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    please explain this question!

    analysis of a compound found it contained:
    76.0% lead (Pb)
    13.0% chlorine (Cl)
    2.2% carbon (C)
    8.8% oxygen (O)

    what is the empirical formula of this compound?

    relative atomic masses: C=12, O=16, Cl=35.5, Pb=207
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    u divide each thing by its relative atomic mass:
    Pb: 76/207 = 0.367
    Cl:13/35.5 = 0.366
    C: 2.2/12 = 0.183
    O:8.8/16 = 0.55

    Then u divide each by the smallest:
    Pb: 0.367/0.183 = 2
    Cl: 0.366/0.183 = 2
    C: 0.183/0.183 = 1
    O:0.55/0.183 = 3
    so the empirical formula is : Pb2Cl2CO3
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    the empirical formula is the simplest whole ratio of the atoms of each element in a compound

    STEPS TO WORK IT OUT:
    1) divide each % by the relative atomic mass (e.g. 76/207)
    2) with the results obtained in 1) ... divide each result by the smallest result obtained
    3) this should give you a ratio in moles of each element!

    hope this helps!
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    Divide the percentages by their relative atomic masses. When you have those answers, find the smallest one and divide all the answers by that.
    Once rounded, those answers will give you your empirical formula.
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    Thanks!!! that really helped, it was easier to understand than my revision guide.
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    :- Yeah it's Pb2Cl2O3
 
 
 
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