Trong $X_1, p=e=n_1$
$\Rightarrow p=e=n_1=\dfrac{18}{3}=6$
$\Rightarrow A_1=p+n_1=6+6=12$
Trong $X_2$:
$p+e+n_2=2p+n_2=19$
$\Rightarrow 2.6+n_2=19$
$\Leftrightarrow n_2=7$
$\Rightarrow A_2=p+n_2=7+6=13$
Vậy nguyên tử khối trung bình là:
$\overline{A}=12.98,9\%+13.1,1\%=12,011$