What is the molecular formula of the compound that contains 39.45% K, 12.12% C, and 48.43% O and has a molecular mass of 396.44 g/mo

Answer 1

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: K₃C₃O₉

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 × (K₃C₃O₉) = K₄C₄O₁₂

Answer 2

Assuming 100 g of compound:

39.45 % K = 39.45 g K x 1 mol K/39 g = 1.01 moles K

12.12% C = 12.12 g C x 1 mol C/12 g = 1.00 moles C

48.43% O = 48.43 g O x 1 mol O/16 g = 3.0 moles O

Empirical formula = KCO₃

To find molecular formula, we need to find how many of the empirical formulae fit into the molar mass of the molecular formula.

The molar mass of KCO₃ = 39 + 12 + 3(16) = 99 g/mol

How many times does this go into 396.44?

396.44 / 99 = 4.00

Molecular formula = K₄C₄O₁₂