Question

1) In urma arderii a 8,2 g hidrocarbura se obtin 26,4 g CO2. Compozitia in procente de masa a hidrocarburii este:
2) Formula moleculara a unui compus organic care contine,in procente masice,53,33℅ C , 11,11℅ H si are masa molara 90g/mol este:
3) Compusul cu formula moleculara C7H8O contine: a) 14,81℅ O
b) 10,26℅ C
c) 10,26℅ H
d) 66,66℅ C.
Am nevoie de ajutor. Multumesc anticipat.

Answer 1

1)
notam formula hidrocarburii: CxH2y
Mhidrocarbura=12x+2y g/mol
nCO2=26,4/44=0,6moli
12x+2y.............................x
CxH2y + (2x+y)/2 O2 -> xCO2 + yH2O
8,2g...............................0,6moli
12x+2y=8,2*44x/26,4=13,6666666x
2y=1,666666666x
x=1,2y
x:y=1,2:1 |*5=x:y=6:5 (am inmultit cu 5 ca sa avem 1,2 numar intreg)
f.m: C6H10
M=12*6+2*5=72+10=82g/mol
%C=MC*100/82=72*100/82=87,8% C
%H=MH*100/82=10*100/82=12,2% H
2)
deoarece procentele adunate nu dau 100% => avem si 100-53,33-11,11=35,56% O
.............................|:2,22
C: 53,33/12=4,44      2
H: 11,11/1=11,11      5
O: 35,56/16=2,22     1
formula bruta: (C2H5O)n
Mbruta=(12*2+5+16)n=45n=90 => n=2
f.m: C4H10O2
3)
Mcompus=12*7+8+16=108g/mol
%C=72*100/108=66,66% C
%H=8*100/108=7,4% H
%O=16*100/108=14,81% O
corecte: a) si d)