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Q8.33:
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M $H_2SO_4$. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Solution :

Step 1: Establishing Given Parameters

  • Mass of organic compound ($m$) = $0.50\text{ g}$
  • Volume of $H_2SO_4$ taken ($V_{total}$) = $50\text{ mL} = 0.050\text{ L}$
  • Molarity of $H_2SO_4$ ($M_{acid}$) = $0.5\text{ M}$
  • Volume of $NaOH$ used for back titration ($V_{NaOH}$) = $60\text{ mL} = 0.060\text{ L}$
  • Molarity of $NaOH$ ($M_{base}$) = $0.5\text{ M}$

Step 2: Calculating Total Moles of Acid Provided

The total millimoles of $H_2SO_4$ added is calculated by: $n = M \times V(\text{in mL})$.

$\text{Total mmol of } H_2SO_4 = 0.5 \times 50 = 25\text{ mmol}$ [Definition of Molarity].

Step 3: Determining Residual Acid Neutralized by NaOH

The reaction between $H_2SO_4$ and $NaOH$ is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.

Millimoles of $NaOH$ used = $M \times V = 0.5 \times 60 = 30\text{ mmol}$.

Per the stoichiometry of the reaction, 2 moles of $NaOH$ neutralize 1 mole of $H_2SO_4$.

$\text{Residual mmol of } H_2SO_4 = \frac{\text{mmol of } NaOH}{2} = \frac{30}{2} = 15\text{ mmol}$.

Step 4: Calculating Acid Consumed by Ammonia

The amount of acid consumed by the ammonia ($NH_3$) evolved from the sample is the difference between total acid and residual acid:

$\text{mmol of } H_2SO_4 \text{ reacted with } NH_3 = 25 - 15 = 10\text{ mmol}$.

Step 5: Stoichiometric Relation to Nitrogen

The neutralization reaction is: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.

Since 1 mole of $H_2SO_4$ reacts with 2 moles of $NH_3$, the moles of $NH_3$ (and thus nitrogen atoms) are:

$\text{mmol of } NH_3 = 2 \times (\text{mmol of } H_2SO_4 \text{ reacted}) = 2 \times 10 = 20\text{ mmol}$.

Mass of Nitrogen = $\text{moles} \times \text{atomic mass} = 20 \times 10^{-3} \text{ mol} \times 14\text{ g/mol} = 0.28\text{ g}$.

Step 6: Calculating Percentage Composition

$\text{Percentage of Nitrogen} = \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \times 100$

$\text{Percentage} = \frac{0.28}{0.50} \times 100 = 56\%$

Total H2SO4: 25 mmol Residual H2SO4: 15 mmol Reacted with NH3: 10 mmol N Content = 56%

Final Solution: The percentage composition of nitrogen in the compound is 56%.


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