Question

In the estimation of nitrogen by Kjeldahl's method 0.933 g of an organic compound 'X' was analyzed. Ammonia evolved was absorbed in 60 mL of 0.1 M H$_2$SO$_4$. The unreacted acid requires 20 mL of 0.1 M NaOH for complete neutralization. The compound 'X' is

• A. C$_6$H$_5$CH$_2$NH$_2$
• B. C$_6$H$_5$NH$_2$
• C. CH$_3$CH$_2$NH$_2$
• D. CH$_3-CO-NH_2$

Hint:

In Kjeldahl's method, the key calculation involves a back titration. 1. Find initial meq of acid. 2. Find meq of base used for back titration (= unreacted acid). 3. Subtract to find meq of acid that reacted with NH$_3$. This equals meq of N. 4. Convert meq of N to mass of N (Mass = meq $\times$ Eq. Wt. = meq $\times$ 14/1000). 5. Calculate the percentage and compare with options.

Answer 1

The Correct Option is B

Step 1: Calculate the millimoles of the initial H$_2$SO$_4$ solution.
\[ \text{mmoles of H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (mL)} = 0.1 \times 60 = 6 \text{ mmol} \] Since H$_2$SO$_4$ is diprotic, meq of acid = $2 \times 6 = 12$ meq.
Step 2: Calculate the milliequivalents of unreacted H$_2$SO$_4$.
\[ \text{meq of NaOH used} = 0.1 \times 20 = 2 \text{ meq} \] Therefore, unreacted meq of H$_2$SO$_4$ = 2 meq.
Step 3: Calculate the meq of H$_2$SO$_4$ that reacted with NH$_3$.
\[ \text{meq reacted} = 12 - 2 = 10 \text{ meq} \] Step 4: Relate this to ammonia and nitrogen.
\[ \text{mmoles of NH}_3 = 10 \text{ mmol} \implies \text{mmoles of N} = 10 \text{ mmol} \] Step 5: Calculate the mass of nitrogen.
\[ \text{Mass of N} = 10 \text{ mmol} \times 14 \text{ mg/mmol} = 140 \text{ mg} = 0.140 \text{ g} \] Step 6: Calculate the percentage of nitrogen in the sample.
\[ %N = \frac{0.140}{0.933} \times 100 \approx 15% \] Step 7: Compare with theoretical % N of options:
(A) C$_6$H$_5$CH$_2$NH$_2$: $14/107 \times 100 \approx 13.1%$
(B) C$_6$H$_5$NH$_2$: $14/93 \times 100 \approx 15.05%$
(C) CH$_3$CH$_2$NH$_2$: $14/45 \times 100 \approx 31.1%$
(D) CH$_3$CONH$_2$: $14/59 \times 100 \approx 23.7%$