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

Chemistry

Zain Dickson

21 December, 17:46

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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Valentin Brady · 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.