100% - 53,33 - 11,11 = 35,56% O (aflat prin diferenta)
notam compusul CaHbOc
100% compus ............ 53,33% C ......... 11,11% H .... 35,56% O
90g ............................... 12a ..................... b .................. 16c
=> a = 4, b = 10, c = 2
=> C4H10O2
N.E. = 2a+2-b/2 = 10-10/2 = 0
=> un compus real saturat
