a, \(\%m_{Al}=\dfrac{54.100\%}{102}=53\%\)
\(\%m_O=100\%-53\%=47\%\)
b, \(\%m_{Ca}=\dfrac{40.100\%}{232}=17,24\%\)
\(\%m_H=\dfrac{2.100\%}{232}=0,86\%\)
\(\%m_P=\dfrac{62.100\%}{232}=26,72\%\)
\(\Rightarrow\%m_O=100\%-17,24\%-0,86\%-26,72\%=55,18\%\)