Principle of Kjeldahl's method

In Kjeldahl's method, nitrogen in the organic compound is converted into ammonia, $NH_3$.

The ammonia neutralizes sulfuric acid: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$

So, $1 \text{ mole of } H_2SO_4 \text{ neutralizes } 2 \text{ moles of } NH_3$

Moles of $H_2SO_4$ used

Given:

• Volume of $H_2SO_4 = 12.5\,\text{mL} = 0.0125\,\text{L}$
• Molarity of $H_2SO_4 = 1\,\text{M}$

Therefore, $\text{moles of } H_2SO_4 = M \times V = 1 \times 0.0125 = 0.0125$

Moles of $NH_3$ formed

From the reaction, $1 \text{ mole } H_2SO_4 \Rightarrow 2 \text{ moles } NH_3$

Hence, $\text{moles of } NH_3 = 2 \times 0.0125 = 0.025$

Moles and mass of nitrogen

Each mole of $NH_3$ contains 1 mole of nitrogen.

So, $\text{moles of nitrogen} = 0.025$

Mass of nitrogen: $m_N = 0.025 \times 14 = 0.35\,\text{g}$

Percentage of nitrogen in the compound

Mass of compound taken $= 0.55\,\text{g}$

$\%N = \frac{0.35}{0.55} \times 100$

$\%N = 63.636\ldots \approx 64$

Final Answer

The percentage of nitrogen in the compound is: $\boxed{64}$