Question

Calculati compozitia procentuala pentru acidul benzoic si acidul butilic

Answer 1

Acidul benzoic : C₆H₅-COOH.
Acidul butilic sau butiric: CH₃-CH₂-CH₂-COOH
M ac. benzoic = M C₇H₇O₂=12×7+6×1+16×2=84+6+32=90+32=122g/mol
M ac. butilic = M C₄H₈O₂=12×4+8×1+16×2=48+8+32=48+40=88g/mol
122g.............100%                                    88g..................100%
  84g................x%                                    48g.....................t%
    6g................y%                                     8g.....................p%
  32g................z%                                    32g....................m%
x=(84×100)÷122=68,85%C                      t=(48×100)÷88=54,54%C
y=(6×100)÷122=4,91%H                         p=(8×100)÷88=9,10%H
z=(32×100)÷122=26,23%O                    m=(32*100)÷88=36,36%O