$0.25 \text{ g}$ of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $\text{CaCl}_2$ tube and potash tube at the end of the experiment. The amount was found to be $0.15 \text{ g}$ and $0.1837 \text{ g}$, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer)(Given: molar mass in $\text{g mol}^{-1}$ $\text{H} : 1, \text{C} : 12, \text{O} : 16$)

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$0.25 \text{ g}$ of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $\text{CaCl}_2$ tube and potash tube at the end of the experiment. The amount was found to be $0.15 \text{ g}$ and $0.1837 \text{ g}$, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer)
(Given: molar mass in $\text{g mol}^{-1}$ $\text{H} : 1, \text{C} : 12, \text{O} : 16$)

Combustion Analysis: Determining Oxygen Percentage

This solution explains how to find the percentage of oxygen in an organic compound using data from combustion analysis.

Step 1: Calculate Mass of Hydrogen (H)

The increase in the mass of the $\text{CaCl}_2$ tube corresponds to the mass of water ($\text{H}_2\text{O}$) produced.

Mass of $\text{H}_2\text{O}$ = $0.15 \text{ g}$
Molar mass of $\text{H}_2\text{O}$ = $(2 \times 1) + 16 = 18 \text{ g mol}^{-1}$
The mass fraction of H in $\text{H}_2\text{O}$ is $\frac{2 \times 1}{18} = \frac{2}{18}$.
Mass of H = $\left( \frac{\text{Mass of H}}{\text{Molar mass of } \text{H}_2\text{O}} \right) \times \text{Mass of } \text{H}_2\text{O}$
Mass of H = $\frac{2}{18} \times 0.15 \text{ g} = \frac{1}{9} \times 0.15 \text{ g} \approx 0.01667 \text{ g}$

Step 2: Calculate Mass of Carbon (C)

The increase in the mass of the potash tube corresponds to the mass of carbon dioxide ($\text{CO}_2$) produced.

Mass of $\text{CO}_2$ = $0.1837 \text{ g}$
Molar mass of $\text{CO}_2$ = $12 + (2 \times 16) = 44 \text{ g mol}^{-1}$
The mass fraction of C in $\text{CO}_2$ is $\frac{12}{44}$.
Mass of C = $\left( \frac{\text{Mass of C}}{\text{Molar mass of } \text{CO}_2} \right) \times \text{Mass of } \text{CO}_2$
Mass of C = $\frac{12}{44} \times 0.1837 \text{ g} = \frac{3}{11} \times 0.1837 \text{ g} \approx 0.05009 \text{ g}$

Step 3: Calculate Mass of Oxygen (O)

The organic compound contains Carbon, Hydrogen, and Oxygen. The mass of Oxygen is the difference between the total mass of the compound and the masses of Carbon and Hydrogen.

Mass of compound "A" = $0.25 \text{ g}$
Mass of O = Mass of compound - Mass of C - Mass of H
Mass of O = $0.25 \text{ g} - 0.05009 \text{ g} - 0.01667 \text{ g}$
Mass of O $\approx 0.18324 \text{ g}$

Step 4: Calculate Percentage of Oxygen

The percentage of oxygen is calculated by dividing the mass of oxygen by the total mass of the compound and multiplying by 100.

Percentage of O = $\left( \frac{\text{Mass of O}}{\text{Mass of compound}} \right) \times 100\%$
Percentage of O = $\left( \frac{0.18324 \text{ g}}{0.25 \text{ g}} \right) \times 100\%$
Percentage of O $\approx 0.73296 \times 100\% \approx 73.30\%$

Rounding to the nearest integer, the percentage of oxygen is 73%.

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