so we've seen that the mass number of an atom refers to the number of protons plus the number of neutrons so for example if you had an isotope of carbon 13 13 is the mass number and that tells us the total number of protons plus neutrons and this is always going to be an integer however this is different from the atomic mass that you'll see written beneath an element symbol on the periodic table so for example if you were to look on your periodic table and find carbon it would look something like this would have an atomic number 6 it would say carbon and then beneath it it would have 12 point 0 1 this is the atomic mass or the average atomic mass so this number has to be experimentally determined and the way that's done is scientists have gone out and they've taken samples of carbon and they've analyzed all of the different isotopes in that sample so depending on the element there may only be one naturally occurring isotope but in some elements there might be two or three or five or seven different naturally occurring isotopes and they typically are not there in an even ratio so if there's two isotopes they're probably not gonna be 50/50 it might be 75% one isotope and only 25% the other one so when you want to calculate the average mass so the way you do that is you take the sum that's the symbol here a capital Greek letter Sigma the sum of the abundance of the first isotope times the mass of that isotope plus the abundance of the second isotope times the mass of that isotope and you keep doing that for as many isotopes as there are so for example if there were two isotopes and their abundances were 75% and 25% well you would take 75% times the mass the first isotope plus 25% times the mass is the second isotope so this is what's known as a weighted average again it's not a simple average where you just add up the two masses and divide by two or you add up the three masses if there's three isotopes and divide by three because they're not equally split 50/50 or a third a third a third so we have to do a weighted average so let me give you an example of this using neon so neons a gas you've probably heard of it's commonly found in neon signs although technically only the bright orange is typically neon gas other colors are actually different gases but neon gas if you were to grab a sample of it it would not all be the same isotope you would find there are three naturally occurring isotopes of neon in that sample there's neon 20 there's neon 21 and neon 22 and of course they all had different masses because they had different numbers of neutrons and they also have different abundances so you can see the most abundant isotope is neon 20 it's ninety point four eight percent next is neon twenty two which is nine point two five percent and the least abundant is neon twenty one which only comprises 0.27% of all the neon and again these abundances are basically stable regardless of where you find the sample of neon so you grab the sample of neon here or in France or the South Pole or anywhere on the planet you're gonna get these same basic abundances so when you want to calculate the average mass of neon well you have to take the abundance of neon twenty times the mass of neon twenty plus the abundance of neon twenty-two times the mass of neon 22 plus the abundance of neon twenty-one times the mass of neon twenty-one so let's just try that here as an example so I've summarized all that data here into this table so you see there's three isotopes and here's the mass numbers 20 21 22 which I went ahead and wrote him here 20 21 22 again these are always integers this is just the protons plus the neutrons now this the atomic mass is the mass of an atom of neon 20 now you'll notice the atomic mass in AMU which again means atomic mass units is very close to the mass number so the mass number is 20 the atomic mass is gonna be very close to 20 and it is 19.99 but again this is the mass if you were to actually weigh out an atom of neon 20c away all the protons neutrons electrons you could put on a super tiny balance it would weigh 19.99 to 4 and of all the neon in nature ninety point four eight percent of it is neon twenty four neon twenty-one again it's got a mass number of twenty one and so Tomic mass is very close to twenty one twenty point nine nine three eight again that's the mass of a single atom of neon twenty-one and it has an abundance of 0.27% and the neon 22 is has a mass close to twenty to twenty one point nine nine one four and this abundance is nine point two five percent so if we want to calculate the average atomic mass we need to take the abundance of neon twenty times the mass of neon twenty plus the abundance of neon twenty-one times the mass of neon twenty-one plus the abundance of neon twenty two times the mass of neon twenty two so let's just try that so let's just go through and do a weighted average where we're gonna take the abundance times the mass of each of the naturally occurring isotopes so ninety point four eight percent well you probably know from a previous math class that when you use a percentage in a calculation we always want to convert those back to a decimal mode so ninety point four eight percent as a decimal is actually written as zero point nine zero four eight so you basically divide the percentage by a hundred to get the decimal or you multiply the decimal by a hundred to get the percentage so we're gonna take the abundance times the mass of neon twenty so ninety point four eight percent was the abundance the mass is nineteen point nine nine to four so nineteen point nine nine to four plus the abundance of neon twenty-one times the mass of neon twenty-one well how would you convert point two seven percent into a decimal again you want to divide it by a hundred so it becomes point zero zero two seven times its mass twenty point nine nine three eight plus the next one nine point two five percent writing that as a decimal is point zero nine two five times the mass of that isotope twenty-one point nine nine one four twenty one point nine nine one four and you add that all up and I got twenty point one eight am you so this is the average atomic mass of neon and if we look over here on our periodic table