Question

Following Kjeldahl's method, 1g of organic compound released ammonia that neutralized 10 mL of 2M H\(_2\)SO\(_4\). The percentage of nitrogen in the compound is _________.%.

Answer 1

Correct Answer: 56

Step-by-step Calculation:
Given:
Volume of \( \text{H}_2\text{SO}_4 \) used = 10 mL = 0.01 L
Molarity of \( \text{H}_2\text{SO}_4 \) = 2M
Moles of \( \text{H}_2\text{SO}_4 \) used:
\[\text{Moles of } \text{H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (in L)} = 2 \times 0.01 = 0.02 \, \text{mol}\]
Reaction between \( \text{NH}_3 \) and \( \text{H}_2\text{SO}_4 \):
\[2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4\]
From the stoichiometry of the reaction, 2 moles of \( \text{NH}_3 \) react with 1 mole of \( \text{H}_2\text{SO}_4 \). Therefore, moles of \( \text{NH}_3 \) released:
\[\text{Moles of } \text{NH}_3 = 2 \times \text{Moles of } \text{H}_2\text{SO}_4 = 2 \times 0.02 = 0.04 \, \text{mol}\]
Mass of nitrogen in \( \text{NH}_3 \):
\[\text{Mass of nitrogen} = \text{Moles of } \text{NH}_3 \times \text{Molar mass of nitrogen (14 g/mol)}\]
\[\text{Mass of nitrogen} = 0.04 \times 14 = 0.56 \, \text{g}\]
Percentage of nitrogen in the compound:
\[\text{Percentage of nitrogen} = \left( \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \right) \times 100\]
\[\text{Percentage of nitrogen} = \left( \frac{0.56}{1} \right) \times 100 = 56\%\]
Conclusion: The percentage of nitrogen in the compound is \( 56\% \).