A compound found in the athletes urine had a percent composition of 80.8% carbon, 8.97% hydrogen and 10.3% oxygen. What is the Empirical for

Question

A compound found in the athletes urine had a percent composition of 80.8% carbon, 8.97% hydrogen and 10.3% oxygen. What is the Empirical formula of the compound? What is the molecular formula of the compound if the molar mass of the compound is 312g/mol?​

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Thiên Ân 5 years 2021-09-05T11:42:00+00:00 1 Answers 18 views 0

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  1. Euphemia
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    2021-09-05T11:43:43+00:00
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    Answer:

    C21H2802

    Explanation:

    C=12g/mol

    H=1g/mol

    O=16g/mol

    Part (C) of compound-80.18%

    (0.8018 x 312)/12=21

    Part (H) of compound-8.97%

    (0.897 x 312)/1=28

    Part (O) of compund-10.3%

    (0.103 x 312)/16 = 2

    Therefore the emp. formula is C12H28O2

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