In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is
(Given : Aqueous tension at 300 K = 15 mm Hg)
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- 15.71%
- 20.95%
- 17.46%
- 7.85%
The Correct Option is A
Approach Solution - 1
Given: Mass of organic compound = 0.4 g
Volume of nitrogen collected = 60 mL = \( 60 \times 10^{-3} \) L
Temperature (T) = 300 K Pressure of wet nitrogen = 715 mm Hg
Aqueous tension (partial pressure of water vapor) = 15 mm Hg
The pressure of dry nitrogen (P) is: \[ P = \text{Pressure of wet nitrogen} - \text{Aqueous tension} \] \[ P = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg} \] To use the ideal gas law (PV = nRT), we need to convert the pressure to atm: \[ P (\text{atm}) = \frac{700}{760} \, \text{atm} \] The ideal gas constant R = 0.0821 L atm mol\(^{-1}\) K\(^{-1}\). Now, we can calculate the number of moles (n) of nitrogen gas evolved using the ideal gas law: \[ n = \frac{PV}{RT} = \frac{\left( \frac{700}{760} \right) \times (60 \times 10^{-3})}{0.0821 \times 300} \] \[ n = \frac{42000 \times 10^{-3}}{760 \times 0.0821 \times 300} = \frac{42}{760 \times 24.63} = \frac{42}{18718.8} \approx 0.002244 \, \text{moles} \] The molar mass of nitrogen gas (N\( _2 \)) is 28 g/mol. The mass of nitrogen evolved is: \[ \text{Mass of N}_2 = n \times \text{Molar mass of N}_2 = 0.002244 \, \text{moles} \times 28 \, \text{g/mol} \approx 0.06283 \, \text{g} \] The percentage composition of nitrogen in the organic compound is: \[ % \text{ Nitrogen} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100 \] \[ % \text{ Nitrogen} = \frac{0.06283 \, \text{g}}{0.4 \, \text{g}} \times 100 \approx 15.7075 % \] Rounding to two decimal places, the percentage composition of nitrogen is 15.71%.
Approach Solution -2
To determine the percentage composition of nitrogen in the organic compound using Dumas' method, we follow these steps:
- First, calculate the pressure of nitrogen gas collected. Atmospheric pressure is given as 715 mm Hg and the aqueous tension (pressure exerted by water vapor at 300 K) is 15 mm Hg. So, the pressure due to nitrogen alone is: \(P_{\text{N}_2} = 715 \text{ mm Hg} - 15 \text{ mm Hg} = 700 \text{ mm Hg}\).
- Convert this pressure into atm for calculation. Given that 1 atm = 760 mm Hg, we have: \(P_{\text{N}_2} = \frac{700}{760} \text{ atm}\).
- The volume of nitrogen collected is given as 60 mL, which we convert to liters: \(V = 60 \times 10^{-3} \text{ L}\).
- Use the ideal gas equation \(PV = nRT\) to find the number of moles of nitrogen, \(n\). Here, the universal gas constant \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\) and the temperature \(T = 300 \text{ K}\). \(n = \frac{P \times V}{R \times T} = \frac{\left( \frac{700}{760} \right) \times 60 \times 10^{-3}}{0.0821 \times 300}\).
- Calculating the above expression: \(n \approx 0.00229 \text{ mol}\).
- The molar mass of nitrogen gas \((N_2)\) is 28 g/mol, so the mass of nitrogen is: \(\text{mass of } N_2 = n \times 28 = 0.00229 \times 28 \approx 0.06412 \text{ g}\).
- Now, calculate the percentage of nitrogen in the compound: \(\text{Percentage of Nitrogen} = \left( \frac{\text{mass of } N_2}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.06412}{0.4} \right) \times 100\).
- This calculation gives: \(\approx 15.71\%\).
Therefore, the percentage composition of nitrogen in the compound is 15.71%.
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