m alkanu = 2,52 g
m H₂O + m CO₂ = 11,43 g
CnH2n+2 + (2n+n+1 )/2O₂ -----> nCO₂ + (2n+2)/2H₂O
CnH2n+2 + (3n+1)/2 O₂ -----> nCO₂ + (n+1)H₂O
wyniku spalania 1 mola alkanu powstaje n moli CO₂ czyli n * 44 g CO₂
oraz (n+1 ) mola H₂O czyli (n+1) * 18 g H₂O
∑ (m CO₂ + m H₂O) = 44 * n + 18 ( n+1) = 44 n + 18n + 18 = 62n + 18
Masa molowa alkanu CnH2n+2 = 12 * n + 2*n + 2 = 14n + 2
z (14n + 2) g alkanu powstaje (62n+18) g (CO₂+H₂O)
2,52 g alkanu ------------------ 11,43 g (CO₂ + H₂O)
2,52 * (62n+18) = (14n+2) * 11,43
156,24 n + 45,36 = 160,02 n + 22,86
3,78n = 22,5
n = 5,95 ≈ 6
wzór alkanu to C₆H₁₄ - HEKSAN