An element has two naturally occurring isotopes. Isotope 1 has a mass of 120.9038 amu and a relative abundance of 57.4%, and isotope 2 has a mass of 122.9042 amu. Find the atomic mass of this element and identify it.

Video transcript

Everyone. Welcome back. Our next problem says an element has two naturally occurring isotopes. Isotope one has a mass of 120.9038 AM U and a relative abundance of 57.4%. And isotope two has a mass of 122.9042 AM U find the atomic mass of this element and identify it. And our answer choices are a 118.7 AM U 10 B 120.6 A U tellurium C 121.8 A U antimony or D 122.1 AM U Indian. So let's think through how we need to do this problem. We need to think about that. We're looking for the average mass as they talk mass of this element. So average mass, but if our isotopes occur in different abundances, we have to take that into account. So for each isotope, we need to multiply its atomic mass by its relative abundance and then add together those amounts to get the average mass of the element as it naturally occurs. So we have two isotopes and I'm gonna use highlighting to separate out. So I don't mix up the values. So I'll highlight in blue for isotope one. So that'll be highlight in blue, I have its mass and a relative abundance. And then for isotope two, I'll highlight in red, we just are given its mass. So it just helps me here to keep the values straight. So I don't actually mix them up. So our formula to calculate this with two isotopes would be that our atomic mass is equal to the mass of element one. So I just wrote mass subscript one multiplied by what I wrote as fa one that would stand for fractional abundance. Because again, we're going to express this as a fractional amount of the total when we add it all together and then add that too. And then in brackets mass subscript two plus fa subscript two. So we have the masses of both. We have the relative abundance as a percent of isotope one which we can easily convert to a fractional abundance just by dividing by 100. But we don't have an abundance of isotope two. That's right fa of one. We're given 57.4% divide that by 100% and we'll get 0.574 for the fractional abundance of two. Well, if we've only got two isotopes making up the, this element naturally occurring, as our problem tells us, then the relative abundance will be 100% minus 57.4%. So just the rest of what's out there in nature, that whole amount divided by 100%. So that's going to equal 42.6% divided by 100% which will be 0.426. So now I have all the values that I need. So we'll do that calculation. So we have atomic mass equals and then in brackets, the mass of one, we look at our problem 120.9038 AM U. And that's going to be multiplied by 0.574. And then add atomic mass of number 2, 122.9042 AM U multiplied by 0.426. So now we'll get the amounts in brackets. Let's pay attention to significant figures here. Our atomic masses are given out to four decimal points. But our uh fractional abundance here is just as three significant figures. So since we're multiplying, we need a total of three significant figures in the parentheses. So that's going to equal 69.4 AM U for the first expression and then 52.4 AM U for the second. So we have three significant figures in each term, add those together we're adding, so we express our answer with just one decimal point. So we have 121.8 AM U. So that is the atomic mass that we get, and we can see that's there in answer choice C and let's just make sure that corresponds to our correct element. So now we refer to the periodic table and when we look at this value, we have 121.76 AM U as corresponding to Antimony or SB as its abbreviation. So that does make sense with our answer choice. It says Antimony. SB. So choice C will be our answer if you want to look just briefly at the other answers. Um In the choice A we have 10 but 10 when we look it up is about 119 AM U that 118.7. So again, that's too far off what we calculate choice B tellurium. When we look up that value, it has 127 0.6. So at 128 AM U. So again, you can see quite a bit too large and choice DDM. When we look at on a periodic table, it's about 115 AM U. So too small. So that's why our other answer choices are incorrect. And the element that we have here and its atomic mass as it occurs in nature, it's choice C 121.8 mu and it's the element antimony. See you in the next video.