Kjeldahl’s method principle

In Kjeldahl’s method, the nitrogen present in the organic compound is converted to $\mathrm{NH_3}$.

The released ammonia neutralises sulfuric acid:

$2\mathrm{NH_3} + \mathrm{H_2SO_4} \rightarrow (\mathrm{NH_4})_2\mathrm{SO_4}$

Calculate moles of $\mathrm{H_2SO_4}$ used

Given:

• Volume of acid $= 10\,\mathrm{mL} = 0.010\,\mathrm{L}$
• Molarity of $\mathrm{H_2SO_4} = 2\,\mathrm{M}$

$\text{moles of } \mathrm{H_2SO_4} = M \times V = 2 \times 0.010 = 0.020$

Calculate moles of $\mathrm{NH_3}$ neutralised

From the reaction,

$1\text{ mol } \mathrm{H_2SO_4} \text{ neutralises } 2\text{ mol } \mathrm{NH_3}$

Therefore,

$\text{moles of } \mathrm{NH_3} = 2 \times 0.020 = 0.040$

Calculate mass of nitrogen present

Each mole of $\mathrm{NH_3}$ contains $1$ mole of nitrogen atoms.

So moles of nitrogen $= 0.040$.

$\text{mass of nitrogen} = 0.040 \times 14 = 0.56\,\mathrm{g}$

Calculate percentage of nitrogen

Mass of organic compound $= 1\,\mathrm{g}$

$\%\,N = \frac{0.56}{1} \times 100 = 56\%$

Final answer

$\boxed{56}$

Comparison with stored answer

Stored correct answer $= 56$

My derived answer matches the stored answer.