Question

questions related to
percentage of isotopes in a sample

Answer 1

Let's start by repeating the solution for nitrogen from the Average Atomic Weight tutorial:

(14.003074) (0.9963) + (15.000108) (0.0037) = 14.007

The solution is laid out like this:

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element

In the other tutorial, the average atomic weight is the unknown value calculated. In this tutorial, the unknown values calculated are the TWO percent abundances.

However, you might protest that we have two unknowns, but only one equation. You would be right, except that there is a trick. Ah, the ChemTeam loves tricks!!

Think about the sum of the percent abundances of the TWO isotopes. That's right, they add up to 100% (or, since we use decimal abundances in the calculation, 1.00). So, the trick is to express both abundances using only one unknown. I'll show how using the above nitrogen example.

Answer 2

we should find the average atomic mass of chlorine