Question

Antimony - 121 has an atomic mass of 120.9038 and Antimony - 123 has an atomic mass of 122.9041. You have 100 gram specimen of each pure isotpe, and youprepare the largest possible sample of antimony with an average atomic mass of 121.500.
Q1. The '%' composition of Antimony - 123 and Antimony - 121 in the mixture respectively is nearly ?
Q2. What would be the final mass of such a mixture ?
Q3. If the average atomic mass is 121.9 then what is the composition of Antimony - 121 and Antiomony - 123 ?

Answer 1

This requires a bit of creative algebra, but it's all about finding the correct equations. You know that there are only 2 isotopes of Sb, let's call them X and Y. So that means \[X + Y = 1\] where X is the fraction of Sb-121, and Y is the fraction of Sb-123. The combined percentages of the 2 isotopes must add up to 100%.

You also know that the mass of the combined pile must be 121.5g, so
\[(121*X) + (123*Y) = 121.5\]
substitution allows you to find X (or Y) first:
\[(121*X) + (123*(1-X)) = 121.5\]Distribute the multiplication and solve for X. That tells you the fraction of the sample that has to be made up of Sb-121 in order to make a weighted average of 121.5. The rest of the sample must be made up of Sb-123. Multiply the fractions by 100 to get the percentages.

Q3 is the same thing, except the average mass is 121.9, instead of 121.5