The complete combustion of 0.492 g of an organic compound containing ‘C’, ‘H’ and ‘O’ gives 0.793 g of CO₂ and 0.442 g of H₂O. The percentage of oxygen composition in the organic compound is ___________. (nearest integer)

Complete Solution

Calculate Moles of Carbon and Hydrogen:

Mass of CO₂ produced: 0.793 g

Molar mass of CO₂: 44 g/mol

Moles of CO₂: 0.793 / 44 = 0.01802 mol

Moles of Carbon: 0.01802 mol

Mass of Carbon: 0.01802 × 12 = 0.21624 g

Mass of H₂O produced: 0.442 g

Molar mass of H₂O: 18 g/mol

Moles of H₂O: 0.442 / 18 = 0.02456 mol

Moles of Hydrogen: 2 × 0.02456 = 0.04912 mol

Mass of Hydrogen: 0.04912 × 1 = 0.04912 g

Calculate Mass of Oxygen in the Compound:

Total mass of the compound: 0.492 g

Mass of Carbon and Hydrogen: 0.21624 + 0.04912 = 0.26536 g

Mass of Oxygen: 0.492 - 0.26536 = 0.22664 g

Calculate Percentage of Oxygen:

Percentage of Oxygen = (0.22664 / 0.492) × 100 = 46.06 ≈ 46% (nearest integer)

Final Answer:

The percentage of oxygen composition in the organic compound is 46%.