. The residual acid required 60 mL of 0.25 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer 1

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

No. of mole of $NaOH=60ml\times 0.25\times {10}^{-3}$

$=15\times {10}^{-3}$

$2mol$ required $1mol{H}_{2}S{O}_4$

$15\times {10}^{-3}$ required $\frac{1}{2}\times 15\times {10}^{-3}$

$7.5\times {10}^{-3}$

Number of mole of $H_{2}S{O}_4$ consumed in

absorbing ammong $=25ml\times 0.5\times {10}^{-3}-7.5\times {10}^{-3}$

$=2.5\times {10}^{-3}-7.5\times {10}^{-9}$

$=5\times {10}^{-3}$

$2N{H}_3+H_{2}S{O}_4\to (N{H}_4{)}_{2}S{O}_4+H_{2}O$

$1mol$ $H_{2}S{O}_4$ required $2molN{H}_3$

$5\times {10}^{-3}{H}_{2}S{O}_4$ required $10\times {10}^{-3}2molN{H}_3$

Mass of $N$ in $N{H}_2=10\times {10}^{-3}\times 14$

$=0.14g$

$\%$ of nitrogen in organic compound

$=\frac{0.14}{0.25}\times 100$

$=56\%$