$d_{hh/C_2H_4}=\dfrac{\overline{M}_{hh}}{M_{C_2H_4}}=0,875$

$\to \overline{M}_{hh}=0,875.28=24,5$

Vì hh chứa hai alkane liên tiếp nên hai alkane là `CH_4` và `C_2H_6`

Sơ đồ đường chéo:

$\rm {M_{CH_4}=16}\hspace{3cm}{30-24,5}\\\hspace{1,4cm}\huge{\searrow}\hspace{2,5cm}\huge{\nearrow}\\\hspace{2,4cm}{\overline{M}_{hh}=24,5}\\\hspace{1,4cm}\huge{\nearrow}\hspace{2,5cm}\huge{\searrow}\\{M_{C_2H_6}=30}\hspace{3cm}{24,5-16}$

`\to V_{CH_4}/V_{C_2H_6}=11/17`

`\to %V_{C_2H_6}=17/(11+17).100%=60,71%`