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

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