# Atomic Mass: How to Calculate Isotope Abundance

here we're going to some problems where we already know what the

## atomic

### mass

of an element is and we have to figure out what the percent#### abundance

of various### isotope

s of that element are here's a typical question there are two stable### isotope

s of chlorine chlorine 35 and chlorine 37 I put this visual form for you so here are the two atoms and here are the### mass

es 34 point nine seven amu and thirty-six point nine seven AMU of these two### isotope

s if the relative## atomic

### mass

of chlorine isthirty 5.45 am you so like that's what it is for the element itself what is the

#### abundance

of each of these### isotope

s so in other words what we're solving for is what's the percent#### abundance

of this one and once the percent#### abundance

of this one they're both unknowns right now so how are we going to go about solving the problem since these are unknowns the first thing let's do is assign some variables to chlorine 35 just for the sake of it let's call that X the the percent#### abundance

of it will be X what about chlorine 37 well we could say that it's Y but that would be a pain because then we'd have two variables to solve for x and y now they make things much more difficult what we really want to be able to do is express chlorine 37's#### abundance

using X in some way let me show you how we're going to do it I want to show you just some random possible#### abundance

values for these two### isotope

s just so you can see a pattern here and then we'll figureout how to solve for it okay let's say just randomly that there could be I don't know 60% chlorine 35 and 40 percent chlorine 36 or if we express these as decimals would be 0.6 0.4 or there could be 25 percent chlorine 35 and 75% chlorine 37 there the decimals point to five point to 0.25 and 0.75 or we could have 80 percent chlorine 35 20 percent chlorine 37 again I'm just making these up and we have point and point two do you see the pattern going on here because we only have two

possible

### isotope

s for this problem when they when we add the two of them together they always have to come out to a hundred percent or if we're expressing their#### abundance

s in decimals here when we add the two#### abundance

s up they have to come out to one we can use this to come up with an expression for chlorine-37 here's the first thing that we've noticed chlorine 35 plus chlorine 37 when they're expressed decimals add them together you'll always get one so we can justrearrange this equation here to solve for chlorine 37 and then we get chlorine 37 equals one minus the amount of chlorine 35 that we have so chlorine 37 here is going to be one minus the amount of chlorine 35 which we are saying is X so here are two variables X the amount of chlorine 35 and one minus X for the amount of chlorine 37 add the two amounts together and we're going to get a hundred percent or we're going to get one if we're expressing them as decimals so now we're

ready to write an equation that sets these variables up what's it going to look like in order to do this let's take a look at a related equation that we'd write when we want to find out the

## atomic

### mass

and we already know the#### abundance

s of the### isotope

s we're starting with okay here I got boron 10 I take its weight 10 amu and I multiply it by its#### abundance

as expressed as a decimal and I do the same thing for boron 11 multiply it by its#### abundance

expressed as a decimal so I'mgoing to do that here except we just already know the

## atomic

### mass

so let's start out with chlorine 35 okay I'll start out with its### mass

just like we have the### mass

of boron 10 here so the### mass

of chlorine 35 is thirty-four point nine seven AMU down here we multiply this by 0.2 because we know the percent but for chlorine-35 it's just going to be X we don't know it yet so we take this and now we move on to the second### isotope

here's the### mass

of boron we're going to take the### mass

of chlorine thirty seven thirty six point nine seven and multiply that by the#### abundance

it's point eight here and for chlorine 37 it's going to be one minus X so we add those two together and unlike down here where we don't know the## atomic

### mass

we already do know the## atomic

### mass

for chlorine that should be 35 point four five now 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'mgoing to rewrite this and the first thing I'm going to do is I'm going to distribute thirty six point nine seven between these parentheses okay so I'm going to get thirty six point nine seven times one minus thirty six point nine seven x equals 35 point four five now I have two X terms this one and this one so I'm going to do 34 point nine seven X minus thirty-six point nine seven X and that's going to give me two point zero zero X negative negative two point zero zero X plus

thirty six point nine seven equals 35 point four five now I want to get rid of this from this side of the equation so I'm going to do I'm going to subtract thirty six point nine seven here minus thirty six point nine seven here and that is going to give me negative two point zero zero x equals negative one point five two these are both negative so I can multiply both sides of the equation by negative one to make them both positive so I'm going to get two point zero zero x equals one

point five two and for the very last step to actually solve for X I'm going to do x equals one point five two divided by two point zero just divided both sides by two point zero I'm going to move up here I've got a little bit more room and now I can finally say that X equals these divided together equals zero point seven six zero or CL 35 because that's what X is after all equals seventy six point zero percent okay now let's do the other part which is chlorine-37 so

that's going to be one minus X is going to turn out to be zero point two four zero and just multiply this by hundred to get the percent I'm going to get chlorine-37 s percentage equals a twenty four point zero percent so these are the two percentages that we were solving for just to review the only really tricky thing you got to do here is set one of them equal to X and one of them to one minus X and just as we saw here that's because when you add the two percentages or decimals

together you're going to get 100% or you're going to get one when expresses decimals here so it kind of took a while doing this problem pausing along the way to look at how the math worked in various situations so now let's do one more problem where we'll just go through it really fast step by step so you can be all ready to solve a problem like this if you see it on the homework or see it on a test there are two naturally occurring

### isotope

s of lithium lithium six with a### mass

ofsix point zero one five amu and lithium-7 with a

### mass

of seven point zero one six am you what is the#### abundance

of each okay so we're going to be calculating#### abundance

there's one piece of information that we don't have that we need and that is the## atomic

### mass

of lithium how can we find it out it's not given in the problem we can look lithium up on the periodic table and this number down here six point nine four one tells us the relative## atomic

### mass

for lithium so that's theone other piece that puzzle that we're going to need okay so we got two things we're solving for here lithium six lithium six to make the

#### abundance

of that equal to X and lithium-7 let's make the#### abundance

of that equal to one minus X now we'll set up an equation that works with the#### abundance

s and the## atomic

### mass

so we'll do the weight of lithium six where I should say the### mass

of lithium six you know people use them interchangeably with this stuff doesn't really matter6.01 five times x plus the

### mass

of lithium seven 7.01 six times one minus x add these together and we should get the## atomic

### mass

of six point nine four one here okay rewrite this distribute this number here so seven point zero one six minus seven point zero one six x equals six point nine four one I have my X terms here I'm going to subtract this from this and that is going to give me negative one one point zero zero one X plus seven point zero one six equals six point nine four one now toget X by itself I'm going to subtract this from both sides of the equation I'll get negative one point zero zero one x equals negative 0.075 I'm going to just multiply both by negative one turn that positive and for the last step I will divide both sides by one point zero zero one which is not really going to do anything because it's so close to one it can be zero point zero seven five divided by one point zero zero one all right so now x over here equals zero point zero seven

five that's a decimal to turn it to a percent we multiply a by a hundred so we're going to get seven point five and that is the amount of lithium-6 that we have and then one minus X is going to be zero point nine to five or to turn that into a percentage we multiply it by 100 we get ninety two point five percent and that tells us the amount of lithium-7 that we have so once again the only really tricky thing is so we had to set one of these

### isotope

s equal to X and the other we set to oneminus X then we work through the math and you want to obviously remember to take these decimals that you get at the end and turn them into percentages by multiplying by a hundred