Question

On combustion, \(0.210\,\text{g}\) of an organic compound containing C, H and O gave \(0.127\,\text{g}\) of \( \mathrm{H_2O} \) and \(0.307\,\text{g}\) of \( \mathrm{CO_2} \). The percentages of hydrogen and oxygen in the given organic compound respectively are:

• A. \(53.41,\; 39.6\)
• B. \(6.72,\; 53.41\)
• C. \(7.55,\; 43.85\)
• D. \(6.72,\; 39.87\)

Hint:

In combustion analysis:
Use \( \mathrm{CO_2} \) to find carbon.
Use \( \mathrm{H_2O} \) to find hydrogen.
Oxygen is always calculated by mass difference.

Answer 1

The Correct Option is B

Concept:

From the mass of \( \mathrm{CO_2} \), carbon content is calculated.
From the mass of \( \mathrm{H_2O} \), hydrogen content is calculated.
Oxygen content is obtained by difference.
Step 1: Calculate mass of hydrogen. \[ \text{Mass of H} = \frac{2}{18} \times 0.127 = 0.01411\,\text{g} \]
Step 2: Calculate mass of carbon. \[ \text{Mass of C} = \frac{12}{44} \times 0.307 = 0.08373\,\text{g} \]
Step 3: Calculate mass of oxygen in the compound. \[ \text{Mass of O} = 0.210 - (0.08373 + 0.01411) \] \[ \text{Mass of O} = 0.11216\,\text{g} \]
Step 4: Calculate percentage of hydrogen. \[ \%H = \frac{0.01411}{0.210} \times 100 = 6.72\% \]
Step 5: Calculate percentage of oxygen. \[ \%O = \frac{0.11216}{0.210} \times 100 = 53.41\% \]