1. Find percentage of oxygen

Given:

• Carbon $= 42.1\%$
• Hydrogen $= 6.4\%$
• Oxygen $= 100 - 42.1 - 6.4 = 51.5\%$

So the composition is: $C = 42.1\%, \quad H = 6.4\%, \quad O = 51.5\%$

2. Convert percentage into moles

Assume $100\,\text{g}$ of compound. Then:

• Mass of C $= 42.1\,\text{g}$
• Mass of H $= 6.4\,\text{g}$
• Mass of O $= 51.5\,\text{g}$

Moles: $\text{Moles of C} = \frac{42.1}{12} \approx 3.51$ $\text{Moles of H} = \frac{6.4}{1} = 6.4$ $\text{Moles of O} = \frac{51.5}{16} \approx 3.22$

3. Find simplest whole number ratio

Divide by the smallest value $3.22$: $\frac{\text{C}}{3.22} = \frac{3.51}{3.22} \approx 1.09$ $\frac{\text{H}}{3.22} = \frac{6.4}{3.22} \approx 1.99$ $\frac{\text{O}}{3.22} = 1$

This is approximately: $1.09 : 1.99 : 1 \approx 1 : 2 : 1$

So the empirical formula is: $\mathrm{CH_2O}$

4. Find empirical formula mass

$\text{Empirical formula mass} = 12 + 2(1) + 16 = 30$

5. Use molecular mass to find molecular formula

Given molecular mass $= 342$.

$n = \frac{342}{30} = 11.4$

This is not an integer, so due to rounding in percentage data, we should check the options directly against the given percentages.

6. Check each option

Option A: $\mathrm{C_{14}H_{20}O_{10}}$

Molar mass: $14(12) + 20(1) + 10(16) = 168 + 20 + 160 = 348$ Not equal to $342$. So incorrect.

Option B: $\mathrm{C_{12}H_{22}O_{11}}$

Molar mass: $12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342$ Matches molecular mass.

Now percentages: $\%C = \frac{144}{342}\times 100 \approx 42.1\%$ $\%H = \frac{22}{342}\times 100 \approx 6.43\%$ $\%O = \frac{176}{342}\times 100 \approx 51.46\%$ These match the given data.

Option C: $\mathrm{C_{12}H_{20}O_{12}}$

Molar mass: $144 + 20 + 192 = 356$ Not equal to $342$. Incorrect.

Option D: $\mathrm{C_{11}H_{18}O_{12}}$

Molar mass: $132 + 18 + 192 = 342$ Mass matches, but check percentages: $\%C = \frac{132}{342}\times 100 \approx 38.6\%$ This does not match $42.1\%$. Incorrect.

7. Final answer

The molecular formula is: $\boxed{\mathrm{C_{12}H_{22}O_{11}}}$ So the correct option is B.