A compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give barium chromate. If a 1.345-g sample of the compound gave 2.012 g bacro4, what is the formula of the compound?

The empiricial formula of the compound is BaO2

Explanation:

Step 1: Data given

Barium and oxygen dissolved in hydrochloric acid gives a solution of barium ion. This was precipitated with an excess of potassium chromate and gives barium chromate.

The original compound weighs 1.345g and gives 2.012g of BaCrO4

Step 2: Calculate moles of BaCrO4

moles of BaCrO4 = mass of BaCrO4 / molar mass of BaCrO4

moles of BaCrO4 = 2.012g / 253.37 g/mol = 0.0079 moles

Step 3: Calculate moles of Ba

Mole ratio for Ba and BaCrO4 is 1:1 so this means for 0.0079 moles of BaCrO4, there are 0.0079 moles of Ba-ion

Step 4: Calculate mass of Ba-ion

Mass of Ba = Moles of Ba / Molar mass of Ba

Mass of Ba = 0.0079 moles * 137.327g/mole = 1.085 g Ba

Step 5: Mass of oxygen

Since the original compound has barium and oxygen, the mass of oxygen is the difference between the original mass and the mass of the Ba-ion

1.345 g - (1.085 g Ba) = 0.260 g O

Step 6: Calculate moles of Oxygen

moles oxygen = mass of oxygen / Molar mass of oxygen

moles oxygen = 0.260g / 16g/mole = 0.01625 moles O

We divide the number of moles by the smallest number of moles which is 0.0079

Ba → 0.0079/0.0079 = 1

O → 0.01625 / 0.0079 ≈ 2

(1.091 g Ba) / (137.3277 g Ba/mol) = 0.00794450 mol Ba

(0.254 g O) / (15.99943 g O/mol) = 0.0158756 mol O

This gives us the empirical formula of BaO2 for this compound