Question

La arderea unei hidrocarburi cu densitatea relativa fata de aer 1.93 s-au obtinut 52.8 g de oxid carbon(IV) si 21.6 apa .Determinati formula hidrocarburii.

Answer 1

CxHy + (4x+y)/4 O2 ⇒ xCO2 + y/2 H2O
M = 1,93·28,9 ≈ 56g/mol 
nCO2 = x·nCxHy = 52,8/44 = 1,2 moli 
nH2O = y/2 ·nCxHy = 21,6/18 = 1,2 moli 
x = y/2 ⇒⇒ y = 2x    CxH2x    14x = 56   x = 4   C4H8 

Answer 2

dgaz = MCxHy/Mgaz => MCxHy = dgaz*Mgaz = 1,93*28,9=56g/mol
dgaz(aer) = 28,9
a=52,8g CO2
b=21,6g H2O
44g CO2............12g C
52,8g CO2..........x=14,4g C
18g H2O...............2g H2
21,6g H2O..........x=2,4g H2
s=2,4+14,4=16,8g
℅C=300a/11s = (300*52,8)/(11*16,8) = 85,714℅
℅H=100b/9s = (100*21,6)/(9*16,8) = 14,28571 ℅
nC=(℅C*MCxHy)/AC*100= (85,714*56)/(12*100) = 4 atomi C
nH=℅H*M/AH*100 = (14,285*56)/(1*100) = 8 atomi H
F.M => C4H8, 1-butena, 2-butena, 2-metil-1-propena, cis si trans 2-butena