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 ]

Step-by-step solution

From AgBr precipitate: 0.75 g AgBr (M = 188) equals 0.00399 mol Br, so the compound has one Br atom. From combustion of 1.0 g: 1.32 g CO₂ (M = 44) equals 0.03 mol C. The molar mass is found from 0.53 g sample: M = (0.53 × 188)/(0.75 × 1) = 133 g/mol. From 1.0 g sample producing 0.03 mol CO₂ We have (1.0/133) × x = 0.03, giving x = 4. Thus the molecular formula is C₄H₅Br with M = 48 + 5 + 80 = 133. Percentage of hydrogen = (5/133) × 100 = 3.76%, which rounds to 4%.