Stvarna i empiriska formula
Submitted by gxxrjj on Sat, 10/17/2015 - 18:41

Forums:
Mento(M=156,3) sadrzi ugljenik, vodonik i kiseonik. Sagorevanjem 0,1595g mentola dobija se 0.446g CO2 i 0.184g H2O. Izracunati stvarnu i empirisku formulu mentola.
2C10H20O+29O2-------------
2C10H20O+29O2------------->20CO2+20H2O
n(mentola)=m/Mr=0.1595/156.3=0.001 mol
n(O)=n(mentola)=0.001 mol
n(CO2)=m/Mr=0.446/44=0.01 mol
n(H2O)=m/Mr=0.184/18=0.01 mol
n(C)=n(CO2)=0.01 mol
n(H)=2*n(H2O)=2*0.01=0.02 mol
n(C):n(H):n(O)=0.01:0.02:0.001/*1000
n(C):n(H):n(O)=10:20:1 i stvarna formula je C10H20O
Mr=10*Ar(C)+20*Ar(H)+x*Ar(O)
156.3=10*12+20*1+16*x
16.3=16*x
x=16.3/16=1 atom kiseonika i empirijska formula je C10H20O