On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g $$H_2O$$ and 0.307 g $$CO_2$$. The percentages of hydrogen and oxygen in the given organic compound respectively are :

The combustion analysis gives the following data:

Mass of organic compound burnt $$= 0.210 \text{ g}$$
Mass of $$CO_2$$ formed $$= 0.307 \text{ g}$$
Mass of $$H_2O$$ formed $$= 0.127 \text{ g}$$

Step 1: Calculate mass of carbon in the sample.

Molar mass of $$CO_2 = 44 \text{ g mol}^{-1}$$ and each mole contains $$12 \text{ g}$$ of C.
Therefore,

$$\text{Mass of C} = 0.307 \times \frac{12}{44} \text{ g}$$

$$\text{Mass of C} = 0.08372 \text{ g}$$

Step 2: Calculate mass of hydrogen in the sample.

Molar mass of $$H_2O = 18 \text{ g mol}^{-1}$$ and each mole contains $$2 \text{ g}$$ of H.
Therefore,

$$\text{Mass of H} = 0.127 \times \frac{2}{18} \text{ g}$$

$$\text{Mass of H} = 0.01411 \text{ g}$$

Step 3: Calculate mass of oxygen in the sample.

Total mass of C and H in the sample:

$$0.08372 + 0.01411 = 0.09783 \text{ g}$$

Hence,

$$\text{Mass of O} = 0.210 - 0.09783 = 0.11217 \text{ g}$$

Step 4: Convert masses to percentages.

Percentage of hydrogen:

$$\%H = \frac{0.01411}{0.210} \times 100 = 6.72\%$$

Percentage of oxygen:

$$\%O = \frac{0.11217}{0.210} \times 100 = 53.41\%$$

Step 5: State the result.

The percentages of hydrogen and oxygen in the organic compound are $$6.72\%$$ and $$53.41\%$$ respectively.

Thus, the correct choice is Option B.