Estimation Of Nitrogen (dumas Method)
In Duma's method, 0.42 g of an organic compound gave 60 mL of nitrogen gas collected over water at 300 K and 715 mm Hg, where the aqueous tension is 15 mm Hg; what is the percentage of nitrogen?
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Solution
15%
Duma's method estimates nitrogen by heating the organic compound with copper oxide in an atmosphere of carbon dioxide, converting the nitrogen into free nitrogen gas, which is collected over a potassium hydroxide solution and measured. Because the gas is collected over water, the observed pressure must be corrected for aqueous tension: the true pressure of nitrogen is 715 - 15 = 700 mm Hg. The collected volume is then reduced to standard temperature and pressure using V(STP) = (P x V x 273) / (760 x T) = (700 x 60 x 273) / (760 x 300). The numerator is 700 x 60 x 273 = 11,466,000 and the denominator is 760 x 300 = 228,000, giving V(STP) = 50.3 mL. Since 22,400 mL of nitrogen at STP weighs 28 g, %N = (28 / 22400) x (V(STP) / mass) x 100 = (28 / 22400) x (50.3 / 0.42) x 100 = 0.00125 x 119.7 x 100 = 15.0%. The value 20% ignores the pressure and temperature correction. The value 28% wrongly uses the uncorrected volume directly. The value 12% applies an incorrect aqueous tension subtraction. This procedure follows the NCERT Duma's calculation. Plausibility check: correcting to STP lowers the volume, giving a sensible 15% nitrogen content.
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About This Question
- Subject
- chemistry
- Chapter
- purification and characterisation of organic compounds
- Topic
- estimation of nitrogen (dumas method)
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
15%
Duma's method estimates nitrogen by heating the organic compound with copper oxide in an atmosphere of carbon dioxide, converting the nitrogen into free nitrogen gas, which is collected over a potassium hydroxide solution and measured. Because the gas is collected over water, the observed pressure must be corrected for aqueous tension: the true pressure of nitrogen is 715 - 15 = 700 mm Hg. The collected volume is then reduced to standard temperature and pressure using V(STP) = (P x V x 273) / (760 x T) = (700 x 60 x 273) / (760 x 300). The numerator is 700 x 60 x 273 = 11,466,000 and the denominator is 760 x 300 = 228,000, giving V(STP) = 50.3 mL. Since 22,400 mL of nitrogen at STP weighs 28 g, %N = (28 / 22400) x (V(STP) / mass) x 100 = (28 / 22400) x (50.3 / 0.42) x 100 = 0.00125 x 119.7 x 100 = 15.0%. The value 20% ignores the pressure and temperature correction. The value 28% wrongly uses the uncorrected volume directly. The value 12% applies an incorrect aqueous tension subtraction. This procedure follows the NCERT Duma's calculation. Plausibility check: correcting to STP lowers the volume, giving a sensible 15% nitrogen content.
This hard difficulty chemistry question is from the chapter purification and characterisation of organic compounds, covering the topic of estimation of nitrogen (dumas method). It appeared in the 2025 exam.
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