Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\,CO\\
{M_{CO}} = 12 + 16 = 28\,dvC\\
\% {m_C} = \dfrac{{12}}{{28}} \times 100\% = 42,86\% \\
\% {m_O} = 100 - 42,86 = 57,14\% \\
C{O_2}:\\
{M_{C{O_2}}} = 12 + 16 \times 2 = 44dvC\\
\% {m_C} = \dfrac{{12}}{{44}} \times 100\% = 27,27\% \\
\% {m_O} = 100 - 27,27 = 72,73\% \\
b)\\
F{e_3}{O_4}\\
{M_{F{e_3}{O_4}}} = 56 \times 3 + 16 \times 4 = 232\,dvC\\
\% {m_{Fe}} = \dfrac{{56 \times 3}}{{232}} \times 100\% = 72,41\% \\
\% {m_O} = 100 - 72,41 = 27,59\% \\
F{e_2}{O_3}:\\
{M_{F{e_2}{O_3}}} = 56 \times 2 + 16 \times 3 = 160dvC\\
\% {m_{Fe}} = \dfrac{{56 \times 2}}{{160}} \times 100\% = 70\% \\
\% {m_O} = 100 - 70 = 30\% \\
c)\\
S{O_2}:\\
{M_{S{O_2}}} = 32 + 16 \times 2 = 64\,dvC\\
\% {m_S} = \dfrac{{32}}{{64}} \times 100\% = 50\% \\
\% {m_O} = 100 - 50 = 50\% \\
S{O_3}:\\
{M_{S{O_3}}} = 32 + 16 \times 3 = 80\,dvC\\
\% {m_S} = \dfrac{{32}}{{80}} \times 100\% = 40\% \\
\% {m_O} = 100 - 40 = 60\%
\end{array}\)
