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23 August, 14:00

A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The following data was collected: Mass of crucible: 38.26 g Mass of crucible and iridium: 39.52 g Mass of crucible and iridium oxide: 39.73 g Show, or explain, your calculations as you determine the empirical formula of the compound.

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  1. 23 August, 17:09
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    1) Compund Ir (x) O (y)

    2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

    3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

    4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

    5) Convert grams to moles

    moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

    moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

    6) Find the proportion of moles

    Divide by the least of the number of moles, i. e. 0.00656

    Ir: 0.00656 / 0.00656 = 1

    O: 0.0131 / 0.00656 = 2

    => Empirical formula = Ir O2 (where 2 is the superscript for O)

    Answer: Ir O2
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