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24 January, 00:00

A 12.95 g sample of Mo2O3 (s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.82 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

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  1. 24 January, 00:30
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    (18.75 g Mo2O3) x [ (1 mole Mo2O3) / (239.88 g) ] x [ (2 moles Mo) / (1 mole Mo2O3) ] = 0.1563 moles

    We know that Mox = 0.1563 moles

    Mo = 15.00 g Mo, which in turn means that there must be (22.50 - 15.00) grams of O or 7.50 g O, which converts to:

    (7.50 g O) x [ (1 mole O) / (15.999 g O) ] = 0.4688 moles O

    Taking the ratio of Mo to O results in:

    O/Mo = [ (0.4688) / (0.1563) ] = 3
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