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 CO 2 si 4,86 g H2O.Determina formula moleculara a substantei

Answer 1

s.o=4,38g

M=146g/mol

m/M=V/Vm ⇒ m=28*2,016/22,4=2,52g CO

m/M=V/Vm ⇒ m=44*3,36/22,4=6,6g CO₂

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

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

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

mO=4,38-1,8-1,08-0,54=0,96g O

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

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

n=m/M=0,54/2=0,27 moli H2

n=m/M=0,96/16=0.06 moli O

0,03 moli s.o.........0,24 moli C........0,27 moli H.......0.06 moli O
1mol.....................x=8 atomi C.........y=18 atomi H........z=2 atomi O

Formula moleculara => C8H18O2