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: \({M_{S{O_3}}} = 32 + 16.3 = 80\)
\( \to \% {m_S} = \frac{{32}}{{80}} = 40\% \to \% {m_O} = 60\% \)
