Ask Question
2 March, 03:44

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?

+2
Answers (1)
  1. 2 March, 05:26
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers