n CO2 = 6,6/44=0,15(mol)
m C = 0,15.12=1,8(g)
n H2O = 2,7/18=0,15(mol)
m H = 0,15.2.1=0,3(g)
Ta có: m C+m H=1,8+0,3< 4,5
-> Trong HCHC gồm C,H,O
m O = 4,5-1,8-0,3=2,4(g)
n O=2,4/16=0,15(mol)
CTTQ: CxHyOz
x:y:z=n C: n H: n O = 0,15:0,3:0,15=1:2:1
-> CTĐG: (CH2O)n
Ta có: 30n=60 ⇔ n = 2
-> CTHH: C2H4O2 hay CH3COOH