100%-53,33% C - 11,11% H = 35,56% O
notam compusul CaHbOc
unde avem 12a g C + b g H + 16c g O
=>
100% .............. 53,33% C ........ 11,11% H ............. 35,56% O
90g ................ 12a g C ........ b g H ................. 16c g O
a = 4, b = 10 si c = 2
=> Fm = C4H10O2