Question

In urma descompunerii termice a 4, 38 g de substanta organica cu masa molara 146 g/mol se obtin 2, 016 dm³ CO, 3,36 dm³ CO₂ si 4,86 g H₂O. Determina formula moleculara a substantei.

Answer 1

n=m/M=4,38/146=0,03 moli s.o

m/M=V/Vm=>m=M*V/Vm=28*2,016/22,4=2,52g CO

m/M=V/Vm=>m=M*V/Vm=44*3,36/22,4=6,6g CO2

28g CO.......................12g C
2,52g CO....................x=1,08g C

44g CO2................12gC
6,6g CO2................x=1,8g C

18g H2O...................2gH
4,86g H2O.................x=0,54g H

mO=4,38-2,88-0,54=0,96g 

n=m/M=2,88/12=0,24 moli C

n=m/M=0,54/2=0,27 moli H

n=m/M=0,96/16=0,06 moli O

0,03 moli s.o........0,24 moli C.....0,27 moli H2......0,06 moli O
1 mol s.o..............x=8 atomi C......y=18 atomi H.....z=2 atomi O

F.M => C8H18O2

Answer 2

22,4l CO.........12gC 2,016lCO.........x x=1,08gC 22,4l CO2.......12gC 3,36l CO2..........y y=1,8gC 18gH2O.........2gH 4,86gH2O.......z z=0,54gH 4,38g-2,88gC-0,54gH=0,96gO 4,38g...2,88gC...0,54gH...0,96gO 146g......a................b................c a=96gC->8 atg C b=18gH->18 atg H c=32gO->2 atg O Formula moleculară:C8H18O2