On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of $$CO_2$$ and 0.567 g of $$H_2O$$. The empirical formula mass of compound (X) is ________ g.

(Given molar mass in g mol$$^{-1}$$ C: 12, H: 1, O: 16)

During combustion, all carbon in the compound converts to $$CO_2$$ and all hydrogen converts to $$H_2O$$. We first calculate the moles of each element obtained from the given products.

Step 1: Moles of carbon
Mass of $$CO_2 = 1.46 \text{ g}$$. Molar mass of $$CO_2 = 44 \text{ g mol}^{-1}$$.
$$\text{Moles of }CO_2 = \frac{1.46}{44} = 0.03318 \text{ mol}$$.
Each mole of $$CO_2$$ contains one mole of C, so $$\text{Moles of C} = 0.03318 \text{ mol}$$.

Step 2: Mass of carbon
$$\text{Mass of C} = 0.03318 \times 12 = 0.398 \text{ g}$$.

Step 3: Moles of hydrogen
Mass of $$H_2O = 0.567 \text{ g}$$. Molar mass of $$H_2O = 18 \text{ g mol}^{-1}$$.
$$\text{Moles of }H_2O = \frac{0.567}{18} = 0.0315 \text{ mol}$$.
Each mole of $$H_2O$$ contains two moles of H, so $$\text{Moles of H} = 2 \times 0.0315 = 0.0630 \text{ mol}$$.

Step 4: Mass of hydrogen
$$\text{Mass of H} = 0.0630 \times 1 = 0.063 \text{ g}$$.

Step 5: Mass and moles of oxygen in the compound
Total mass of sample = 1.00 g.
Mass of C + H already determined = $$0.398 + 0.063 = 0.461 \text{ g}$$.
Remaining mass is oxygen: $$\text{Mass of O} = 1.00 - 0.461 = 0.539 \text{ g}$$.
Moles of O: $$\text{Moles of O} = \frac{0.539}{16} = 0.0337 \text{ mol}$$.

Step 6: Find simplest mole ratio
Divide each mole value by the smallest, $$0.03318$$:

$$\frac{\text{C}}{0.03318} = 1.00$$
$$\frac{\text{H}}{0.03318} = \frac{0.0630}{0.03318} \approx 1.90 \approx 2$$
$$\frac{\text{O}}{0.03318} = \frac{0.0337}{0.03318} \approx 1.02 \approx 1$$

Hence the empirical formula is $$CH_2O$$.

Step 7: Empirical formula mass
$$\text{E.F. mass} = 12 + 2(1) + 16 = 30 \text{ g mol}^{-1}$$.

Therefore, the empirical formula mass of compound (X) is $$30 \text{ g}$$.

Option A is correct.