\[n_{CO_2}=\dfrac{13,2}{44}=0,3(mol)\]
\[\to n_C=0,3(mol)\]
\[\to m_C=0,3\times 12=3,6(g)\]
\[n_{H_2O}=\dfrac{7,2}{18}=0,4(mol)\]
\[\to n_H=0,4.2=0,8(mol)\]
\[\to m_H=0,8.1=0,8(g)\]
Bảo toàn mol O
$\to n_{O(X)}+2.n_{O_2}=2.n_{CO_2}+n_{H_2O}$
$\to n_{O(X)}=0,1(mol)$
$\to$ X có dạng $C_xH_yO_z$
$\to x:y:z=0,3:0,8:0,1=3:8:1$
$\to C_3H_8O$