On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2 and 0.567 g of H2O. The empirical formula…

🧠 Find C, H, then oxygen by leftover. Then ratio. The compound has C, H, and possibly O. Find C from $\ce{CO2}$ and H from $\ce{H2O}$. Whatever weight is unaccounted for must be oxygen. Then convert all three to moles and find the simplest ratio.

🗺️ Step by step Moles of C $= 1.46/44 \approx 0.0332\,\text{mol}$ → mass C $= 0.398\,\text{g}$. Moles of H atoms $= 2 \times 0.567/18 \approx 0.063\,\text{mol}$ → mass H $= 0.063\,\text{g}$. Mass of O $= 1.0 - (0.398 + 0.063) = 0.539\,\text{g}$. Moles of O $= 0.539/16 \approx 0.0337\,\text{mol}$.

Mole ratio: C ($0.0332$) : H ($0.063$) : O ($0.0337$) $\approx 1 : 2 : 1$.

So empirical formula is $\ce{CH2O}$, with mass: $12 + 2 + 16 = 30\,\text{g/mol}.$

⚠️ Common mistake Forgetting to subtract C and H weights to get oxygen is the classic slip. The compound has O even though no oxygen weight was given directly.

$\boxed{\text{Answer: (a)}}$