CxHyOz
M= 74
nr de moli subst = 0.0222/74 = 0.3 mmoli
nr de moli CO2 = 0.0528÷44 = 1.2mmoli
nr de moli H2O = 0.0270÷18= 1.5 mmoli
2....................................2x.............y
2CxHyOz + aO2 = 2xCO2 + yH2O
0.3..................................1.2...........1.5
x = 4
y =10
C4H10Oz
M = 4×12 + 10×1 + 16 × z = 74
74-58 = 16z
z=1
C4H10O
c%C = 48÷74 ×100 =64.86%
c%H = 10÷74×100= 13.51%
c%O = 16÷74×100 = 21.63%