Question

In urma analizei a 6,32 g substanta organica care contine in molecula atomi de carbon,hidrogen,oxigen si un otom de sulf s-au obtinut 2,688 litri dioxid de carbon, 2,16 g apa si 9,56 g sulfura de plumb.Determinati formula moleculara a substantei.

Answer 1

m/M=V/Vm=>m=M*V/Vm=44*2,688/22,4=5,28g CO2

MCO2=44g/mol

In 44g CO2 se gasesc............12g C
In 5,28g CO2...........................x=1,44g C

MH2O=18g/mol

In 18g H2O se gasesc............2g H
In 2,16g H2O..........................x=0,24g H

MPbS=APb+AS=207,2+32=239g/mol

In 239,2g PbS se gasesc.................32g S
In 9,56g PbS...............................x=1,2789gS

mO=6,32-1,44-0,24-1,28=3,36g O

n=m/M=1,44/12=0,12 moli C

n=m/M=0,24/1=0,24 moli H2

n=m/M=1,28/32=0,04 moli S

n=m/M=3,36/32=0,105 moli O2

Dupa impartire => 3 atomi C, 6 atomi H, 1 atom S, 2,625 atomi O

F. B = (C3H6SO2,625)n ?

Imi scapa ceva :(