JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 22)
On complete combustion of $$0.492 \mathrm{~g}$$ of an organic compound containing $$\mathrm{C}, \mathrm{H}$$ and $$\mathrm{O}$$, $$0.7938 \mathrm{~g}$$ of $$\mathrm{CO}_{2}$$ and $$0.4428 \mathrm{~g}$$ of $$\mathrm{H}_{2} \mathrm{O}$$ was produced. The % composition of oxygen in the compound is ___________.
Answer
46
Explanation
$\%$ of $\mathrm{H}=\frac{2}{18} \times \frac{\text { wt. of } \mathrm{H}_{2} \mathrm{O}}{\text { wt. of organic compound }} \times 100$
$$ \begin{aligned} &=\frac{2}{18} \times \frac{0.4428}{0.492} \times 100 \\\\ &=0.11 \times 0.9 \times 100 \\\\ &=0.099 \times 100=9.9 \end{aligned} $$
$\%$ of $C=\frac{12}{44} \times \frac{0.7938}{0.492} \times 100$
$$ \begin{aligned} &=0.27 \times 1.61 \times 100 \\\\ &=43.47 \end{aligned} $$
$$\% \text { Oxygen }=100-(43.47 + 9.9)$$
$=100-53.37 \simeq 46$
$$ \begin{aligned} &=\frac{2}{18} \times \frac{0.4428}{0.492} \times 100 \\\\ &=0.11 \times 0.9 \times 100 \\\\ &=0.099 \times 100=9.9 \end{aligned} $$
$\%$ of $C=\frac{12}{44} \times \frac{0.7938}{0.492} \times 100$
$$ \begin{aligned} &=0.27 \times 1.61 \times 100 \\\\ &=43.47 \end{aligned} $$
$$\% \text { Oxygen }=100-(43.47 + 9.9)$$
$=100-53.37 \simeq 46$
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