Atomic Mass: How to Calculate Isotope Abundance

Here we are going to do some problems in which we already know what the

atomic

mass

of an element is and we have to

calculate

what the percentage

abundance

of various

isotope

s of that element is. Here is a typical question. There are two stable

isotope

s of chlorine. 35 and chlorine 37. I put this in visual form, so here are the two atoms and here are the

mass

es 34.97 AMU and 36.97 AMU of these two isotopes if the relative

atomic

mass of chlorine is 35.45 AMU, so that's what it is for the element itself what is the

abundance

of each of these isotopes, in other words, what we're solving for is what is the percentage abundance of this and what is the percentage abundance of this.

They're both unknowns right now, so how are we going to solve the problem since these are unknowns, the first thing we're going to do is assign them some chlorine 35 variables just for the sake of calling it X, uh, the percentage abundance will be x what happens with chlorine 37? Well, we could say it's y, but that would be a pain because then we'd have two variables to solve for x and y and that would make things much more difficult. What we really want to be able to do is express the abundance of chlorine 37 using X in some way. let me show you how we're going to do it.

I want to show you just some possible random abundance values ​​for these two isotopes just so you can see a pattern here and then we'll figure out how to solve it, well, let's say. just randomly that there could be I don't know 60% chlorine 35 and 40% chlorine 36 or if we express them as decimals they would be 6 and 4 or there could be 25% chlorine 35 and 75% % chlorine 37 there are the decimals .25 and 2.25 and 75 or we could have 80% chlorine 35 20% chlorine 37 again. I'm making this up and we have 8 and. 2 Do you see the pattern that follows here because we only have two possible isotopes for this problem when when we add them they always have to come out to 100% or if we are expressing their abundances in decimals here when we add the two abundances that have to result in one we can use this to get an expression for chlorine 37 here is the first thing we notice chlorine 35 plus chlorine 37 when expressed in decimals add them together We will always get one so we can rearrange this equation here to solve for chlorine 37 and then we get chlorine 37 = 1 minus the amount of chlorine 35 that we have, so the chlorine 37 here will be 1 minus the amount of chlorine 35. what we say is amount of chlorine 37, we add the two amounts and we will obtain 100%.

We will get one if we are expressing them as decimals, so now we are ready to write an equation that sets these variables. To do this, let's take a look at a related equation that we would write down when we want to find out. the atomic mass and we already know the abundances of the isotopes that we are starting with, well, here I have Boron 10, I take its weight 10 AMU and I multiply it by its abundance expressed as a decimal and I do the same with Bor in 11 multiply it by its abundance expressed as a decimal, so I'm going to do that here except we already know the atomic mass, so let's start with chlorine. 35 okay, I'll start with your dough just like we have the dough. of boron 10 here, so the mass of chlorine 35 is 34.97 AMU down here, we multiply this by 2 because we know the percentage, but for chlorine 35 it will be forward.

For the second isotope, here is the mass of boron, we are going to take the mass of chlorine 37 36.97 and multiply it by the abundance, here it is 8 and for chlorine 37 it will be 1 - x, so we add those two, unlike from here. where we don't know the atomic mass we already know the atomic mass of chlorine and it should be 35 5.45 now that we've set up our equation and all we have to do is rearrange multiply and divide a little bit and we can solve for x let's do that so I'm going to rewrite this and the first thing I'm going to do is distribute 36.97 between these parentheses, okay, so I'm going to get 36.97 * 1 minus 36.97 .97 x and that will give me 2.x a negative negative 2.x + 36.97 = 35.45 now I want to get rid of this this side of the equation, so I'm going to subtract 36 .97 here - 36.97 here and that will give me -2.00 1.52 divided by 2.0.

I just divided both sides by 2.0 I'm going to go up here where I have a little more room and now I can finally say that x is equal to these divided is equal to 0.76 0 or cl35 because that's what x is after all it's equal to 76.0%, okay, now do the other part which is chlorine 37, so it will be 1 - x will turn out to be 0.240 and just multiply this by 100 to get the percentage that I will get. Chlorine 37 percent equals 24.0%, so these are the two percentages we were working out just to review, the only really complicated thing you have to do here is set one of them equal to X and the other to 1 - x and such as we saw here, that's because when you add the two percentages or decimals together you will get 100% or you will get one when they express decimals here, so it took me a while to do this problem pausing along the way to see how the math worked in several situations, so Now let's do one more problem in which we will go over it very quickly step by step so that you can now solve a problem like this.

If you look at it on homework or on a test, there are two natural isotopes. of lithium lithium 6 with a mass of 6.01 5 AMU and lithium 7 with a mass of 7.06 AMU, what is the abundance of each? Okay, so let's

calculate

abundance. There is information that we do not have and that we need. and that is the atomic mass of lithium, how can we find out? It is not given in the problem. We can look up lithium on the periodic table and this number down here, 6941, tells us the relative atomic mass of lithium, so that's the other piece. puzzle that we're going to need, okay, so we have two things that we're solving here: lithium 6, lithium 6, let's make the abundance of that equal to X and lithium 7, let's make the abundance of that equal to 1 - x now We'll set up an equation that works with the abundances and atomic mass, so we'll use the weight of lithium 6 or should I say the mass of lithium 6.

You know people use them interchangeably with this, it doesn't really matter 6.01 5 * X plus the mass of lithium 7 7.1 6 * 1 - x add these and we should get the atomic mass so 6941 here okay rewrite this distribute this number here so 7.01 6 minus 7.01 16x = 6941 I have my just multiply both by uh ne1, make it positive and for the last step I'll divide both sides by 1.001, which isn't actually going to do anything because it's so close to one that it's going to be 0.075 divided by 1.001, so now X here is equal to 0.075, that's a decimal to convert it to a percentage, we multiply it by 100, so we will get 7.5% and that is the amount of lithium 6 that we have and then 1 - x will be uh 0.925 or to convert that to a percentage we multiply it by 100 and we get 92.5% and that tells us how much lithium 7 we have, so again, the only really complicated thing was that we had to set one of these isotopes equal to X. and the other one we set to 1 minus X, then we work with the calculations and obviously you have to remember to take these decimals that you get at the end and convert them into percentages by multiplying by 100.