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3 December, 16:36

A 10.81 g sample of a compound contains 3.45 g of potassium, K, 3.13 g of chlorine, Cl, and oxygen, O. Calculate the empirical formula. Insert subscripts as needed. Empirical formula:

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  1. 3 December, 16:54
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    The empirical formula is KClO3

    Explanation:

    Step 1: Data given

    Mass of the sample = 10.81 grams

    Mass of potassium = 3.45 grams

    Mass of chlorine = 3.13 grams

    molar mass O = 16.0 g/mol

    Molar mass K = 39.1 g/mol

    Molar mass of chlorine = 35.5 g/mol

    Step 2: Calculate mass of oxygen

    Mass of oxygen = mass of sample - mass of potassium - mass of chlorine

    Mass oxygen = 10.81 grams - 3.45 grams - 3.13 grams

    Mass oxygen = 4.23 grams

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles K = 3.45 grams / 39.1 g/mol

    Moles K = 0.0882 moles

    Moles Cl = 3.13 grams / 35.5 g/mol

    Moles Cl = 0.0882 moles

    Moles O = 4.23 grams / 16.0 g/mol

    Moles O = 0.264 moles

    Step 4: Calculate mol ratio

    We divide by the smallest amount of moles

    K: 0.0882 / 0.0882 = 1

    Cl: 0.0882 / 0.0882 = 1

    O: 0.264/0.0882 = 3

    for 1 mol K we have 1 mol Cl and 3 moles O

    The empirical formula is KClO3
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