Question

Determinati formula moleculara a substantei organice daca din arderea 3,12g substanta cu M=104g/mol se obtin 3,36 dm3 CO2 si 3340mg H2O.

Answer 1

1L=1dm3
1g=1000mg
m/M=V/Vm=>m=M*V/Vm=44*3.36/22.4=6.6g CO2
mH2O=3.24g H2O
s.o=3.12g
a=cantitatea de CO2
s=cantitatea de s.o
b=cantitatea de H2O
℅C=300a/11s=300*6.6/11*3.12=1980/34.32=57.6923℅
℅H=100b/9s=100*3.24/9*3. 12=324/28.08=11.538
℅O=100-(℅C+℅H)=100-69.23=30.769
nC=℅C*M/AC*100=57.6923*1 04/1200=5
nH=11.538*104/100=12
nO=30.769*104/1600=2
Formula moleculară este C5H12O2

Answer 2

Teorie: Masa molara a unei substante cuprinde 22.4 l in conditii normale
Formula bruta se poate determina cunoscand masa de elemente din fiecare component
formula moleculara=n*formula bruta
Calcule:
44 g dioxid contine 22,4 l dioxid
x g dioxid contine 3.36 l dioxid
x=6.6 g dioxid
44 g dioxid contin 12 g carbon
6.6 g dioxid contin y g carbon
y=1.8 g carbon
18 g apa contin 2 g hidrogen
3.24 g apa contin z g hidrogen
z=0.36 g hidrogen
3.12-1.8-0.36=0.96 g oxigen
1.8/12=0.15
0.36/1=0.36
0.96/16=0.06
Prin impartire la 0.06=>formula bruta:C2.5H6O
formula moleculara:(C2,5H6O)n
masa molara=104=2.5*12n+6n+16n=>52n=104=>n=2=>formula moleculara: C5H12O2
Raspuns:
Substanta are formula moleculara C5H12O2