and we find neon we'll see a look the average atomic mass of neon is twenty point one eight so that's where all of these average atomic masses come from scientists have gone out and taken samples of elements all over the world and then they have determined what is the percent breakdown of all the isotopes and then they did this calculation of abundance times mass plus abundance times mass for every isotope and they found all of these average atomic masses written beneath the element symbols okay well let's practice a few of these together so go ahead and pull out your chapter for a handout and let's do number eight a together so a says the natural abundance for boron isotopes is 19.9% boron tin with a mass of ten point zero one three AMU and 80.1% boron eleven with the mass of 11.00 nine a.m. you calculate the atomic mass of boron and report it to two decimal places now for most of these average atomic mass calculations it is going to tell you how many decimal places to report it to so you won't need to worry about significant digits because it can get confusing since you're multiplying and adding okay well let's try this out so the average is gonna be the sum of the abundance times the mass plus the abundance times the mass just go ahead and do that for boron so 19.9% how would you write that as a decimal will you would divide it by a hundred and get point one nine nine basically just move that decimal places two times and then times the mass of boron ten which is around ten as it should be and then plus the abundance times the mass of the next isotope so 80.1% becomes point eight zero one times the mass of boron eleven eleven point zero zero nine amu and we put that in our calculator and it says report it to two decimal places and I got ten point eight one amu the nice thing is we can check this on our periodic table and see what it actually is on here it is boron is in fact ten point eight one of course you might be off by a hundred but that's okay because you recall from significant digits that last digits the uncertain place anyway but you shouldn't get something that's drastically different like you shouldn't get nine point two or twelve or something it should be pretty close to ten point eight one now one thing that I always like to do on these kind of calculations is just to ask myself is this answer reasonable so even without doing any calculations you can get a rough idea of what the answer should approximately be so in this case we know that basically twenty percent of the boron weighs ten and 80 percent of the boron weighs 11 so we know the mass is gonna be somewhere between ten and eleven well if it's 80 percent boron 11 we know the mass is gonna be much closer to eleven than it is to ten and we calculated it out and sure enough it is it's basically 80 percent of the way between ten and eleven which is what we expected it should be 80 percent boron 11 only 20 percent boron 10 so let's look at B it says if silver is 51.84 percent silver 107 with the mass of 106.9 zero five one amu and the rest is silver 109 with the mass of 108 point nine zero for 8:00 a.m. you calculate Silver's atomic mass to two decimal places so we need to take the abundance times the mass of each isotope now the first little challenge is to figure out what is the abundance of silver 109 it says 51.84 percent of silver 107 the rest of it is silver 109 so how would you figure out how much this is what is the rest well we know that both the silver 107 and the silver 109 have to add up to be a hundred percent so if silver 107 is 51.84 percent then silver 109 is just the difference from a hundred so it would be 100 - 51.84 so that would give us 48 point one six so that must be the percentage of silver 109 okay now that we have two abundances and two masses we can go ahead and solve this so let's take the abundance of silver 107 again convert this percentage to a decimal so 51.84 percent becomes zero point five one eight four times the mass of silver 107 which is 106.9 zero five one amu plus the next abundance which we just found was 48 point one six percent which as a decimal is point four eight one six times the mass of silver 109-108 point nine zero four eight so we put that in our calculators and it says to report it to two decimal places and I got 107 point eight seven am you let's go ahead and check our periodic table see if we're on the right track and here's silver and sure enough it's 107 point eight seven now imagine this wasn't silver imagine it just said element X well how would you determine if this answer is reasonable well again let's go back and look at our data so it's basically 50% silver 107 and 50 percent silver 109 so half of it is roughly 107 half of it is roughly 109 so we know it's going to be somewhere in the middle between 107 and 109 probably around 108 and in fact our answer is right around 108 so this answer seems reasonable now I really do encourage you to get in the habit of thinking through your answers like this this is one of the most valuable tools a scientist can develop what's known as critical thinking the ability just to think through problems and get an idea of is this answer even reasonable because what can happen otherwise is you just get used to blindly trusting your calculator and whatever it spits out you don't even stop to think is this answer even a reasonable and you might have missed a number or you might have switched a couple of digits if you just took a few seconds just to think through the answer you would be able to realize whether or not the answer is reasonable okay well go ahead and try C on your own so pause the video calculate the average atomic mass of copper out to two decimal places and then come back here and hit play and we'll work it out together okay so this one says copper is sixty nine point one seven percent copper 63 with the mass of sixty-two point nine three nine six AMU and the rest of it again it's the rest is copper 65 with the mass of sixty-four point nine two seven eight amu fine copper is atomic mass out to two decimal places okay so we just saw on B that if this one is 69 point one seven percent the abundance of the other isotope must just be a hundred minus that so 100 