BTKL, \(\text{mX=8,8+3,6-6,4=6g}\)
nC=nCO2=\(\frac{8,8}{44}\)=0,2 mol \(\rightarrow\) \(\text{mC=0,2.12=2,4g}\)
nH=2nH2O=\(\frac{2.3,6}{18}\)=0,4 mol \(\rightarrow\) mH=0,4g
\(\rightarrow\)\(\text{mO=6-2,4-0,4=3,2g}\)\(\rightarrow\) nO=0,2 mol
X có C, H, O
Ta có nC:nH:nO=0,2:0,4:0,2= 2:4:2
\(\rightarrow\) CTĐGN (C2H4O2)n
n=\(\frac{1t}{m}\)\(\rightarrow\) CTPT C2H4O2