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9 February, 04:36

A 1.345-g sample of 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 2.012 g of barium chromate, BaCrO4. What is the formula of the compound

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  1. 9 February, 06:40
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    The empirical formula is BaO2

    Explanation:

    Step 1: Data given

    Mass of compound = 1.345 grams

    Mass of barium chromate produced = 2.012 grams

    Molar mass BaCrO4 = 253.37 g/mol

    Step 2: Calculate moles BaCrO4

    Moles BaCrO4 = mass BaCrO4 / molar mass BaCrO4

    Moles BaCrO4 = 2.012 grams / 253.37 g/mol

    Moles BaCrO4 = 0.00794 moles

    Step 3: Calculate moles Ba

    In 1 mol BaCrO4 we have 1 mol Ba

    In 0.00794 moles BaCrO4 we have 0.00794 moles Ba

    Step 4: Calculate mass Ba

    Mass Ba = moles Ba * molar mass Ba

    Mass Ba = 0.00794 moles * 137.33 g/mol

    Mass Ba = 1.090 grams

    Step 5: Calculate mass O

    Mass O = mass compound - mass Ba

    Mass O = 1.345 grams - 1.090 grams

    Mass O = 0.255 grams

    Step 6: Calculate moles O

    Moles O = 0.255 grams / 16.0 g/mol

    Moles O = 0.0159 moles

    Step 7: Calculate mol ratio

    We divide by the smallest amount of moles

    Ba: 0.00794 / 0.00794 = 1

    O: 0.0159 / 0.00794 = 2

    This means for 1 Ba atom we have 2 O atoms

    The empirical formula is BaO2
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