A 0.250 g sample of a hydrocarbon is burned completely in excess O2. The combustion products are passed through a dehydrating tube (gains 0.306 g) followed by a CO2 absorber (gains 0.747 g). The hydrocarbon contains only C and H.
What is the empirical formula?
- A
CH3
- B
CH
- C
C2H3
- Dcheck_circle
CH2
Explanation
mol C = 0.747/44.01 = 0.01698; mass C = 0.2039 g. mol H = (0.306/18.02)*2 = 0.03396; mass H = 0.0342 g. Sum = 0.238 g (consistent with 0.250 within error). Ratio C:H = 0.01698:0.03396 = 1:2 -> CH2.