1. Assume 100 g of the compound so that the percentages become masses:

• Carbon $= 54.2\text{ g}$
• Hydrogen $= 9.2\text{ g}$
• Oxygen $= 36.6\text{ g}$

2. Convert masses into moles using atomic masses $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$:

$n_\mathrm{C} = \frac{54.2}{12} = 4.5167$

$n_\mathrm{H} = \frac{9.2}{1} = 9.2$

$n_\mathrm{O} = \frac{36.6}{16} = 2.2875$

3. Find the simplest mole ratio by dividing by the smallest value $2.2875$:

$\frac{n_\mathrm{C}}{2.2875} = \frac{4.5167}{2.2875} \approx 1.975 \approx 2$

$\frac{n_\mathrm{H}}{2.2875} = \frac{9.2}{2.2875} \approx 4.02 \approx 4$

$\frac{n_\mathrm{O}}{2.2875} = 1$

So the empirical formula is:

$\mathrm{C}_2\mathrm{H}_4\mathrm{O}$

4. Calculate empirical formula mass:

$2(12) + 4(1) + 16 = 24 + 4 + 16 = 44$

5. Use the molar mass to find the molecular formula:

Given molar mass $=132\,\text{g mol}^{-1}$

$\text{Multiplier} = \frac{132}{44} = 3$

Therefore molecular formula is:

$(\mathrm{C}_2\mathrm{H}_4\mathrm{O})_3 = \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_3$

6. Check options:

• A: $\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$ has molar mass $= 88$ ❌
• B: $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ has molar mass $= 180$ ❌
• C: $\mathrm{C}_4\mathrm{H}_9\mathrm{O}_3$ has molar mass $= 105$ ❌
• D: $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_3$ has molar mass $= 132$ ✅

Hence, the correct answer is Option D.