\(M_{Mg\left(NO_3\right)_2}=148\left(g/mol\right)\)
Trong 1 mol h/c có 1 mol nguyên tử Mg, 2 mol n tử N, 6 mol n tử O
\(\%m_{Mg}=\dfrac{1.M_{Mg}}{M_{Mg\left(NO_3\right)_2}}.100\%=\dfrac{24}{148}.100\%=16\%\)
\(\%m_N=\dfrac{2.M_N}{M_{Mg\left(NO_3\right)_2}}.100\%=\dfrac{2.14}{148}.100\%=19\%\)
\(\%m_O=100\%-\left(\%m_{Mg}+\%m_N\right)=100\%-\left(16\%+19\%\right)=65\%\)