An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω (Br) = 43.66% : 100%.

ω (Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m (Br) = ω (Br) · 100 g.

m (Br) = 0.4366 · 100 g.

m (Br) = 43.66 g; mass of bromine.

n (Br) = m (Br) : M (Br).

n (Br) = 43.66 g : 79.9 g/mol,

n (Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n (Br) = n (K).

m (K) = 0.55 mol · 39.1 g/mol.

m (K) = 21.365 g; mass of potassium in the compound.

m (O) = 100 g - 21.365 g - 43.66 g.

m (O) = 34.97 g; mass of oxygen.

n (O) = 34.97 g : 16 g/mol.

n (O) = 2.185 mol.

n (K) : n (Br) : n (O) = 0.55 mol : 0.55 mol : 2.185 mol / : 0.55 mol.

n (K) : n (Br) : n (O) = 1 : 1 : 4.