$n_{CO_2}=\dfrac{2,64}{44}=0,06(mol)$
$n_{H_2O}=\dfrac{1,44}{18}=0,08(mol)>n_{CO_2}$
$\Rightarrow$ ancol no, đơn chức $C_nH_{2n+2}O$
Ta có: $n_Y=\dfrac{n_{CO_2}}{n}=\dfrac{n_{H_2O}}{n+1}$
$\Rightarrow 0,08n=0,06(n+1)$
$\Leftrightarrow n=3$
Vậy Y là $C_3H_8O$
CTCT:
$CH_3-CH_2-CH_2OH$ (ancol propylic)
$CH_3-CH(OH)-CH_3$ (ancol isopropylic)