Đáp án:

 \(\begin{array}{l}
a)\\
CTPT:{C_2}{H_4},{C_3}{H_6}\\
b)\\
\% {m_{{C_2}{H_4}}} = 57,14\% \\
\% {m_{{C_3}{H_6}}} = 42,86\% 
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
{C_{\overline n }}{H_{2\overline n }} + B{r_2} \to {C_{\overline n }}{H_{2\overline n }}B{r_2}\\
{n_{anken}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{M_{{C_{\overline n }}{H_{2\overline n }}}} = \dfrac{{4,9}}{{0,15}} = 32,67\,g/mol\\
 \Rightarrow 14\overline n  = 32,67 \Rightarrow \overline n  = 2,33\\
 \Rightarrow CTPT:{C_2}{H_4},{C_3}{H_6}\\
b)\\
hh:{C_2}{H_4}(a\,mol),{C_3}{H_6}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,15\\
28a + 42b = 4,9
\end{array} \right.\\
 \to a = 0,1;b = 0,05\\
\% {m_{{C_2}{H_4}}} = \dfrac{{0,1 \times 28}}{{4,9}} \times 100\%  = 57,14\% \\
\% {m_{{C_3}{H_6}}} = 100 - 57,14 = 42,86\% 
\end{array}\)