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

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 $56\%$.