minus 69 point one seven is thirty point eight three percent by the way the hundred is exact so we got two decimals we're going to have two decimals down here okay let's plug that into our equation here so 69 point one seven percent as a decimal is zero point six nine one seven multiply that times the mass of copper 63 which is sixty two point nine three nine six AMU plus the abundance of copper 65 which we just calculated here so thirty point eight three is point three zero eight three times the mass of copper 65 which is sixty four point nine two seven eight AMU so put that in our calculator and I got sixty three point five five amu so before we check it on the periodic table just ask ourselves that question does this answer even seem reasonable based off the data that we plugged in here is this answer reasonable well we see that basically 70 percent of it is 63 and roughly thirty percent of it is 65 well we know the average mass has to be somewhere on the scale between 63 and 65 so where's it gonna fall do you think it's going to be closer to 63 or 65 well it's mostly copper 63 right at 69 point one seven percent so it ought to be closer to 63 than 65 and we see that in fact it is the mass is closer to 63 than it is to 65 and in our case since we know it's copper we can even check it out and look right there 63 point five five okay we'll go ahead and try this one out this is the same thing but now we have three isotopes so you're gonna do what you did before for two but that's gonna do it three times so abundance times mass plus abundance times mass plus abundance times mass three times so pause the video try it out and then come back here to check your answer all right well let's work it out so ninety two point two three percent becomes zero point nine two two three times the mass of silicon twenty eight is twenty seven point nine seven six nine AMU okay plus the abundance of silicon twenty nine times the mass of silicon twenty nine so four point six seven percent becomes point zero four six seven times the mass of silicon twenty nine which is 28 point nine seven six five am you thus the abundance times the mass of silicon 30 so 3.10% becomes zero point zero three one zero times the mass of silicon thirty twenty nine point nine seven three eight AMU so put all that in our calculator and I got twenty eight point zero nine amu well once again I hope you stopped for a minute just to see does this answer seem reasonable or as you may be a fat finger something putting in in the calculator well let's just look at the rough masses so we know that roughly 92 percent of it weighs about twenty-eight four and a half percent weighs 29 and three percent weighs thirty so the mass is somewhere between 28 and 30 but it's very heavily weighted towards 28 again 92 point two three percent of it basically weighs 28 so it's gonna be close to 28 and look right here it is in fact close to 28 so that answer does seem reasonable and of course we can always check ourself out on the periodic table for these and we see that it is twenty-eight point zero nine but again on an exam I will not give you an actual element I'll just make up an element or I'll call it element X so you won't be able to check your answers quite so easily so it is a good habit to develop to just stop and see if your answer seems reasonable rather than just blindly trusting your calculator okay now I want you to stop and think about ease so go ahead and pause the video try it out and come back here to check your answer so this problem says bromine has too naturally occurring isotopes bromine 79 and bromine 81 and the atomic mass or the average atomic mass is seventy nine point nine zero four so that's the atomic mass of just bromine so that's what you'd find on the periodic table the mass of bromine 81 is 80 point nine one six three amu and it's abundance is forty-nine point three one percent what is the mass and the abundance of bromine seventy-nine so now on this question is not asking us to solve the average atomic mass it actually gave us that instead it wants to know what's the mass and abundance of one of the two isotopes let's go ahead and fill in what we know so we know seventy nine point nine zero four amu that's the average that has to be equal to the abundance times the mass of bromine 79 plus the abundance times the mass of bromine 81 well we know the abundance and the mass of bromine 81 we don't know the abundance in the mass of bromine 79 but it was there one of these that we can figure out and the answer is yes we can figure out the abundance because recall when there's two isotopes well they have to add up to be a hundred percent so this one is 49 point three one well the other one must be 100 minus forty-nine point three one so bromine 79 must have an abundance of fifty point six nine percent okay so now we know the abundance the only thing we're missing now is the mass of bromine 79 so we can just solve for that we'll just call that X so let's go ahead and plug in what we know so abundance of bromine 79 is 0.5 zero six nine again that's that percentage we found here converted to a decimal times the mass which is our unknowns what is called that X plus the abundance of bromine 81 which was forty-nine point three one percent so that becomes point four nine three one times its mass which was eighty point nine one six three eighty point nine one six three am you so now it's just algebra we got a solve for X here let's go ahead and work a few things out so I'm going to drop the units for now just to make it a little easier so seventy nine point nine zero four equals 0.5 zero six nine x plus 0.493 one times eighty point nine one six three is thirty nine point eight nine nine eight and as for our significant digits go this one has four this one has six I'm just gonna mark this true my myself this should really have four well now I want to group all the non x terms so I want to get this number over here on the left hand side so I need to subtract both sides by 39 point eight nine nine eight so I'll do that real quick and this will help me to isolate the X term so that gave me 40 point zero zero four equals point five zero six nine X so now to get X by itself we need