Tỉ lệ số nguyên tử $C$, $H$, $N$:
$\dfrac{53,33}{12}:\dfrac{15,55}{1}:\dfrac{31,12}{14}$
$=4,44: 15,55: 2,22$
$=2:7:1$
$\Rightarrow$ CTĐGN $C_2H_7N$
Đặt CTPT A là $C_2H_7N$
$\Rightarrow (12.2+7+14)n=46$
$\Leftrightarrow n=1,02\approx 1$
Vậy A là $C_2H_7N$