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    Boron occurs naturally as a mixture of two isotopes Boron-10 and Boron-11. The relative isotopic masses are Boron-10 10.00 and Boron-11 11.00. Calculate the percentage abundance in samples with relative atomic masses of a.) 10.80 b.) 10.83. Please help.
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    (Original post by zara_ruby)
    Boron occurs naturally as a mixture of two isotopes Boron-10 and Boron-11. The relative isotopic masses are Boron-10 10.00 and Boron-11 11.00. Calculate the percentage abundance in samples with relative atomic masses of a.) 10.80 b.) 10.83. Please help.
    Write it as an equation relating the isotopic abundances to those averages.
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    (Original post by zara_ruby)
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    (Original post by zara_ruby)
    Boron occurs naturally as a mixture of two isotopes Boron-10 and Boron-11. The relative isotopic masses are Boron-10 10.00 and Boron-11 11.00. Calculate the percentage abundance in samples with relative atomic masses of a.) 10.80 b.) 10.83. Please help.
    For part b let's say we want to find the percentage of Boron 10 and then by elimination we can find the percentage of Boron 11. So let's use "x" as the percentage multiplier for Boron 10. Thus, the multiplier for Boron 11 would have to be "1-x" as the sum of isotopic abundance percentages is equal to 1 or 100%.

    The equation would then be:

    11.00(1-x) + 10.00(x) = 10.83

    Multiplying out the brackets:

    11.00 - 11.00x + 10.00x = 10.83

    Rearranging:

    11.00 - 1.00x = 10.83

    Therefore x = 0.17 therefore the abundance of Boron10 is 17% and Boron11 is 83%


    Can you use this method to work out part a?
 
 
 
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