Question

For the estimation of nitrogen, $1.4\, g$ of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in $60 \,mL$ of $M/10$ sulphuric acid. The unreacted acid required $20\, mL$ 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 $=1.4\, g$
Let it contain $x m$ mole of $N$ atom.
Organic compound $\rightarrow \underset{x\,m\,mole}{NH_3}$
$2NH_3 + \underset{6\,m\text{mole initially taken}}{ H_2SO_4} \rightarrow (NH_4)_2SO_4.....$ (i)
$\underset{\text{2 m mole NaOH reacted}}{H_2SO_4 + 2NaOH} \rightarrow Na_2SO_4 + 2H_2O ....$ (ii)
Hence, $m$ moles of $H _{2} SO _{4}$ reacted in Eq. (ii) $=1$
$\Rightarrow m$ moles of $H _{2} SO _{4}$ reacted from Eq. (i) $=6-1$ $=5 m$ moles
$\Rightarrow m$ moles of $NH _{3}$ in Eq. (i) $=2 \times 5=10 m$ moles
$\Rightarrow m$ moles of $N$ atom in the organic compound $=10 m$ moles
$\Rightarrow$ Mass of $N =10 \times 10^{-3} \times 14=0.14 g$
$\%$ of $N =\frac{\text { Mass of } N \text { present in organic compound }}{\text { Mass of organic compound }} \times 100$
$\Rightarrow \%$ of $N =\frac{0.14}{1.4} \times 100=10 \%$