stehiometrija
Submitted by lejla93 on Sat, 03/22/2014 - 23:04

Forums:
elementarnom analizom je ustanovljeno da jedan spoj sadrzi 15,4% C, 3,2%H, i 81,4% J.Kakvu formulu ima ovo jedinjenje ako 12,5 cm3 na 20 stepeni C i 752 mmHg ima masu od 0.081 g?
p=752*133.322=100.258 kPa
p=752*133.322=100.258 kPa
n(X)=V*p/R*T
n(X)=12.5*10(-6)*100.258*10(3)/293.15*8.314
n(X)=5.14*10(-4) mol
Mr=m/n(X)=0.081/5.14*10(-4)=157.6 g/mol
w(C)=x*Ar(C)/Mr
x*12=157.6*0.154x=24/12=2 atoma C
w(H)=y*Ar(H)/Mr
y*1=157.6*0.032
y=5/1=5 atoma H
w(J)=z*Ar(J)/Mr
z*127=157.6*0.814
z=128/127=1 atom J i formula je C2H5J odnosno CH3CH2J