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)

Solution

Chemistry ma’am · 4 steps

All the carbon ends up in CO2 and all the hydrogen in water. Oxygen is whatever mass is left over.

1. Carbon from CO2. Every 44 g of CO₂ carries 12 g of carbon. So the carbon mass is 0.793 × 12/44 = 0.2163 g, which is 0.2163/0.492 × 100 = 43.96%.
2. Hydrogen from water. Every 18 g of H₂O carries 2 g of hydrogen, because there are two H atoms. So the hydrogen mass is 0.442 × 2/18 = 0.0491 g, which is 0.0491/0.492 × 100 = 9.98%.
3. Oxygen by difference. Some of the oxygen in the products came from the air, so you cannot read it off the products. Take it as the remainder instead.

   % O = 100 - 43.96 - 9.98 = 46.06

4. So. The compound is about 46.06% oxygen, which rounds to 46.

The common slip: 51. Taking hydrogen as 1/18 of the water mass instead of 2/18 halves %H to about 4.99 and gives %O of 51, not 46.

The idea to keep. In combustion analysis, %C comes from 12/44 of the CO₂ mass and %H from 2/18 of the water mass. Oxygen is always found by difference.

Answer from the NTA final answer key. Solution by PrepWiser.