$n_C=n_{CO_2}=\dfrac{0,792}{44}=0,018(mol)$
$n_H=2n_{H_2O}=\dfrac{0,234.2}{18}=0,026(mol)$
Khi phân huỷ $0,366g$ A thu được: $V_{N_2}=\dfrac{37,42.0,366}{0,549}=\dfrac{1871}{75}(cm^3)=0,025l$
Áp suất $P=750mmHg=\dfrac{75}{76}(atm)$
$\Rightarrow n_{N_2}=\dfrac{\dfrac{75}{76}.0,025}{0,082.(27+273)}=0,001(mol)$
$\Rightarrow n_N=2n_{N_2}=0,002(mol)$
$\Rightarrow m_O=0,366-0,018.12-0,026-0,002.14=0,096g$
$\Rightarrow n_O=0,006(mol)$
$n_C : n_H : n_O: n_N=0,018:0,026:0,006:0,002=9:13:3:1$
A chứa $1N$ nên CTPT A là: $C_9H_{13}O_3N$