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Em tham khảo nha:

\(\begin{array}{l}
a)\\
\% {m_C} = \dfrac{{12 \times 2}}{{12 \times 2 + 6 + 16}} \times 100\%  = 52,17\% \\
\% {m_H} = \dfrac{6}{{12 \times 2 + 6 + 16}} \times 100\%  = 13,04\% \\
\% {m_O} = 100 - 52,17 - 13,04 = 34,79\% \\
b)\\
{n_{C{O_2}}} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
{n_C} = {n_{C{O_2}}} = 0,15\,mol\\
{n_{{C_2}{H_6}O}} = \dfrac{{0,15}}{2} = 0,075\,mol\\
{m_{{C_2}{H_6}O}} = 0,075 \times 46 = 3,45g
\end{array}\)