Question

While estimating the nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.25g of the compound neutralized 2.5mL of 2M H₂SO₄. The percentage of nitrogen present in organic compound is ________

Hint:

Use the equivalents method: normality of the acid equals its molarity times basicity, and at the neutralization point, milliequivalents of acid always equal milliequivalents of the base. From there, convert moles of nitrogen straight into mass.

Answer 1

Correct Answer: 56

 NH₃ gas is neutralized by 2⋅5mL of 2M H₂SO₄

∴ Moles of NH₃ neutralized = \(2⋅5 × 2 × 2\) millimole
= \(10×10^{–3}\) moles

∴ Weight of N present in compound will be
= \(10 × 10^{–3}\times14\)
= 0⋅14g

% of ‘N’ in compound
=\(\frac{0.14}{0.25}\)×100=56%

Answer 2

Concept:

• In a neutralization reaction, the number of equivalents of acid used always equals the number of equivalents of base used, regardless of how many moles of each are actually present.
• Normality of an acid equals its molarity multiplied by its basicity (the number of replaceable $H^+$ ions per formula unit).
• Since $NH_3$ accepts only one proton, its equivalents are the same as its moles.

Step 1: Find the normality of the $H_2SO_4$ used.
$H_2SO_4$ is diprotic, so its basicity is 2.
$N = M \times \text{basicity} = 2 \times 2 = 4\ N$

Step 2: Find the milliequivalents of $H_2SO_4$ used.
$\text{Milliequivalents} = N \times V(\text{mL}) = 4 \times 2.5 = 10$

Step 3: Equate equivalents of acid and base at neutralization.
Milliequivalents of $NH_3$ neutralized $=$ Milliequivalents of $H_2SO_4$ used $= 10$
Since $NH_3$ has a basicity of 1, its millimoles also equal 10.

Step 4: Convert to mass of nitrogen.
Moles of $N$ = Moles of $NH_3$ = $10 \times 10^{-3}$ mol
Mass of $N$ = $10\times10^{-3} \times 14 = 0.14$ g

Step 5: Calculate the percentage of nitrogen.
$\%N = \dfrac{0.14}{0.25}\times 100 = 56\%$

Final Answer: $56\%$