Estimation Of Carbon And Hydrogen
In the combustion method, 0.20 g of an organic compound on complete combustion produced 0.366 g of carbon dioxide; what is the percentage of carbon in the compound (C = 12, O = 16)?
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Solution
50%
Carbon and hydrogen are estimated by burning a known mass of the organic compound in a stream of oxygen so that carbon is converted into carbon dioxide and hydrogen into water, which are absorbed and weighed. The percentage of carbon is obtained from %C = (12 / 44) x (mass of CO2 / mass of compound) x 100, where 44 is the molar mass of carbon dioxide and 12 is the atomic mass of carbon, since each mole of carbon dioxide contains one mole of carbon. Substituting the data: %C = (12 / 44) x (0.366 / 0.20) x 100 = 0.2727 x 1.83 x 100 = 50%. The value 18.3% wrongly omits the 12/44 carbon fraction and uses only the mass ratio. The value 8.33% inverts part of the calculation by misplacing the molar masses. The value 27.3% reports only the 12/44 fraction as a percentage without multiplying by the mass ratio of carbon dioxide to compound. This is the NCERT Liebig combustion relation for carbon. Plausibility check: a carbon content of 50% is typical for many organic compounds, and the magnitude of carbon dioxide produced is consistent with it, confirming the result.
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About This Question
- Subject
- chemistry
- Chapter
- purification and characterisation of organic compounds
- Topic
- estimation of carbon and hydrogen
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
50%
Carbon and hydrogen are estimated by burning a known mass of the organic compound in a stream of oxygen so that carbon is converted into carbon dioxide and hydrogen into water, which are absorbed and weighed. The percentage of carbon is obtained from %C = (12 / 44) x (mass of CO2 / mass of compound) x 100, where 44 is the molar mass of carbon dioxide and 12 is the atomic mass of carbon, since each mole of carbon dioxide contains one mole of carbon. Substituting the data: %C = (12 / 44) x (0.366 / 0.20) x 100 = 0.2727 x 1.83 x 100 = 50%. The value 18.3% wrongly omits the 12/44 carbon fraction and uses only the mass ratio. The value 8.33% inverts part of the calculation by misplacing the molar masses. The value 27.3% reports only the 12/44 fraction as a percentage without multiplying by the mass ratio of carbon dioxide to compound. This is the NCERT Liebig combustion relation for carbon. Plausibility check: a carbon content of 50% is typical for many organic compounds, and the magnitude of carbon dioxide produced is consistent with it, confirming the result.
This medium difficulty chemistry question is from the chapter purification and characterisation of organic compounds, covering the topic of estimation of carbon and hydrogen. It appeared in the 2025 exam.
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