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Naturally occuring boron is composed of only two isoptopes 10/5B and 11/B with isotopic masses of 10.0129 and 11.00931 respectively. To account for the atomic mass of 10.811 what must be the percentage abundance of each isotope?

  • chemistry -

    Let 10B abundance = x
    Then 11B abundance = 1-x
    -------------------------
    x(10.0129) + (1-x)(11.00931) = 10.811.
    Solve for x and 1-x, then convert to percent.

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