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28 August, 22:20

Calculate the empirical formula the compound with the following percent composition: 27.59%C 1.15%H 16.09%N 55.17%O

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  1. 29 August, 01:00
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    Answer: C2HNO3

    Explanation: C = 27.59/12.011 = 2.297

    H = 1.15/1.008 = 1.1409

    N = 16.09/14.007 = 1.1487

    O = 55.17/15.999 = 3.4483

    Divide by smallest result:

    C = 2

    H=1

    N=1

    O = 3

    Empirical formula = C2HNO3
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