Question

Complete combustion of 1.80 g of an oxygen containing compound (C$_x$H$_y$O$_z$) gave 2.64 g of CO$_2$ and 1.08 g of H$_2$O. The percentage of oxygen in the organic compound is :

• A. 50.33
• B. 53.33
• C. 63.53
• D. 51.63

Hint:

In combustion analysis, always find the masses of C and H first. If the sum is less than the total sample mass, the remainder is Oxygen (assuming no other elements are present).

Answer 1

The Correct Option is B

To find the percentage of oxygen in the compound C_xH_yO_z, we need to determine the mass of carbon and hydrogen in the compound first and then find the oxygen percentage by difference. Follow these steps:

1. Calculate the moles of CO₂ produced:
   n(\text{CO}_2) = \frac{\text{mass of CO}_2}{\text{molar mass of CO}_2} = \frac{2.64 \, \text{g}}{44 \, \text{g/mol}} = 0.06 \, \text{mol}
2. Calculate the mass of carbon from CO₂:
   The molar mass of carbon is 12 g/mol. Hence, the mass of carbon is calculated as follows:
   \text{mass of C} = n(\text{CO}_2) \times \text{molar mass of C} = 0.06 \, \text{mol} \times 12 \, \text{g/mol} = 0.72 \, \text{g}
3. Calculate the moles of H₂O produced:
   n(\text{H}_2\text{O}) = \frac{\text{mass of H}_2\text{O}}{\text{molar mass of H}_2\text{O}} = \frac{1.08 \, \text{g}}{18 \, \text{g/mol}} = 0.06 \, \text{mol}
4. Calculate the mass of hydrogen from H₂O:
   The molar mass of hydrogen is 1 g/mol (since 1 molecule of H₂O gives 2 moles of H atoms).
   \text{mass of H} = 2 \times n(\text{H}_2\text{O}) \times \text{molar mass of H} = 2 \times 0.06 \, \text{mol} \times 1 \, \text{g/mol} = 0.12 \, \text{g}
5. Determine the mass of oxygen in the compound:
   Given that the total mass of the compound is 1.80 g.
   \text{mass of O} = \text{total mass} - (\text{mass of C} + \text{mass of H}) = 1.80 \, \text{g} - (0.72 \, \text{g} + 0.12 \, \text{g}) = 0.96 \, \text{g}
6. Calculate the percentage of oxygen in the compound:
   \% \text{O} = \left( \frac{\text{mass of O}}{\text{total mass}} \right) \times 100 = \left( \frac{0.96 \, \text{g}}{1.80 \, \text{g}} \right) \times 100 \approx 53.33\%

Therefore, the percentage of oxygen in the organic compound is approximately 53.33%.