Đáp án:

 58,18% và 41,82%

Giải thích các bước giải:

\(\begin{array}{l}
CTTQ:{C_{\overline n }}{H_{2\overline n  + 1}}OH(\overline n  > 1)\\
{C_{\overline n }}{H_{2\overline n  + 1}}OH + \dfrac{{3\overline n }}{2}{O_2} \xrightarrow{t^0} \overline n C{O_2} + (\overline n  + 1){H_2}O\\
{n_{{O_2}}} = \dfrac{{38,4}}{{32}} = 1,2\,mol\\
{n_X} = 1,2 \times \dfrac{1}{{\frac{{3\overline n }}{2}}} = \dfrac{{0,8}}{{\overline n }}(g/mol)\\
{M_X} = \dfrac{{22}}{{\dfrac{{0,8}}{{\overline n }}}} = 27,5\overline {n\,} g/mol \Leftrightarrow 14\overline n  + 18 = 27,5\overline n \\
 \Leftrightarrow \overline n  = 1,333 \Rightarrow hh:C{H_3}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
2C{H_3}OH + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 4{H_2}O\\
{C_2}{H_5}OH + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
\left\{ \begin{array}{l}
32a + 46b = 22\\
1,5a + 3b = 1,2
\end{array} \right.\\
 \Rightarrow a = 0,4;b = 0,2\\
\% {m_{C{H_3}OH}} = \dfrac{{0,4 \times 32}}{{22}} \times 100\%  = 58,18\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 58,18 = 41,82\% 
\end{array}\)