# A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.

Question:

A sample of $0.50 \mathrm{~g}$ of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in $50 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$. The residual acid required $60 \mathrm{~mL}$ of $0.5 \mathrm{M}$ solution of $\mathrm{NaOH}$ for neutralisation. Find the percentage composition of nitrogen in the compound.

Solution:

Given that, total mass of organic compound = 0.50 g

60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.

$60 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{NaOH}$ solution $=\frac{60}{2} \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=30 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$

$\therefore$ Acid consumed in absorption of evolved ammonia is $(50-30) \mathrm{mL}=20 \mathrm{~mL}$

Again, $20 \mathrm{~mL}$ of $0.5 \mathrm{MH}_{2} \mathrm{SO}_{4}=40 \mathrm{~mL}$ of $0.5 \mathrm{MNH}_{3}$

Also, since $1000 \mathrm{~mL}$ of $1 \mathrm{MNH}_{3}$ contains $14 \mathrm{~g}$ of nitrogen,

$\therefore 40 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{NH}_{3}$ will contain $\frac{14 \times 40}{1000} \times 0.5=0.28 \mathrm{~g}$ of $\mathrm{N}$

Therefore, percentage of nitrogen in $0.50 \mathrm{~g}$ of organic compound $=\frac{0.28}{0.50} \times 100=56 \%$