Question

A compound contains 69.4% C, 5.8% H, and the rest N and O. Nitrogen is estimated by Kjeldahl method. Find empirical formula.

• A. C$_3$H$_3$NO
• B. C$_6$H$_6$N$_2$O
• C. C$_7$H$_7$NO
• D. C$_8$H$_7$N$_2$O

Hint:

Kjeldahl method directly gives nitrogen amount from NH$_3$.

Answer 1

The Correct Option is C

Step 1: Nitrogen estimation by Kjeldahl:
0.30 g sample, NH$_3$ formed neutralizes acid.

Step 2: Moles of acid initially: \[ 0.05 \times 0.05 = 0.0025\ \text{mol} \]
Step 3: Excess acid neutralized: \[ 0.025 \times 0.1 = 0.0025\ \text{mol} \]
Step 4: Total acid = 0.005 mol → corresponds to NH$_3$ produced: \[ n(N) = 0.005\ \text{mol} \]
Step 5: Mass of N: \[ = 0.005 \times 14 = 0.07\ g \]
Step 6: %N: \[ = \frac{0.07}{0.30}\times100 = 23.3% \]
Step 7: Now composition: C = 69.4%, H = 5.8%, N = 23.3%, O = remaining small (~1.5%)
Step 8: Convert to moles: \[ C \approx \frac{69.4}{12} = 5.78 \quad H = 5.8 \quad N = \frac{23.3}{14} = 1.66 \]
Step 9: Divide by 1.66: \[ C:H:N \approx 3.5:3.5:1 \]
Step 10: Multiply by 2: \[ 7:7:2 \]
Step 11: Empirical formula: \[ C_7H_7N_2O \]