$n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)$
Bảo toàn $C\to n_C=n_{CO_2}=0,05(mol)$
$→m_C=0,05.12=0,6g$
Bảo toàn $H\to n_H=2n_{H_2O}=0,12(mol)$
$→m_H=0,12.1=0,12g$
$m_A-m_C-m_H=0,72-0,6-0,12=0$
$→X$ là hidrocacbon
$n_{H_2O}>n_{CO_2}$
$→X$ là ankan
$→n_X=n_{H_2O}-n_{CO_2}=0,06-0,05=0,01(mol)$
Số $C=\dfrac{n_{CO_2}}{n_X}=\dfrac{0,05}{0,01}=5$
$→$ CTPT: $C_5H_{12}$
CTCT: $CH_3-CH_2-CH_2-CH_2-CH_3$