A combustion analysis of 2.00 g of an unknown hydrocarbon C_xH_y produced 6.16 g CO₂ and 2.52 g H₂O.
What is the empirical formula?
- A
C₂H₅
- B
CH
- Ccheck_circle
CH₂
- D
CH₄
Explanation
mol C = 6.16/44.01 = 0.140; mol H = (2.52/18.02)·2 = 0.280. Ratio C:H = 1:2. Empirical = CH₂.