Răspuns:
%C=40: 12 (A C) = 3,33:3,33= 1 atg
%H=6,67: 1 (A H)= 6,67:3,33= 2 atg
%O= 100-(40+6,67)=53,33: 16=3,33:3,33= 1 atg
=> F. chimica bruta e (CH₂O)ₙ
M= 12n+2n+16n=30n
M= 180 g/mol => n=180/30=6
=> f.chim Moleculara: C₆H₁₂O₆
NE= 2*6+2-12/2=1 ∈ Z₊
nr.covalente= 4*6+12*1+6*2=48 - nr par