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Chem

Bromine has two natural occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu. The mass of Br-81 is 80.9163 amu, and its natural abundance is 49.31%. Calculate the mass and natural abundance of Br-79.

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  1. If 81 has abundance of 49.31%, then Br79 must be 100-49.31. Then
    80.916(0.4931) + x(100-49.31) = 79.904. Solve for x.

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  2. My teacher said the answer is 78.92 amu and 50.69%. But i keep getting 81.13amu and i don't know how to get the percentage.

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  3. Do you see in my post where I have 100-0.4931? That is where you get the 0.5069 = 50.69%.
    Post your work and I'll find your error.

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  4. Your teacher is right. The correct answer is 78.92.

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  5. (79.904)=(80.9163)(0.4931) + x(0.4931)
    79.904=39.90 + 0.4931x
    40.004=0.4931x
    x=81.13amu

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  6. My first response was:
    80.916(0.4931) + x(100-49.31) = 79.904. Solve for x.

    (79.904)=(80.9163)(0.4931) + x(0.4931)
    Note that I made a typo. It should have been 1.00-0.4931 (and not 100-49.31), then you should have
    79.904 = (80.9163) + x(0.5069) and go from there.

    79.904=39.90 + 0.4931x
    Should be
    79.904 = 39.900 + 0.5069x etc.

    40.004=0.4931x
    x=81.13amu

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  7. oh okay i get it now, thank you! Why did you subtract 1.00 from the abundance though? how do you know?

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  8. There are two Br isotopes. One is 49.31% so the other one must be 100%-49.31% = 50.69% or 0.5069 as a fraction.

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  9. Br-79 & Br-81 have a total amu of 79.904
    Br-81 = 80.9163amu & 49.31% abundance
    100-49.31=50.69 which would be the abundance of Br-79
    Convert the % into decimals.
    50.69%=0.5069 49.31%= 0.4931
    x= total amu of Br-79
    0.5069(x) + 0.4931(80.9163)= 79.904
    0.5069x + 39.89982753 = 79.904
    0.5069x = 40.00417247
    x= 78.92 amu

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