Đáp án:

$n_{N_2}=\dfrac{112.10^{-3}}{22,4}=0,005$ mol 

$\Rightarrow n_N=2n_{N_2}=0,005.2=0,01$ mol 

$\Rightarrow \%m_N=\dfrac{0,01.14.100\%}{1,07}=13,08\%$ 

Đặt CTTQ hợp chất $X$ là $C_xH_yN$ 

$\Rightarrow \dfrac{14.100\%}{12x+y+14}=13,08\%$ 

$\Rightarrow 12x+y=93$ 

$\Rightarrow (x;y)=(7;9)$ 

Vậy CTPT $X$ là $C_7H_9N$ 

CTCT:

$C_6H_5-CH_2NH_2$: benzylamin 

$C_6H_5-NH-CH_3$: N-metylanilin 

$o-CH_3-C_6H_4-NH_2$: 2-metylanilin 

$m-CH_3-C_6H_4-NH_2$: 3-metylanilin 

$p-CH_3-C_6H_4-NH_2$: 4-metylanilin