Question

3. When 10.0 g of a pure compound was analyzed, it was found to contain 3.65 g of K, 3.33 g of CI,
and 3.02 g of O. Determine the empirical formula of the compound.

Answer 1

Convert all the grams to moles first.

K - 3.65 g / 39.1 g/mol = 0.0934 mol
Cl - 3.33 g / 35.45 g/mol = 0.0939 mol
O - 3.02 g / 16 g/mol = 0.189 mol

Divide them all by the smallest number of moles (0.0934 moles of K) to get whole numbers.

K - 0.0934/0.0934 = 1
Cl - 0.0939/0.0934 ≈ 1
O - 0.189/0.0934 ≈ 2

KClO2