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16 February, 13:22

A compound contains 57.2 percent carbon, 6.1 percent hydrogen, 9.5 percent nitrogen, and 27.2 percent oxygen. What the empirical formula of the compound?

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  1. 16 February, 16:14
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    So the empirical formula is C14H18N2O5

    Explanation:

    C = 57.2% = 12g/mol

    H = 6.1% = 1g/mol

    N = 9.5% = 14g/mol

    O = 27.2% = 16g/mol

    Empirical Formula for compound hmm

    Assume

    C = 57.2g

    H = 6.1g

    N = 9.5g

    O = 27.2g

    So we have

    C = 57.2g/12g = 4.76 moles

    H = 6.1g/1g = 6.10 moles

    N = 9.5g/14g = 0.68 moles

    O = 27.2g/16g = 1.70 moles

    Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number.

    C = 4.76 moles / 0.68 moles = 7

    H = 6.10 moles / 0.68 moles = 9

    N = 0.68 moles / 0.68 moles = 1

    O = 1.70 moles / 0.68 moles = 2.5

    Ok so we now have the ratios but for O it's 2.5, have to be whole numbers so we will need to double all the numbers.

    C = 14

    H = 18

    N = 2

    O = 5

    So the empirical formula is C14H18N2O5
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