On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H2O and 0.307 g CO2. The percentages of…

🧠 The C-H-O split In a CHO compound, all the carbon ends up in $\ce{CO2}$ and all the hydrogen ends up in $\ce{H2O}$. Oxygen is what's left after subtracting C and H from the original sample. So you find C and H directly, and back-calculate O.

🗺️ Step-by-step Hydrogen: $\ce{H2O}$ is $\frac{2}{18}$ hydrogen by mass. $\text{Mass H} = \frac{2}{18} \times 0.127 \approx 0.0141\,\text{g}.$ $\%\text{H} = \frac{0.0141}{0.210} \times 100 \approx 6.72\%.$

Carbon: $\ce{CO2}$ is $\frac{12}{44}$ carbon by mass. $\text{Mass C} = \frac{12}{44} \times 0.307 \approx 0.0837\,\text{g}.$ $\%\text{C} = \frac{0.0837}{0.210} \times 100 \approx 39.87\%.$

Oxygen by difference: $\%\text{O} = 100 - 6.72 - 39.87 = 53.41\%.$

So H = $6.72\%$, O = $53.41\%$.

⚠️ Common mistake Don't confuse the masses. Option (a) flips the values — students who rush write the C% as H%. Always tag each percentage with its element before picking.

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