A )
C = 52.1 %
H = 13.1%
O = 100-52.1-13.1 = 34.8 %
[tex]C : \frac{52.1}{12} =4.34\\H : 13.1\\O : \frac{34.8}{16} = 2.17 \\[/tex]
impartim la 2.17 si rezulta 2 atomi de C , 6 atomi de H , 1 atom de O
(C2H6O) n
B )
CH3 - CH2 - OH etanol (alcool etilic )
CH3-O-CH3 dimetil eter
C ) CH3-CH2-OH