Răspuns:
Fie CₐHₓ hidrocarbura.
1 mol H ..... 44 g => μ H= 44 g/mol => 12a+x=44 ( in procente masice, in 44 g H exista 12a g C si x g H)
stim ca 44 g CO₂ (μ).......... 12 g C
33 g CO₂ ...... 9 g C
11 g H..... 9 g C
μ=44 g H.... 12a g C
=> a=3
12a+x=44 <=> 36+x=44 => x=44-36=8
=> F.ch.M C₃H₈ (propan)