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7 May, 12:19

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

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  1. 7 May, 15:29
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    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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