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10 July, 07:03

A compound is 53.31 % C, 11.18 % H, and 35.51 % O by mass. What is its empirical formula? Insert subscripts as needed.

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  1. 10 July, 07:28
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    The empirical formula is C2H5O

    Explanation:

    Step 1: Data given

    Suppose the mass of the compound = 100.0 grams

    The compound contains:

    53.31 % C = 53.31 grams C

    11.18 % H = 11.18 grams H

    35.51 % O = 35.51 grams O

    Step 2: Calculate moles

    moles = mass / molar mass

    Moles C = 53.31 grams / 12.01 g/mol

    Moles C = 4.439 moles

    Moles H = 11.18 grams / 1.01 g/mol

    Moles H = 11.07 moles

    Moles O = 35.51 grams / 16.0 g/mol

    Moles O = 2.22 moles

    Step 3: Calculate mol ratio

    We divide by the smallest amount of moles

    C: 4.439 / 2.22 = 2

    H: 11.07 / 2.22 = 5

    O: 2.22/2.22 = 1

    C2H5O

    The empirical formula is C2H5O
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