Concept:Combustion analysis reveals the empirical formula of an organic compound by capturing all its Carbon in \( \text{CO}_2 \) and all its Hydrogen in \( \text{H}_2\text{O} \). Oxygen is found by subtraction.
Formula:$$ \%\text{C} = \left( \frac{\text{Mass of CO}_2}{\text{Mass of compound}} \right) \times \left( \frac{12}{44} \right) \times 100 $$
Solution:- Calculate %C: \( \left( \frac{1.039}{0.5439} \right) \times \left( \frac{12}{44} \right) \times 100 \approx 52.11\% \).
- Calculate %H: \( \left( \frac{0.6369}{0.5439} \right) \times \left( \frac{2.016}{18} \right) \times 100 \approx 13.11\% \).
- Calculate %O by difference: \( 100 - (52.11 + 13.11) = 34.78\% \).
- Find molar ratios (divide by atomic mass): C = \( 52.11/12 = 4.34 \); H = \( 13.11/1.008 = 13.01 \); O = \( 34.78/16 = 2.17 \).
- Divide by the smallest ratio (2.17): C = \( 4.34/2.17 = 2 \); H = \( 13.01/2.17 = 6 \); O = \( 2.17/2.17 = 1 \).
- Empirical formula is \( \text{C}_2\text{H}_6\text{O} \).
Why other options are incorrect:- \( \text{CH}_3\text{O} \) & \( \text{CH}_4\text{O} \): Result from mathematical errors in the division steps.
- Option C: Is structurally and logically invalid as an empirical formula format.
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