$n_{CO_2}=\dfrac{3,52}{44}=0,08(mol)$
$n_{H_2O}=\dfrac{1,62}{18}=0,09(mol)$
$0,08<0,09\to$ hỗn hợp có ankan
$n_{\text{ankan}}=n_{H_2O}-n_{CO_2}=0,01(mol)$
$\to n_{\text{hidrocacbon còn lại}}=n_{\text{ankan}}=0,01(mol)$
$n_{hh}=0,01+0,01=0,02(mol)$
$\to C=\dfrac{n_{CO_2}}{n_{hh}}=4$
Vậy CTPT ankan là $C_4H_{10}$
Đặt CTPT hidrocacbon còn lại là $C_4H_x$
Bảo toàn $H$: $5.0,01+0,5x.0,01=0,09$
$\to x=8$
Vậy CTPT hidrocacbon còn lại là $C_4H_8$
* CTCT:
- $C_4H_{10}$:
$CH_3-CH_2-CH_2-CH_3$
$CH_3 - CH -CH_3$
$|$
$CH_3$
- $C_4H_8$:
$CH_2=CH-CH_2-CH_3$
$CH_3-CH=CH-CH_3$
$CH_2 = C -CH_3$
$|$
$CH_3$
$CH_2 - CH_2$
$|$ $|$
$CH_2 - CH_2$
$CH - CH_3$
$/$ $|$
$CH_2 - CH_2$
