- Concept Videos
- Solutions 58

Naturally occuring iodine has an atomic mass of 126.9045. A 12.3849-g sample of iodine is accidentally contaminated with an additional 1.00070 g of ^{129}l, a synthetic radioisotope of iodine used in the treatment of certain diseases of the thyroid gland. The mass of ^{129}l, is 128.9050 amu. Find the apparent "atomic mass" of the contaminated iodine.

Hey guys, in this question they're talking about a sample of Iodine isotopes so they're talking about naturally occurring Iodine, naturally occurring Iodine is Iodine 127 we're told that it has an atomic mass of 126.9045 AMU, now in this question we're dealing with a 12.3849 gram sample of Iodine, but here's the thing with in this we've accidentally added 1.0070 grams of Iodine 129 so this was after accidently added to it and now we're told that Iodine 129 has its own atomic mass of 128.9050 AMU and when we need to do here is figure out what is the atomic mass of this contaminated or tainted Iodine sample, alright so what we need to do here is realize that we need to figure out the atomic mass or average mass remember the formula for this and I'm going to abbreviate it as AM so atomic mass or AM equals mass of isotope 1 times its fractional abundance plus mass of isotope 2 times its fractional abundance which I'll just abbreviate as FA, alright so we know the masses of each Isotope but we don't know their fractional abundances that's what we need to figure out so how do we do that? Well, we're going to take that amount of the sample 12.3849 grams, we're going to subtract from it the mass of the Iodine 129 that was accidently added that would give me the amount of Iodine 127 in there. Now that I have those 2 numbers I can finally figure out their fractional abundances so fractional abundance here but let's first do it for iodine 129 so the fractional abundance here would be the mass of iodine which is 1.00070 grams divided by the total so here that gives us 0.0808 and then we'll do the same thing for Iodine 127 so it will be the mass of Iodine 127 divided by the total so that give me 0.9192 so now we have a fractional abundances of both isotopes so all we do now is we plug it into the formula so the mass of natural occurring Iodine is 126.9045 AMU we just figured out its fractional abundance so plug that in plus the mass of Iodine 129 then times its fractional abundance which we found with this number here when we multiply all that out that gives us 127.066 AMU so this right here represents the average or atomic mass of our tainted Iodine sample so this is a little bit different from previous questions where we've seen we have to use the Isotope formula but this is the approach we would have to take here in order to figure out both the fractional abundances once we do that we plug into the equation and we'll get our answer.