1 mol aminoacid...........x*46 g etanol
1,2 moli aminoacid...........55,2 g etanol
x = 1 => se aditioneaza 1 mol etanol => aminoacid monocarboxilic
1 mol acid.............x g compus
1.2 moli acid..............140.4 g compus
x = 117 g
CnH2n+1NO2
%C= m C/masa produs*100 => 12n/14n+47*100=51.28
=>1200n=717.92n+2410.16=>n=5 => C5H11NO2 => 3 atomi de carbon are aminoacidul => alanina