Ta có:
\(\left\{{}\begin{matrix}n_{CO2}=\frac{9,68}{44}=0,22\left(mol\right)\\n_{H2O}=\frac{5,76}{18}=0,32\left(mol\right)\end{matrix}\right.\)
\(n_{CO2}< n_{H2O}\Rightarrow X:\) hỗn hợp anken
\(n_X=n_{H2O}-n_{CO2}=0,32-0,22=0,1\left(mol\right)\)
\(\overline{C}=\frac{n_{CO2}}{n_X}=\frac{0,22}{0,1}=2,2\)
Vậy CTPT là C2H6 và C3H8
Gọi \(\left\{{}\begin{matrix}n_{C2H6}:a\left(mol\right)\\n_{C3H8}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}2a+3b=0,22\\3a+4b=0,32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,08\\b=0,02\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C2H8}=\frac{0,08.30}{0,08.30+0,02.44}.100\%=73,17\%\\\%m_{C3H8}=100\%-73,17\%=26,83\%\end{matrix}\right.\)