Establish the empirical formula of the substance of the mass fraction of potassium, sulfur

Establish the empirical formula of the substance of the mass fraction of potassium, sulfur and oxygen in which are equal, respectively: 44.91%, 18.39%, 36.78%

Given:
KxSyOz
ω (K) = 44.91%
ω (S) = 18.39%
ω (O) = 36.78%

To find:
KxSyOz -?

Solution:
1) Let m (KxSyOz) = 100 g;
2) m (K) = ω (K) * m (KxSyOz) / 100% = 44.91% * 100/100% = 44.91 g;
3) m (S) = ω (S) * m (KxSyOz) / 100% = 18.39% * 100/100% = 18.39 g;
4) m (O) = ω (O) * m (KxSyOz) / 100% = 36.78% * 100/100% = 36.78 g;
5) n (K) = m (K) / M (K) = 44.91 / 39 = 1.152 mol;
6) n (S) = m (S) / M (S) = 18.39 / 32 = 0.575 mol;
7) n (O) = m (O) / M (O) = 36.78 / 16 = 2.299 mol;
8) x: y: z = n (K): n (S): n (O) = 1.152: 0.575: 2.299 = 2: 1: 4;
Unknown substance – K2SO4 – potassium sulfate.

Answer: Unknown substance – K2SO4 – potassium sulfate.