Problem: Boron has only two naturally occurring isotopes. The mass of boron-10 is 10.01294 amu and the mass of boron-11 is 11.00931 amu .Use the atomic mass of boron to calculate the relative abundance of boron-11.

FREE Expert Solution

Atomic Mass=[mass×f.a.]isotope 1+[mass×f.a.]isotope 2


We’re given the following values:

Average Atomic Mass = 10.821 amu  (found in books or the internet)

Boron-10: Mass = 10.01294 amu

Boron-11: Mass = 11.00931 amu



We don’t know the percent abundance of boron but we can use the fact that the f.a. of all isotopes of an element add up to 1

Since there are only two isotopes, this means:

Boron-10:      f.a. = x

Boron-11:    f.a. = 1 – x


Solving for x:


Atomic Mass = [mass×f.a.]isotope 1+[mass×f.a.]isotope 2


10.821 amu=10.01294x+11.00931(1-x)


10.821=10.01294x+11.00931-11.00931x10.821-11.00931=10.01294x-11.00931x-0.18831-0.99637=-0.99637x-0.99637


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Problem Details

Boron has only two naturally occurring isotopes. The mass of boron-10 is 10.01294 amu and the mass of boron-11 is 11.00931 amu .

Use the atomic mass of boron to calculate the relative abundance of boron-11.

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