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.

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