120\,g of an organic compound (only C and H) gives 330\,g CO2 and 270\,g H2O on combustion. % of C and H respectively…

🧠 Burn it, then split the products The compound has only C and H. So every atom of carbon ends up in $\ce{CO2}$ and every atom of hydrogen ends up in $\ce{H2O}$. Find the mass of C and H from the products, then take their share of $120\,\text{g}$.

🗺️ Two clean steps Carbon from $\ce{CO2}$: $\ce{CO2}$ is $44\,\text{g/mol}$, of which $12\,\text{g}$ is carbon. So $\text{Mass C} = \frac{12}{44} \times 330 = 90\,\text{g}.$ Hydrogen from $\ce{H2O}$: $\ce{H2O}$ is $18\,\text{g/mol}$, of which $2\,\text{g}$ is hydrogen. So $\text{Mass H} = \frac{2}{18} \times 270 = 30\,\text{g}.$

Now divide by the sample mass: $\%\,\text{C} = \frac{90}{120} \times 100 = 75\%, \quad \%\,\text{H} = \frac{30}{120} \times 100 = 25\%.$

⚡ Fast check $90$ is three-quarters of $120$, so C is $75\%$ straight away. Then H must be $25\%$ since there is nothing else.

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