Organic Chemistry Basics Result Question 15
15. For the estimation of nitrogen, $1.4 \mathrm{~g}$ of an organic compound was digested by Kjeldahl’s method and the evolved ammonia was absorbed in $60 \mathrm{~mL}$ of $M / 10$ sulphuric acid. The unreacted acid required $20 \mathrm{~mL}$ of $M / 10$ sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is
(a) $6 %$
(b) $10 %$
(c) $3 %$
(d) $5 %$
(2014 Main)
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Answer:
Correct Answer: 15. ( b )
Solution:
- This problem is based on the estimation of percentage of $\mathrm{N}$ in organic compound using Kjeldahl’s method. Use the concept of stoichiometry and follow the steps given below to solve the problem.
(a) Write the balanced chemical reaction for the conversion of $\mathrm{N}$ present in organic compound to ammonia, ammonia to ammonium sulphate and ammonium sulphate to sodium sulphate.
(b) Calculate millimoles ( $m$ moles) of $\mathrm{N}$ present in organic compound followed by mass of $\mathrm{N}$ present in organic compound using the concept of stoichiometry.
(c) At last, calculate $%$ of $\mathrm{N}$ present in organic compound using formula
$$ \begin{gathered} % \text { of } \mathrm{N}=\frac{\text { Mass of } \mathrm{N} \times 100}{\text { Mass of organic compound }} \\ \text { Mass of organic compound }=1.4 \mathrm{~g} \\ \text { Let it contain } x \mathrm{~m} \text { mole of } \mathrm{N} \text { atom. } \\ \text { Organic compound } \longrightarrow \begin{array}{c} \mathrm{NH}_3 \\ x \mathrm{~m} \text { mole } \end{array} \\ 2 \mathrm{NH}_3+\begin{array}{c} \mathrm{H}_2 \mathrm{SO}_4 \text { mole } \\ \text { initially taken } \end{array} \\ \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \\ \mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O} \end{gathered} $$
Let it contain $x \mathrm{~m}$ mole of N atom. $$ \text { Organic compound } \longrightarrow \underset{x \mathrm{~m} \text { mole }}{\mathrm{NH}_3} $$ 2 m mole NaOH reacted. $$ \begin{aligned} & \text { Hence, } \mathrm{m} \text { moles of } \mathrm{H}_2 \mathrm{SO}_4 \text { reacted in Eq. (ii) }=1 \\ & \begin{array}{c} \Rightarrow \mathrm{m} \text { moles of } \mathrm{H}_2 \mathrm{SO}_4 \text { reacted from Eq. (i) }=6-1 \\ \quad=5 \mathrm{~m} \text { moles } \end{array} \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \mathrm{m} \text { moles of } \mathrm{NH}_3 \text { in Eq. (i) }=2 \times 5=10 \mathrm{~m} \text { moles } \\ & \Rightarrow \mathrm{m} \text { moles of } \mathrm{N} \text { atom in the organic compound } \\ & \qquad \quad=10 \mathrm{~m} \text { moles } \\ & \Rightarrow \text { Mass of } \mathrm{N}=10 \times 10^{-3} \times 14=0.14 \mathrm{~g} \end{aligned} $$ $$ \begin{aligned} % \text { of } \mathrm{N} & =\frac{\begin{array}{c} \text { Mass of } \mathrm{N} \text { present in } \\ \text { organic compound } \end{array}}{\text { Mass of organic compound }} \times 100 \\ \Rightarrow \quad % \text { of } \mathrm{N} & =\frac{0.14}{1.4} \times 100 \\ & =10 % \end{aligned} $$
BETA
