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    Not sure how to work this out even though it's only worth one mark.
    Bromine has two isotopes BR-79 and Br-81 the relative atomic mass of br is 79.9 calculate the percentage of br-79 atoms in a sample of bromine.
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    Method 1:

    So the equation we'll use is:

    Atomic mass x abundancy (%) (for each isotope) divided by total abundance = relative atomic mass of sample.

    We'll call the abundance of Br-79 atoms 'y'. The total abundance of atoms is 100, so the abundance of Br-81 atoms is 100 - y.

    This gives us: 79y + 81(100-y) ÷ 100 = 79.9
    -2y + 8100 = 7990
    2y = 110
    y = 55%

    Obviously this is a somewhat time-consuming method for 1 mark but is 100% effective.


    Method 2:

    79.9 - 79 = 0.9
    81 - 79.9 = 1.1

    Ratio Br-79 : Br-81 = 1.1 : 0.9 (i know this is a bit confusing but remember that there must be more 79 atoms than 81 atoms, because 79.9 is closer to 79 than 81)

    (Converting ratio to percentage) 1.1/(1.1+0.9) x 100 = 55%

    Hope this helped
 
 
 
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