Question

Urgent va rog!! 6,7 g dintr-un fenol, formeaza prin ardere 10,08 L (c.n) CO2 si 4,5 g H2O. Scrie formulele structurale ale izomerilor fenolului, care prezinta izomerie geometrica

Answer 1

n=m/M=V/Vm => m=M*V/Vm=44*10,08/22,4=19,8g CO2

44g CO2............12g C
19,8g CO2.........x=5,4g C

18g H2O..................2g H2
4,5g H2O...............x=0,5g H2

mO = 6,7-5,4-0,5=0,8g

n=m/M=5,4/12=0,45 moli C

n=m/M=0,5/2=0,25 moli H2

n=m/M=0,8/16=0,05 moli O

=> 9 atomi C, 10 atomi H, 1 atom O

F.M => C9H10O. M=134g/mol

NE=5

℅C=300a/11s=300*19,8/11*6,7=80,597

℅H=100b/9s=100*4,5/9*6,7=7,462

℅O=100-℅C-℅H=11,94

nC = ℅C*M/AC*100=80,597*134/1200=9 atomi C

nH=10 atomi

nO=1 atom

F.M => C9H10O


F.S => CH2=CH-CH2-C6H4-OH, CH3-CH=CH-C6H4-OH, etc