\(M_{FeO}=56+16=72\left(DvC\right)\\ \%Fe=\dfrac{56}{72}.100\%=77,7\%\\ \%O=100\%-77,7\%=22,3\%\)
\(M_{FeO}=20+16=36\) (g/mol)
\(\Rightarrow\%m_{Fe}=\dfrac{20.100\%}{36}=55,5\%\)
\(\Rightarrow\%m_O=\dfrac{16.100\%}{36}=44,5\%\)
\(\%m_{Fe}=\dfrac{56.100\%}{72}=77,7\%\)
\(\%m_O=100\%-77,7\%=22,3\%\)