\(\begin{array}{l}
nhh = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
Mhh = \dfrac{{28}}{{0,6}} = 46,67\\
= > 14\overline n = 46,67 = > \overline n = 3,33\\
= > CTPT:{C_3}{H_6},{C_4}{H_8}\\
b)\\
{C_3}{H_6}:C{H_2} = CH - C{H_3}\\
{C_4}{H_8}:C{H_3} - CH = CH - C{H_3}
\end{array}\)