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    b)Iron Oxide (11.16g of iron reacts to form 15.96g of Iron Oxide)

    What is the empirical formula ???

    This is all I managed to get to

    b) First we work out how many moles of Fe = 11.2/55 = 0.2 moles
    Then for Oxygen = 15.96-11.2 = 4.74g/16 = 0.3 moles

    So where do I go from here ... ??

    any help is greatly appreciated.

    kind regards


    Sunny
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    Multiply both number of moles by 10 as we want to get them in whole numbers so it will be 2 and 3.
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    Divide by the smallest moles
    Fe- 0.2/0.2=1
    O- 0.3/0.2=1.5
    Times by 2 to get whole number
    Fe- 1x2=2
    O- 1.5x2=3

    Empirical formula = Fe2O3
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    Basically Fe2O3

    Iron comes in different forms. It can come in fe(III) or fe(II) so to balance out the charges iron is in its III form.
 
 
 
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