Atomic weight of an isotope?

Question

The atomic weight of a newly discovered element is 110.352 amu. It has two naturally occurring isotopes. One has a mass of 111.624 amu. The other has an isotopic mass of 109.75 amu. What is the percent abundance of the last isotope (109.75 amu)?

Please teach me how to do this problem. I can't find an example in my textbook.

Answer 1

Let x be the fraction of the 109 isotope. Then 1 - x is the fraction of the 111 isotope.

109.75x + (1-x)(111.624) = 110.352

109.75x + 111.624 - 111.624x = 110.352

-1.874x = -1.272

x = 0.6788 = 67.88% is the percent abundance of the "last isotope."

Answer 2

Let x = % abundance of the 1st isotope and let y = % abundance of the 2nd isotope

x + y = 100

x = 100-y

110.352 = 111.624 ( 100-y) + 109-75 y / 100

solve for y  

Answer 3

Call the unknown % 'a'

Then to calculate atomic weight you multiply all the isotopic weights by their percentage abundance and add them up (like 25/100 x 37 + 75/100 x 35 = 35.5 for chlorine)

You only have two isotopes so the % of the last you have called 'a', which means that the % of the other must be (100 - a)

So to calculate the atomic weight in this case:

(100-a)/100 x 111.624  +  a/100 x 109.75 = 110.352

So now you have to rearrange and solve for 'a'

(67.9% I think - but check that!)