Question

For the estimation of nitrogen, $1.4g$ of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in $60mL$ of $M/10$ sulphuric acid. The unreacted acid required $20mL$ of $M/10$ sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is

• A. $6\%$
• B. $10\%$
• C. $3\%$
• D. $5\%$

Answer 1

Answer: (B)

(a) Write the balanced chemical reaction for the conversion of $N$ present in organic compound to ammonia, ammonia to ammonium sulphate and ammonium sulphate to sodium sulphate.
(b) Calculate millimoles ( $m$ moles) of $N$ present in organic compound followed by mass of $N$ present in organic compound using the concept of stoichiometry.
(c) At last, calculate $\%$ of $N$ present in organic compound using formula
$\%$ of $N=\frac{\text{ Mass of }N\times 100}{\text{ Mass of organic compound }}$
Mass of organic compound $=1.4g$
Let it contain $xm$ mole of $N$ atom.
Organic compound $\to \underset{xmmole}{N{H}_3}$
$2N{H}_3+\underset{6m\text{mole initially taken}}{H_{2}S{O}_4}\to (N{H}_4{)}_{2}S{O}_4.....$ (i)
$\underset{\text{2 m mole NaOH reacted}}{H_{2}S{O}_4+2NaOH}\to N{a}_{2}S{O}_4+2H_{2}O....$ (ii)
Hence, $m$ moles of $H_{2}S{O}_4$ reacted in Eq. (ii) $=1$
$\Rightarrow m$ moles of $H_{2}S{O}_4$ reacted from Eq. (i) $=6-1$ $=5m$ moles
$\Rightarrow m$ moles of $N{H}_3$ in Eq. (i) $=2\times 5=10m$ moles
$\Rightarrow m$ moles of $N$ atom in the organic compound $=10m$ moles
$\Rightarrow$ Mass of $N=10\times 10^{-3}\times 14=0.14g$
$\%$ of $N=\frac{\text{ Mass of }N\text{ present in organic compound }}{\text{ Mass of organic compound }}\times 100$
$\Rightarrow \%$ of $N=\frac{0.14}{1.4}\times 100=10\%$