On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is _____________. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

All the carbon ends up in CO₂, so find how much of that 0.20 g is carbon and compare it with 0.30 g.

1. Carbon goes to CO2. On combustion every carbon atom of the compound ends up in CO₂. One mole of CO₂ (44 g) contains 12 g of carbon, so the carbon fraction of CO₂ is 12/44. The water only tells you about hydrogen, so you do not need it here.

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

2. Put in numbers. Mass of carbon = 12/44 × 0.20 g = 0.0545 g. Then % C = 0.0545/0.30 × 100 = 18.18.
3. So. The compound is 18.18% carbon, which rounds to 18.

The common slip: 67. Taking the whole 0.20 g of CO₂ as carbon, without the 12/44 factor, gives 66.7, which rounds to 67, not 18.

The idea to keep. In combustion analysis, %C comes only from CO₂ using 12/44, and %H comes only from H₂O using 2/18. Always divide by the mass of the original compound.

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