a)

\(CO\)

Ta có: \({M_{CO}} = 12 + 16 = 28\)

\( \to \% {m_C} = \frac{{12}}{{28}} = 42,86\%  \to \% {m_O} = 57,14\% \)

\(CO_2\)

Ta có: \({M_{CO_2}} = 12 + 16.2 = 44\)

\( \to \% {m_C} = \frac{{12}}{{44}} = 27,27\%  \to \% {m_O} = 72,73\%\)

b) 

\(Fe_3O_4\)

Ta có: \({M_{F{e_3}{O_4}}} = 56.3 + 16.4 = 232\)

\( \to \% {m_{Fe}} = \frac{{56.3}}{{232}} = 72,41\%  \to \% {m_O} = 27,59\% \)

\(Fe_2O_3\)

Ta có: \({M_{F{e_2}{O_3}}} = 56.2 + 16.3 = 160\)

\( \to \% {m_{Fe}} = \frac{{56.2}}{{160}} = 70\%  \to \% {m_O} = 30\% \)

c)

\(SO_2\)

Ta có: \({M_{S{O_2}}} = 32 + 16.2 = 64\)

\( \to \% {m_S} = \frac{{32}}{{64}} = 50\%  \to \% {m_O} = 50\% \)

\(SO_3\)

Ta có: \({M_{S{O_3}}} = 32 + 16.3 = 80\)

\( \to \% {m_S} = \frac{{32}}{{80}} = 40\%  \to \% {m_O} = 60\% \)