Determine the empirical formula of a compound containing 83% potassium and 17.0% oxygen.

KO

Explanation:

Assuming those are weight percentages, we first convert to moles.

83 g K * (1 mol / 39 g) ≈ 2.128 mol K

17 g O * (1 mol / 8 g) ≈ 2.125 mol K

Normalize by dividing by the smallest:

2.128 / 2.125 ≈ 1.00

2.125 / 2.125 = 1.00

So the empirical formula is:

KO

(75°f − 32) × 5/9 = 23.889°c

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