The element boron has three naturally occuring isotopes. One has the atomic mass 10.0129(54.4%), another 11.0093(12.75%). What is the atomic mass of the third isotope if the average atomic mass of boron is 10.811?

Answer 1

So the third isotope has an abundance of
100%-54.4%-12.75% = ? = about 32.85% or 0.3285 as a decimal).

If we let Z = atomic mass of the third isotope we have\
10.0129(0.544) + 11.0093(0.1275) + Z(0.3285) = 10.811
Solve for Z.

Answer 2

Would the correct answer be 12.167?

Answer 3

I came up with 12.0556 which I would round to 12.056 and that's too many significant figures. By the way, I don't see but two isotopes for B; i.e., B-10 and B11. This may be a made up problem. See
http://www.webelements.com/boron/isotopes.html