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 _________.%.
Correct Answer: 56
Approach Solution - 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 \( 56\% \).
Approach Solution -2
Given:
The chemical reaction is:
\[ H_2SO_4 + 2 NH_3 \rightarrow (NH_4)_2SO_4 \]
Step 1: The millimoles of \( H_2SO_4 \) are given as:
\[ \text{Millimole of } H_2SO_4 = 10 \times 2 = 20 \, \text{millimoles} \]
Step 2: From the reaction, the millimoles of \( NH_3 \) required are twice that of \( H_2SO_4 \), so:
\[ \text{Millimole of } NH_3 = 20 \times 2 = 40 \, \text{millimoles} \]
Step 3: Now, for the organic compound (which is ammonia, \( NH_3 \)) with 40 millimoles:
\[ \text{Mole of } N = \frac{40}{1000} \, \text{moles} \]
Step 4: The molecular weight of nitrogen (\(N\)) is 14, so the weight of nitrogen in the compound is:
\[ \text{w.t. of N} = \frac{40}{1000} \times 14 = 0.56 \, \text{g} \]
Step 5: The percentage composition of nitrogen in the organic compound is:
\[ \% \text{composition of N} = \frac{40 \times 14}{1000 \times 1} \times 100 = 56\% \]
Final Answer:
The percentage composition of nitrogen in the organic compound is \( \boxed{56\%} \).
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