Đáp án:
`a)`
$\rm M_{Fe_2O_3}=56.2+16.3=160 \ (g/mol)$
$\to \begin{cases} \rm \%m_{Fe}=\dfrac{56.2}{160} . 100\%=70\% \\\rm \\ \%m_O=100\%-70\%=30\% \end{cases}$
`b)`
$\rm M_{SO_2}=32+16.2=64 \ (g/mol)$
$\to \begin{cases}\rm \%m_{S}=\dfrac{32}{64} . 100\%=50\% \\ \\ \rm \%m_O=100\%-50\%=50\% \end{cases}$
`c)`
$\rm M_{CuSO_4}=64+32+16.4=160 \ (g/mol)$
$\to \begin{cases} \rm \%m_{Cu}=\dfrac{64}{160} . 100\% = 40\% \\ \\ \rm \%m_S=\dfrac{32}{160} . 100\% = 20\% \\ \\\rm \%m_O=100\%-40\%-20\%=40\% \end{cases}$
