Y có dạng CxHyOz
Ta có: nCH4=\(\dfrac{1,344}{22,4}\)=0,06 mol;nY=\(\dfrac{2,688}{22,4}\)=0,12 mol
→0,06.16+0,12.MY=4,56→MY=30=12x+y+16z
Đốt cháy hỗn hợp Z:
CH4+2O2to→CO2+2H2O
CxHyOz+(x+\(\dfrac{y}{4}-\dfrac{z}{2}\))O2to→xCO2+y2H2O
→nCO2=0,06+0,12x=\(\dfrac{4,032}{22,4}\)=0,18
→x=1
→y+16z=18
→y=2;z=1
Vậy Y là CH2O