The weight percentage of hydrogen in Q, formed in the following reaction sequence, is ______.

[Given : Atomic mass of H = 1, C = 12, N = 14, O = 16, S = 32, Cl = 35]

The reaction sequence (conc. $$H_2SO_4$$ followed by conc. $$HNO_3$$) converts phenol first into its sulphonic-acid derivative and finally into 2,4,6-trinitrophenol (picric acid). Hence the compound $$Q$$ is 2,4,6-trinitrophenol having the molecular formula $$C_6H_3N_3O_7$$.

Step 1 : Find the molecular mass of $$Q$$.
$$\begin{aligned} M &= 6\times12 \;(\text{for }C) + 3\times1 \;(\text{for }H) + 3\times14 \;(\text{for }N) + 7\times16 \;(\text{for }O) \\ &= 72 + 3 + 42 + 112 \\ &= 229 \text{ g mol}^{-1} \end{aligned}$$

Step 2 : Calculate the mass percentage of hydrogen.
Number of hydrogen atoms = 3, so mass of hydrogen in one mole = $$3\times1 = 3$$ g.

Weight % of H $$= \dfrac{\text{mass of H}}{\text{molar mass}}\times100$$
$$\frac{3}{229}\times100 = 1.31\%$$

Thus, the weight percentage of hydrogen in $$Q$$ is 1.31 %.