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21 July, 16:31

A 35.161 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 62.637 mg of carbon dioxide and 25.641 mg of water. In another experiment, 31.321 mg of the compound is reacted with excess oxygen to produce 13.54 mg of sulfur dioxide. Add subscripts to the formula provided to correctly identify the empirical formula of this compound. Do not change the order of the elements.

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  1. 21 July, 19:35
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    The empirical formulae is C6H12S02

    Explanation:

    1. First we need to obtain the mass of each element in the sample and compound formed

    Carbon = (62.637 mg * 12.011 g/mol / 44.009 g/mol) = 17.094 mg of Carbon

    Hydrogen = (25.641 mg * (2 * 1 ... 008 g/mol) / 18.015 g/mol) = 2.869 mg of Hydrogen

    Sulphur = (13.54 mg * 32.066 g/mol / 64.066 g/mol) = 6.777 mg of Sulphur

    2. Next is to determine the percentage composition. Here we divide the respective mass by the mass of the sample

    Carbon = 17.094 / 35.161 * 100 = 48.62 %

    Hydrogen = 2.869 / 35.161 * 100 = 8.16 %

    Sulphur = 6.777 / 31.321 * 100 = 21.64 %

    Oxygen = (100 - (48.62 + 8.16 + 21.64)) = 21.58 %

    3. Next is to divide the mass assuming there are 100 mg by the respective atomic masses to obtain the number of moles

    Carbon = 48.62 / 12.011 = 4.048 mol

    Hydrogen = 8.16 / 1.008 = 8.095 mol

    Sulphur = 21.64 / 32.066 = 0.675 mol

    Oxygen = 21.58 / 16.000 = 1.348 mol

    Next is to divide by the smallest value

    Carbon = 4.048 / 0.675 = 5.997 = 6

    Hydrogen = 8.095 / 0.675 = 11.993 = 12

    Sulphur = 0.675 / 0.675 = 1

    Oxygen = 1.348 / 0.675 = 1.997 = 2

    So therefore the empirical formulae of the sample is C6H12SO2
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