1.) Assume 100 grams of a compound. Find the moles for each atom.

39.45% K → 39.45 g K × (1 mol K / 39.1 g K) = 1.01 mol K

12.12% C → 12.12 g C × (1 mol C / 12.01 g C) = 1.01 mol C

48.43% O → 48.43 g O × (1 mol O / 16.00 g O) = 3.03 mol O

2.) Find the smallest whole-number ratio of atoms.

mol K: 1.01/1.01 = 1 × 3 = 3

mol C: 1.01/1.01 = 1 × 3 = 3

mol O: 3.03/1.01 = 3 × 3 = 9

Empirical Formula: K_{3}C_{3}O_{9}

3.) Use the empirical formula and molar mass to find the molecular formula.

3(39.1) + 3(12.01) + 9(16.00) = 117.3 + 36.03 + 144 = 297.33 g/mol

Multiplication Factor = molar mass of molecular formula/molar mass of empirical formula

= (396.44 g/mol) / (297.33 g/mol)

= 1.33... or 4/3

Molecular Formula: 4/3 × (K_{3}C_{3}O_{9}) = **K**_{4}**C**_{4}**O**_{12}

J.R. S.

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