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19 January, 10:39

What is the empirical formula of a compound that is 62.0% c, 10.4% h, and 27.5% o by mass?

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  1. 19 January, 11:03
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    C=62.0/12 H=10.4/1 O=27.5/16, C=5.1667/1.71875 H=10.4/1.71875 O=1.71875/1.71875, C=3 H=6 O=1 Empirical formula=C3H6O
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