An organic compound weighing 500 mg , produced 220 mg of CO₂, on complete combustion. The percentage composition of carbon in the compound is _________ %. (nearest integer)
(Given molar mass in g mol⁻¹ of C: 12, O: 16 )

Solution

Chemistry ma’am · 4 steps

Every carbon atom ends up in CO₂, so pull the carbon's share out of that 220 mg first.

1. Carbon goes to CO2. On complete combustion, all the carbon in the compound turns into CO₂. One mole of CO₂ (44 g) holds exactly one mole of carbon (12 g), so carbon is 12/44 of the CO₂ by mass.

   % C = 12/44 × (mass of CO₂)/(mass of compound) × 100

2. Mass of carbon. Carbon in the sample is 12/44 × 220 mg = 60 mg.
3. Percentage. 60/500 × 100 = 12. Both masses are in mg, so no unit conversion is needed.
4. So. The compound is 12% carbon by mass.

The common slip: 44. Dividing the CO₂ mass straight by the compound mass gives 44, not 12. Only 12/44 of that CO₂ is carbon.

The idea to keep. In Liebig's method, convert the product back to the element first: carbon is 12/44 of the CO₂ mass, and hydrogen is 2/18 of the H₂O mass.

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