Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.

Question

Jamari Mckay

Chemistry

4 April, 22:06

Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.

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Answers (1)

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Tobias Smith · Answer 1

You will have to do some math.

80.2 * 11.01 + 19.8 * x = 100 * 10.81.

x is the mass in amu of the second isotope.

Solve for x.

19.8 * x = 100 * 10.81 - 80.2 * 11.01

19.8 * x = 1081 - 883.00 = 198.00

x = 1964.00 / 19.8 = 10.00

The mass of the other isotope is 10.00 amu.