Ta có
n CO2=6,6/44=0,15(mol)
---->n C=0,15(mol)--->m C=0,15.12=1,8(g)
n H2O=3,6/18=0,2(mol)
--->n H=0,4(mol)-------->m H=0,4(g)
m H+m C(2,2g)<m A(3g)
----> A gồm 3 nguyên tố C,H,O
m O=3-2,2=0,8(g)
n O=0,8/16=0,05(mol)
n C:n H:n O=0,15:0,4:0,05
=3:8:1
CTĐG: C3H8O
PTK A=60
-->(C3H8O)n=60
-->n=1
CTPT:C3H8O