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5 August, 09:31

A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empirical formula of the sample? Give your answer in the form C#H#N#O# where the number following the element's symbol corresponds to the subscript in the formula. (Don't include a 1 subscript explicitly). For example, the formula CH 22 O would be entered as CH2O.

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  1. 5 August, 09:55
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    Let us assume that there is a 100g sample present. The respective mass of each element will then be:

    C: 74 g

    H: 7.4 g

    N: 8.6 g

    O: 10 g

    Now, we divide each constituent's mass by its Mr to obtain the moles of each

    C: (74 / 12) = 6.17

    H: (7.4 / 1) = 7.4

    N: (8.6 / 14) = 0.61

    O: (10 / 16) = 0.625

    Dividing by the smallest number:

    C: 10

    H: 12

    N: 1

    O: 1

    Thus, the empirical formula is

    C10H12NO
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