[vanalm]

I know this is probably so simple, but I have been trying this for a week and I'm just not seeing it. I feel that there is not enough info.. I try to plug the numbers into the equation I have, but it seems I am missing numbers. Will someone please give me steps or a different formula for this?

Antimony has 2 naturally occurring isotopes. 121 Sb (isotopic mass 120.94 amu) and 123 Sb (isotopic mass 122.904 amu). If the atomic mass of antimony is 121.76 amu, what is the percent natural abundance of each isotope?

[enahs]

Atomic Weight =  ( % Abundance isotope 1 / 100 ) * (atomic weight of isotope 1) +
( % Abundance isotope 2 / 100 ) * (atomic weight of isotope 2)

You have all of those but % abundances. When you divide a percent by 100 you get the fractional percent.
I.E. 80% = 0.80. And so 80% of 50 = 50 * 0.80 = 40.

That said, when you have two percentages, they must add up to 100, or 1 in fractional percent.

In this case and in the formula I gave at top one of the % Abundance becomes “x” and the other becomes “1-x”, and then just solve for x.

Search the internet for "weighted averages" for more.

[vanalm]

Ok, I got it. Thanks so much. I knew it would be something simple.