Đáp án:
\(\begin{array}{l}
a){C_8}{H_{10}}\\
b){C_9}{H_{12}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{7,04}}{{44}} = 0,16\,mol\\
{n_{{H_2}O}} = \dfrac{{1,8}}{{18}} = 0,1\,mol\\
{n_C} = {n_{C{O_2}}} = 0,16\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,2\,mol\\
{n_C}:{n_H} = 0,16:0,2 = 4:5\\
\Rightarrow CTDGN:{C_4}{H_5}\\
{M_A} = 3,65 \times 29 = 106g/mol\\
\Rightarrow 53n = 106 \Rightarrow n = 2\\
CTPT:{C_8}{H_{10}}\\
b)\\
{n_{C{O_2}}} = \dfrac{{3,7}}{{44}} = 0,084\,mol\\
{n_{{H_2}O}} = \dfrac{1}{{18}} = 0,056mol\\
{n_C} = {n_{C{O_2}}} = 0,084\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,112\,mol\\
{n_C}:{n_H} = 0,084:0,112 = 3:4\\
\Rightarrow CTDGN:{C_3}{H_4}\\
{M_B} = 4 \times 29 = 116g/mol\\
\Rightarrow 40n = 116 \Rightarrow n = 3\\
CTPT:{C_9}{H_{12}}
\end{array}\)
