Question

o substanta organinca care contine carbon,hidrogen si clor cu (miu)=127 g/mol este supusa analizei.Din 1,27g de subst. 1-au obtinut 2,82 g AgCl si 1,76 CO2.Det
a)f moleculara
b)f de str. posibile

va rog orice rezolvare,chiar daca este gresita

Answer 1

nAgCl = 2,87/143,5 = 0,02 moli = nCl   mCl = 0,02··35,5 = 0,71g
nCO2 = 1,76/44 = 0,04 moli = nC         mC = 0,04·12 = 0,48 g
mH = 1,27 - (0,71 + 0,48) = 0,08g   nH = 0,08 atomi
n subst. = 1,27/127 = 0,01 moli
0,01moli subst......contin........0,04 at. C...........0,08 at.H............0,02 at.Cl
1mol..................................x = 4.......................y = 8....................z = 2
a) C4H8Cl2
b) Cl2CH-CH2-CH2-CH3;  Cl-CH2-CHCl-CH2-CH3; Cl-CH2-CH2-CHCl-CH3; Cl-CH2-CH2-CH2-CH2-Cl;  CH3-CHCl2-CH2-CH3; CH3-CHCl-CHCl-CH3;
Cl2CH-C(CH3)-CH3; Cl-CH2-(CH3)CCl-CH2-CH3; ClCH2-C(CH3)-CHCl-CH3;
ClCH2-C(CH3)-CH2-CH2Cl; CH3-(CH3)CCl-CHCl-Ch3