0.25 g of an organic compound " A " containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of CaCl ₂ tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is %. (Nearest integer) (Given: molar mass in g mol ⁻¹ H : 1, C : 12, O : 16 )

Step-by-step solution

In combustion analysis: CaCl₂ tube absorbs H₂O and potash tube absorbs CO₂. Mass increase of CaCl₂ = 0.15 g gives moles of H₂O = 0.15/18 = 0.00833 mol, so moles of H = 0.01667 mol and mass of H = 0.01667 g. Mass increase of potash = 0.1837 g gives moles of CO₂ = 0.1837/44 = 0.00417 mol, so moles of C = 0.00417 mol and mass of C = 0.05 g. Mass of O = 0.25 - 0.05 - 0.01667 = 0.18333 g. Percentage of O = (0.18333/0.25) × 100 = 73.33% ≈ 73% (nearest integer).