notam cei doi oxizi ai elementului E
E2On
E2Om
unde n si m = valenetele lui E in oxizi
M.E2On = 2Ae+16n
M.E2Om = 2Ae+16m
2Ae+16n/2Ae+16m = 4/5
=> 10Ae+80n = 8Ae+64m
=> 2Ae+80n = 64m
m-n = 2 => m = n+2
=> 2Ae+80n = 64(n+2)
=> 2Ae+16n = 128
notam cu a = nr moli din fiecare oxid
100% ................................................... 44,44% E
(2Ae+16n)a+(2Ae+16m)a .................. 4aAe
-----------------------------------
4aAe+32na+32a
=> 9,091aAe = 4aAe+32na + 32a
=> 0,16Ae = n
=> 2Ae + 2,545Ae = 128
=> Ae = 28 u.a.m.
=> E = Si