1, $M_{C_2H_4}=12.2+1.4=28$
$\Rightarrow \begin{cases}\%C=\dfrac{2.12}{28}.100\%=85,71\%\\\%H=100\%-85,71\%=14,29\%\end{cases}$
2, $M_{CO_2}=12+16.2=44$
$\Rightarrow \begin{cases}\%C=\dfrac{12}{44}.100\%=27,27\%\\\%O=100\%-27,27\%=72,73\%\end{cases}$