Presupun formula moleculara CxHyOz
0.244g compus......0,616gCO2........0,108gH2o?
122g.................m= 302g.................m= 54g
-calculez moli din masele obtinute, stiind M,CO2=44g/mol M,H2O=18g/mol
n(niu)= m/M
n,CO2=302g/44g/mol= 7 molCO2= 7molC =x
n,H2o= 54g/18g/mol= 3molH2O==> 6molH=y
Compusul ar fi C7H6Oz---. >deduc z din masa molara
122= 7x12gC+6gH+zx16gO---------> z=2
Compusul este C7H6O2
-deduc compozitia procentuala
122g compus.......84gC.......6gH......32gO
100g........................%C?.....%H?.....%O?