Sa se determina formula moleculara:
Acid monocarboxilic provenit de la o alchena si despre care se stie ca 21,6 g de acid reactioneaza cu 12g de NAOH


Răspuns :

CnH2n-2O2+NaOH--->CnH2nONa+H2O

M NaOH=40g/mol
n NaOH=m/M=12/40=0,3 moli NaOH
Reactia e mol la mol===> 0,3 moli acid
n=m/M=>M=m/n=21,6/0,3=72
12n+2n+32-2=72
14n+30=72==>n=3 C3H4O2