Question

A sample of $0.50g$ of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in $50mL$ of $0.5M$ $H_2{SO}_4$. the residual acid required $60mL$ of $0.5M$ solution of $NaOH$ for neutralization. Find the percentage composition of nitrogen in the compound.

Answer 1

$StepI:$To determine the volume of $H_2{SO}_4$ used, volume of acid taken =$50$ ml of $0.5$ (m) $H_2{SO}_4$
=$25$ ml of $0.1$ (m) $H_2{SO}_4$

Volume of alkali used for neutralization of excess of acid= $60$ ml of $0.5$ (m) $NaOH$

=$30$ ml of $1.0$ (m) $NaOH$

Now, $1$ mole of $H_2{SO}_4$ neutralizes 2 moles of $NaOH$

$2NaOH+H_2{SO}_4⟶{Na}_2{SO}_4+2H_{2}O$

Therefore, $30$ ml of 1(m) $NaOH$ =$15$ml of 1(m) $H_2{SO}_4$

Therefore Volume of acid used by ${NH}_3$

=$25-15=10$ml

$StepII:$ To determine the percentage of nitrogen,

In the compound 1 mole of $H_2{SO}_4$ neutralises $2$ moles of ${NH}_3$

Therefore $10$ ml of (M) $H_2{SO}_4$ = $20$ ml of 1(M)${NH}_3$

But 1000ml of 1 M ${NH}_3$ contains $14$g of nitrogen

Therefore $20$ ml of 1(M) ${NH}_3$ will contain $N=\frac{14}{1000}\times 20$
This is the amount of N present in $0.5$ g of compound
%$N=\frac{14}{1000}\times \frac{20}{0.5}\times 100=56.0$