Question

16,12 g de trigliceridă simplă se saponifică cu 80 mL de solutie NaOH de concentratie 1 M.Excesul de NaOH este neutralizat de 10 mL solutie HCl de concentratie 2M.Determinati formula de structura a trigliceridei.

Answer 1

CH2-O-CO-(CH2)n-CH3          
CH-O-CO-(CH2)n-CH3      + 3 NaOH  ---> 3 CH3-(CH2)n-COONa
CH2-O-CO-(CH2)n-CH3
cM=nr moli/V => nr moli = 0,01*2 = 0,02 moli HCl
NaOH + HCl ---> NaCl + H2O
Reactia este mol la mol => 0,02 moli NaOH exces
cM=nr moli/V=> nr moli = 0,08*1 = 0,08 moli NaOH
nr moli reactionat = 0,08-0,02 = 0,06 moli
masa molara triglicerida = 73*3-1+42n = 42n+218
(42n+218) g triglicerida..........3 moli NaOH
16,12 g triglicerida................0,06 moli NaOH   
=> 13,08+2,52n=48,36=>2,52n=35,28=>n=14=>
CH2-O-CO-(CH2)14-CH3
CH-O-CO-(CH2)14-CH3
CH2-O-CO-(CH2)14-CH3         
Triglicerida este tripalmitina