JEE Main 2022 · 28 July · morning · Numerical answer · +4 / −0
On complete combustion of 0.492 g of an organic compound containing C, H and O, 0.7938 g of CO₂ and 0.4428 g of H₂ O was produced. The % composition of oxygen in the compound is ___________.
Answer and solution
Answer 46
Solution
Chemistry ma’am · 3 steps
Oxygen never shows up as its own product, so find C and H from the products and take O as what is left.
- Carbon from CO2. All the carbon ends up in CO₂, and 12 g of every 44 g of CO₂ is carbon. Mass of C = 0.7938 × 12/44 = 0.2165 g, which is 0.2165/0.492 × 100 = 44%.
- Hydrogen from water. All the hydrogen ends up in H₂O, and each 18 g of water holds 2 g of hydrogen. Mass of H = 0.4428 × 2/18 = 0.0492 g, which is 0.0492/0.492 × 100 = 10%.
% O = 100 - % C - % H = 100 - 44 - 10
- So. The compound contains only C, H and O, so oxygen makes up the rest: 0.492 - 0.2165 - 0.0492 = 0.2263 g, which is 46%.
The common slip: 51. Using 1/18 instead of 2/18 for hydrogen gives only 5% H, so oxygen comes out as 51 instead of 46. Water has two H atoms.
The idea to keep. In combustion analysis, C comes from CO₂ (12/44) and H from H₂O (2/18). Oxygen is never measured directly: it is always 100 minus the rest.
Answer from the NTA final answer key. Solution by PrepWiser.