Question

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:

• A. 6.72, 39.87
• B. 6.72, 53.41
• C. 7.55, 43.85
• D. 53.41, 39.6

Hint:

- For combustion analysis: \begin{itemize} \item All H converts to \( \text{H}_2\text{O} \) (2 g H per 18 g \( \text{H}_2\text{O} \)) \item All C converts to \( \text{CO}_2 \) (12 g C per 44 g \( \text{CO}_2 \)) \item Oxygen % is obtained by difference \end{itemize} - Always verify that percentages sum to 100%

Answer 1

The Correct Option is B

To find the percentages of hydrogen and oxygen in the given organic compound, we follow these steps:

1. 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.
2. 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.
3. 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: \(0.01412 \, \text{mol} \times 1 \, \text{g/mol} = 0.01412 \, \text{g}\)
4. 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: \(0.210 \, \text{g} - 0.09788 \, \text{g} = 0.11212 \, \text{g}\)
5. 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

Answer 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 \[ \% \text{O} = 100 - (\% \text{C} + \% \text{H}) = 100 - (39.87 + 6.72) = 53.41\% \]