Question

Suppose the isotopic ratio of the two boron isotopes 10B (10.013 amu) and 11B (11.009 amu) in a sample has been altered from the ratio found in nature and now contains 18.67% 10B in the sample. Determine the atomic weight of this sample of this new boron element.

Answer 1

(0.1876 * 10.013) + ((1 - 0.1876) * 11.009))

taking the known percent and multiply by 10b

subtract from 100 percent(or 1) to get remaining percent and multiply by 11B

Answer 2

Simply when we have 2 types of isotopes the total abundance of both the elements should be equal to 100%
If you have 18.67% of 10B then you will have (100-18.67%) of 11B
So you have 81.33% of 11B.
\[a.m.u. = \frac{ (18.67 * 10.013) + (81.33 * 11.009) }{ 100 }= 10.823\]