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3 January, 16:09

What is the empirical formula for a compound which contains 67.1 zinc and the rest is oxygen

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  1. 3 January, 18:58
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    The empirical formula is ZnO2

    Explanation:

    What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

    Step 1: Data given

    Suppose the compound has a mass of 100.0 grams

    A compound contains:

    67.1 % Zinc = 67.1 grams

    100 - 67.1 = 32.9 % oxygen = 32.9 grams

    Molar mass of Zinc = 65.38 g/mol

    Molar mass of O = 16 g/mol

    Step 2: Calculate moles of Zinc

    Suppose the compound is 100 grams

    Moles Zn = 67. 10 grams / 65.38 g/mol

    Moles Zn = 1.026 moles

    Step 3: Calculate moles of O

    Moles O = 32.90 grams / 16.00 g/mol

    Moles O = 2.056 moles

    Step 4: Calculate mol ratio

    We divide by the smallest amount of moles

    Zn: 1.026/1.026 = 1

    O: 2.056/1.026 = 2

    The empirical formula is ZnO2

    To control this we can calculate the % Zinc for 1 mol

    65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
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