Răspuns:
Propunere de rezolvare
R-COO-CH3
unde, R= radical alchil CnH2n+1
M KOH = 56 g/ mol
m ester =1 g
m KOH=0,2616g
Rezulta:
1g ester.....0,2616 g KOH
RCOOCH3 + KOH ->> RCOOK + CH3OH
M....................56g/ mol
de unde:
M ester = 1g.56 g/mol/ 0,2616 g =214 g/ mol
CnH2n+1 COOCH3, cu M=214 g/mol
M = n.AC+ 2nAH+AH+2, AC+2.AO+3AH
M=12n+2n+1+2.12+2.16+3.1
214-60=14n
n=11
Deci radicalul acil are 12 C.
R: 12 atomi C