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Each isotope of an element has a different Atomic Mass, so an average is taken of all the isotopes, but the average is weighted because the natural abundance (%) of each isotope is factored in. If hydrogen-1 is much more abundant than deuterium and tritium, then the weighted average will be closer to 1 than 2 or 3 but not a whole number.

The following equation shows how percent abundance factors into the weighted average.

(Atomic Mass A)(X% abundance) + (atomic mass B)(Y% abundance)...=(weighted average of all isotopes of the element)(100% abundance)

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Pearline Blick

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3y ago

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From what I remember of chemistry, the amu of each element on the periodic chart is a weighted average of all the isotopes of that element. So, as you indicated - 80.2% of Boron exists as B-11 and 19.8% of boron exists as some other isotope. In that case you can get an estimate of the amu of the unknown through simple math: 0.802*(11.01 amu) + 0.198*(X amu) = 10.81 Solving for x, you get 10.00 amu