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7 August, 14:33

A compound has a percent composition of 24.74% potassium (molar mass = 39.10 g/mol), 34.76% manganese (molar mass = 54.94 g/mol), and 40.50% oxygen (molar mass = 16.00 g/mol). Assuming that the mass of the compound is 100 g, what is the compound’s empirical formula?

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  1. 7 August, 14:36
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    The empirical formua is KMnO4

    Explanation:

    Step 1: Data given

    Mass of the compound = 100 grams

    24.74% = potassium

    Molar mass of K = 39.10 g/mol)

    34.76% = manganese

    Molar mass of Mn = 54.94 g/mol

    40.50% = oxygen

    Molar mass = 16.00 g/mol

    Step 2: Calculate moles of potassium

    Moles K = mass K / molar mass K

    Moles K = 24.74 grams / 39.10 g/mol

    Moles K = 0.6327 moles

    Step 3: Calculate moles of Mn

    Moles Mn = 34.76 grams / 54.94 g/mol

    Moles Mn = 0.6327 moles

    Step 4: Calculate moles of O

    Moles O = 40.50 grams / 16.00 g/mol

    Moles O = 2.53

    Step 5: Calculate the mol ratio

    We divide by the smallest amount of mol

    K: 0.6327 / 0.6327 = 1

    Mn: 0.6327/0.6327 = 1

    O: 2.53 / 0.6327 = 4

    The empirical formua is KMnO4
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