Răspuns:
mA= 3,12g
miuA= 104 g/mol
Vco2 = 3,36 dm3 -> 3,36 L co2
mH2O =3240 mg-> 3,24 g H2O
Explicație:
Rezolvare:
niuCO2= V/V molar = 3,36/22,4 =0,15 moli CO2
mCO2= niuCO2 * miuCO2 = 0,15 * 44= 6,6 g CO2
44g CO2 …………12g C
6,6 g CO2………. X g C
X= 6,6*12/44= 1,8g C
miuH2O = 2+16=18 g/mol
18g H2O ………. 2gH
3,24g H2O …….. y g H
Y= 3,24*2/18= 0,36 g H
mc+mH = 1,8+0,36=2,16
mO= 3,12-2,16=0,96 g O
3,12g A…..1,8g C…..0,36g H…..0,96g O
100g……….a%…………b%………….c%
a=100*1,8/3,12=57,69% C
b=100*0,36/3,12=11,53% H
c=100*0,96/3,12= 30,76% O
57,69\12=4,807
11,53/1=11,53
30,76/16= 1,92
4,807/1,92= 2,5 at de C
11,53/1,92= 6 at de H
1,92/1,92= 1 at O
(C2,5 H6 O) n= 104
(12*2,5+6*1+16) n=104
52n=104 -> n=104:52 = 2
FORMULA MOLECULARA : C5H12O2 (alcan)
OH-CH2-CH2-CH2-CH2-CH2-OH
