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17 December, 10:35

What is the calculated empirical formula of the copper oxide if a 0.626 g sample of the copper oxide was reduced to 0.500 g of copper metal by heating in a stream of hydrogen gas?

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  1. 17 December, 10:42
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    CuO

    Explanation:

    Empirical formula is the simplest ratio of atoms presents in a compound.

    The reduction of copper oxide to copper metal in hydrogen is:

    CuₐOₓ + x H₂ → a Cu + x H₂O

    Moles of copper metal after reaction (Cu = 63.546g/mol) are:

    0.500g * (1mol / 63.546g) = 7.868x10⁻³mol Cu

    Now, if mass of copper is 0.500g, mass of oxygen in the oxide is:

    0.626g - 0.500g = 0.126g O. In moles:

    0.126g * (1mol / 16g) = 7.875x10⁻³mol O

    Thus, ratio between moles of Cu and moles of O is:

    7.868x10⁻³mol Cu / 7.875x10⁻³mol O = 0.999 ≅ 1

    That means empirical formula of copper oxide is Cu₁O₁ = CuO
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