Question

In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas ($ N_2 $) at 300 K temperature and 715 mm Hg pressure. The percentage composition of 'N' in the organic compound is _____ % (Nearest integer) (Aqueous tension at 300 K = 15 mm Hg)

Hint:

In Dumas' method, remember to correct the measured gas volume for aqueous tension and convert it to STP conditions before calculating the mass of nitrogen. The percentage of nitrogen is then calculated based on the mass of nitrogen and the mass of the organic compound.

Answer 1

Correct Answer: 18

Step 1: Calculate the pressure of dry nitrogen gas.
$$P_{dry \, N_2} = P_{total} - \text{Aqueous tension} = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg}$$
Step 2: Convert the volume of \( N_2 \) to STP.
Using the combined gas law:
$$V_{STP} = \frac{P_{dry \, N_2} \times V_{measured} \times T_{STP}}{P_{STP} \times T_{measured}} = \frac{700 \, \text{mm Hg} \times 50 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mm Hg} \times 300 \, \text{K}} \approx 41.91 \, \text{mL}$$
Step 3: Calculate the mass of nitrogen gas.
Using the molar volume of \( N_2 \) at STP (22400 mL/mol):
$$\text{Moles of } N_2 = \frac{41.91 \, \text{mL}}{22400 \, \text{mL/mol}} \approx 0.001871 \, \text{mol}$$ $$\text{Mass of } N_2 = \text{moles} \times \text{molar mass} = 0.001871 \, \text{mol} \times 28 \, \text{g/mol} \approx 0.05239 \, \text{g} = 52.39 \, \text{mg}$$
Step 4: Calculate the percentage composition of nitrogen.
$$\text{Percentage of Nitrogen} = \frac{\text{Mass of } N_2}{\text{Mass of organic compound}} \times 100 = \frac{52.39 \, \text{mg}}{292 \, \text{mg}} \times 100 \approx 17.94 %$$
Rounding to the nearest integer, the percentage of nitrogen is 18%.

Answer 2

Step 1:
Given data:
\[ \text{Mass of compound} = 292 \, \text{mg} = 0.292 \, \text{g} \]
\[ \text{Volume of N}_2 = 50 \, \text{mL} = 0.05 \, \text{L} \]
\[ T = 300 \, \text{K} \]
\[ P = 715 - 15 = 700 \, \text{mm Hg} = \frac{700}{760} \, \text{atm} = 0.921 \, \text{atm} \]

Step 2:
Using the ideal gas equation:
\[ PV = nRT \]
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n(\text{N}_2) = \frac{0.921 \times 0.05}{0.0821 \times 300} = 0.00187 \, \text{mol} \]

Step 3:
Mass of nitrogen gas:
\[ \text{Mass of N}_2 = n \times M = 0.00187 \times 28 = 0.05236 \, \text{g} \]

Step 4:
Percentage of nitrogen in compound:
\[ \% \, \text{of N} = \frac{\text{Mass of N}}{\text{Mass of compound}} \times 100 \]
\[ \% \, \text{of N} = \frac{0.05236}{0.292} \times 100 = 17.94\% \approx 18\% \]

Final Answer:
\[ \boxed{18\%} \]