Question

In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _______ % (nearest integer). (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )

Hint:

Use the ideal gas law to calculate the moles of nitrogen and then determine the percentage composition.

Answer 1

Correct Answer: 20

To find the percentage composition of nitrogen in the compound using Dumas' method, follow these steps:

Step 1: Calculate the number of moles of nitrogen gas.

We use the ideal gas law: PV = nRT.

P (pressure) = 900 mm Hg - 15 mm Hg (aqueous tension) = 885 mm Hg.

Convert pressure to atm: P = 885 mm Hg × (1 atm / 760 mm Hg) ≈ 1.164 atm.

V (volume) = 150 mL = 0.150 L.

R (gas constant) = 0.0821 L·atm/(mol·K).

T (temperature) = 300 K.

Plug these values into the ideal gas equation: n = (1.164 atm × 0.150 L) / (0.0821 L·atm/(mol·K) × 300 K).

Calculate n, the number of moles of nitrogen: n ≈ 0.0071 mol.

Step 2: Calculate the mass of nitrogen.

Molar mass of nitrogen (N₂) = 28.02 g/mol.

Mass of nitrogen = 0.0071 mol × 28.02 g/mol ≈ 0.198762 g.

Step 3: Calculate the percentage composition of nitrogen.

Percentage of nitrogen = (mass of nitrogen / mass of compound) × 100%.

Percentage of nitrogen = (0.198762 g / 1 g) × 100% ≈ 19.8762%.

Thus, the percentage composition of nitrogen is approximately 20%.

Conclusion: The calculated percentage composition of nitrogen aligns with the expected range of 20%.

Answer 2

1. Calculate the partial pressure of $\mathrm{N}_{2}$: \[ \mathrm{p}_{\mathrm{N}_2} = 900 \mathrm{~mm~Hg} - 15 \mathrm{~mm~Hg} = 885 \mathrm{~mm~Hg} \]
2. Calculate the moles of $\mathrm{N}_{2}$: \[ \text{Moles of } \mathrm{N}_2 = \frac{885 \mathrm{~mm~Hg} \times 0.15 \mathrm{~L}}{0.0821 \mathrm{~L~atm/mol} \times 300 \mathrm{~K}} = 0.0071 \mathrm{~mol} \]
3. Calculate the percentage of nitrogen: \[ \text{Percentage of nitrogen} = \frac{0.0071 \mathrm{~mol} \times 28 \mathrm{~g/mol}}{1 \mathrm{~g}} \times 100 = 19.85% \approx 20% \] Therefore, the correct answer is (20).