0.53 g of an organic compound (x) when heated with excess of nitric acid ( concentrated) and then with silver nitrate gave 0. 75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of $$CO_{2}$$ gas on combustion. The percentage of hydrogen in the compound (x) is __ %. [Nearest Integer]
[Given: Molar mass in g $$mol^{-1}$$ H : 1, C : 12, Br: 80, Ag: 108, O : 16 ; Compound (x) :$$C_{x}H_{y}Br_{z}$$]

We need to find the percentage of hydrogen in the organic compound $$C_xH_yBr_z$$.

First, the percentage of bromine is determined by precipitating bromide as silver bromide. From 0.53 g of the compound, 0.75 g of AgBr precipitate is produced. The molar mass of AgBr is 108 + 80 = 188 g/mol, so the moles of AgBr are $$\text{Moles of AgBr} = \frac{0.75}{188} = 0.003989 \text{ mol}$$. Since each mole of AgBr contains one mole of Br, the mass of bromine is $$\text{Mass of Br} = 0.003989 \times 80 = 0.31915 \text{ g}$$, corresponding to $$\% Br = \frac{0.31915}{0.53} \times 100 = 60.22\%$$.

Next, the percentage of carbon is found from the combustion analysis: burning 1.0 g of the compound yields 1.32 g of $$CO_2$$. With the molar mass of $$CO_2$$ being 44 g/mol and each mole of $$CO_2$$ containing one mole of carbon (12 g), the mass of carbon is $$\text{Mass of C from 1.0 g} = \frac{12}{44} \times 1.32 = 0.36 \text{ g}$$, giving $$\% C = \frac{0.36}{1.0} \times 100 = 36\%$$.

Finally, since the compound contains only carbon, hydrogen, and bromine, the percentage of hydrogen is obtained by difference: $$\% H = 100 - \% C - \% Br = 100 - 36 - 60.22 = 3.78\%$$. Rounding to the nearest integer yields $$\% H \approx 4\%$$.

Therefore, the percentage of hydrogen in the compound is 4.