On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g $ \text{H}_2\text{O} $ and 0.307 g $ \text{CO}_2 $. The percentages of hydrogen and oxygen in the given organic compound respectively are:
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- 6.72, 39.87
- 6.72, 53.41
- 7.55, 43.85
- 53.41, 39.6
The Correct Option is B
Approach Solution - 1
To find the percentages of hydrogen and oxygen in the given organic compound, we follow these steps:
- Firstly, we calculate the moles of carbon in the compound from the carbon dioxide produced during combustion:
The molecular weight of \(\text{CO}_2\) is 44 g/mol.
\(\text{Moles of } \text{CO}_2 = \frac{0.307 \, \text{g}}{44 \, \text{g/mol}} = 0.00698 \, \text{mol}\)
As each mole of \(\text{CO}_2\) contains 1 mole of carbon, the moles of carbon in the sample is also 0.00698 mol. - The moles of hydrogen are calculated from the water produced:
The molecular weight of \(\text{H}_2\text{O}\) is 18 g/mol.
\(\text{Moles of } \text{H}_2\text{O} = \frac{0.127 \, \text{g}}{18 \, \text{g/mol}} = 0.00706 \, \text{mol}\)
Since each mole of water contains 2 moles of hydrogen, the moles of hydrogen in the sample is \(2 \times 0.00706 = 0.01412\) mol. - Next, we calculate the mass of carbon and hydrogen in the sample:
- Mass of Carbon: \(0.00698 \, \text{mol} \times 12 \, \text{g/mol} = 0.08376 \, \text{g}\)
- Mass of Hydrogen: \(0.01412 \, \text{mol} \times 1 \, \text{g/mol} = 0.01412 \, \text{g}\) - We then determine the combined mass of carbon and hydrogen, and subtract from the total mass of the sample to find the mass of oxygen:
- Total mass of \(C\) and \(H\): \(0.08376 \, \text{g} + 0.01412 \, \text{g} = 0.09788 \, \text{g}\)
- Mass of Oxygen: \(0.210 \, \text{g} - 0.09788 \, \text{g} = 0.11212 \, \text{g}\) - Finally, we calculate the percentages of hydrogen and oxygen in the compound:
- Percentage of Hydrogen:
\(\frac{0.01412}{0.210} \times 100\% = 6.72\%\)
- Percentage of Oxygen:
\(\frac{0.11212}{0.210} \times 100\% = 53.41\%\)
Thus, the percentages of hydrogen and oxygen in the organic compound are 6.72% and 53.41% respectively, which matches the correct option:
6.72, 53.41
Approach Solution -2
Step 1: Calculate mass of hydrogen in \( \text{H}_2\text{O} \) \[ \text{Mass of H} = \frac{2}{18} \times 0.127\, \text{g} = 0.0141\, \text{g} \] \[ % \text{H} = \left( \frac{0.0141}{0.210} \right) \times 100 = 6.72% \]
Step 2: Calculate mass of carbon in \( \text{CO}_2 \) \[ \text{Mass of C} = \frac{12}{44} \times 0.307\, \text{g} = 0.0837\, \text{g} \] \[ % \text{C} = \left( \frac{0.0837}{0.210} \right) \times 100 = 39.87% \]
Step 3: Calculate percentage of oxygen \[ \% \text{O} = 100 - (\% \text{C} + \% \text{H}) = 100 - (39.87 + 6.72) = 53.41\% \]
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