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
