Bài 3:

$\%m_S=\dfrac{32}{32+16.2}.100\%=50\%$

$\%m_O=100\%-50\%=50\%$

Bài 4:

Trong 1 mol hợp chất:

$m_C=50,5.23,8\%\approx 12(g)$

$\Rightarrow n_C=\dfrac{12}{12}=1(mol)$

$m_H=50,5.5,9\%\approx 3(g)$

$\Rightarrow n_H=\dfrac{3}{1}=3(mol)$

$m_{Cl}=50,5.70,3\%\approx 35,5(g)$

$\Rightarrow n_{Cl}=\dfrac{35,5}{35,5}=1(mol)$

Vậy CTHH của HC là $CH_3Cl$

Em tham khảo nha :

\(\begin{array}{l}
CTHH:{C_x}{H_y}C{l_z}\\
M = 50,5dvC\\
\% C = 23,8\% \\
 \Rightarrow x = \dfrac{{50,5 \times 23,8\% }}{{12}} = 1\\
\% H = 5,9\% \\
 \Rightarrow y = \dfrac{{50,5 \times 5,9\% }}{1} = 3\\
\% Cl = 70,3\% \\
 \Rightarrow z = \dfrac{{50,5 \times 70,3\% }}{{35,5}} = 1\\
CTHH:C{H_3}Cl
\end{array}\)