to divide both sides by 0.5 0 six nine so do that on both sides if you have any questions on this basic algebra I encourage you to check out some of the free tutoring options so that gives me that X is 78.92 a em u so X was what well X was the mass of bromine 79 so let's just stop for a minute does this answer seem reasonable the 78.92 seem like a reasonable mass for bromine 79 well yes if it's bromine 79 it ought to be pretty close to 79 and it is 78.92 now the other thing you can do when you get done is a way to check yourself is plug this number back in and just take point five zero six nine times seventy eight point nine two plus 0.49 three is 18.9 163 and make sure that you get seventy nine point nine zero four so there are a few different ways for you to check yourself okay let's look at F now F is a another one that's a bit of a thinker so when you go ahead and pause the video see if you can figure out how to approach this problem and come back here and hit play after you've given it a shot okay so this one says naturally-occurring lithium has two isotopes lithium six with a mass of six point zero one five one amu and lithium-7 with the mass of seven point zero one six zero amu what is the percent abundance of lithium six so now we have two masses but we don't have either abundance but what other piece of information do you know about lithium well if you look and find lithium on your periodic table you do know the average atomic mass is six point nine four amu let's go ahead at least put that down so six point nine four amu that's the average that's gonna equal the abundance of lithium six times the mass of lithium six plus the abundance of lithium seven times a mass of lithium seven but we don't know either abundance but what do we know has to be true about the abundances if there's only two isotopes what do I know about these two abundances I know they have to add up to be a hundred percent so if one of these is seventy-five the other one has to be twenty five if one of these is thirty the other one has to be seventy so we can put that in the equation as an algebraic form so if we call the abundance of lithium six x that means lithium 7 has to be 100% minus X or in algebraic terms again we don't use a hundred percent we divide that by 100 to find the decimal so we'd call that 1 minus X so now we can solve for x so the abundance of lithium 6 I'll just call that x times the mass of lithium six which is six point zero one five one AMU and then plus the abundance of lithium seven which we said has to be one minus X because again they have to add up to be 100 percent times the mass of lithium seven which is seven point zero one six zero amu so now we're going to do a little algebra here so I'll just drop the unit's for now to make it a little easier so six point nine four equals six point zero one five one X and now we have to do a little bit of distributing so we have to multiply one minus X by whatever's outside so I'm going to multiply seven point zero one six zero by one and by negative X so that will give me seven point zero one six zero minus seven point zero one six zero X so now we're going to group our non X terms together and then our X terms together so I'm going to move this over to the other side so I'm basically going to subtract both sides by seven point zero one six zero so subtract by seven point zero one six zero and that gave me negative zero point zero seven six equals and then when I group these X terms together I got negative one point zero zero zero nine X so now to get X by itself now I divide both sides by negative one point zero zero zero nine negative one point zero zero zero nine and X came out to be basically point zero seven six so how does that help me determine the percent abundance of lithium six so what is X again well X is what we call the percent abundance of lithium six because remember all of this was lithium six and over here all of this was lithium seven so X is the decimal form of the abundance so what's the percent abundance where we just multiply by a hundred so the percent abundance would be seven point six percent lithium six well what if the question had also asked for the abundance of lithium-7 how would you find that well lithium sevens abundance is 1 minus X or 100% minus 7.6 so it would be ninety two point four percent now one last thing to do as always is to kind of check our answer and ask ourselves does this seem reasonable does it seem reasonable that the abundance of lithium 6 is so small well let's look at the two isotopes we have lithium-6 and lithium-7 so we know the mass has to be somewhere on the spectrum between basically 6 and 7 well which one should it be closer to how do we know well if I look at the average the average is 6 point 9 4 so I know there must be much more lithium 7 compared to lithium six and in fact there is there's only seven point six percent lithium 6 but there's 92.4% lithium 7 so this answer does seem reasonable okay last one so letter G is very similar to letter F so once you go ahead pause the video try it out on your own to ensure it makes sense and then come back here and hit play to check your answer okay copper exists as two isotopes copper 63 with a mass of 62 point nine two nine eight and copper 65 with a mass of sixty-four point nine two seven eight what are the percent abundances of the isotopes so it wants to know what's the percent abundance of both of these isotopes copper 63 and copper 65 well what else do we know well we know it's copper so we can find the average atomic mass of copper right here off the periodic table and it's 63 point five five so 63 point five five amu is the average and that must equal the abundance times the mass of the two different isotopes so once again we're going to call one of these X and the other one has to be a hundred percent minus X or in decimal terms one minus X so I'll call copper 63 X and I'll call copper 65 one minus X so x times the mass of copper 63 is 60 - point 9 - 9 8 a.m. you plus copper 65 the abundance is 1 minus X times its mass which is 64 point nine two seven eight AMU so now we're going to do some distributing and multiply 1 minus x times 64 point nine two seven eight and sixty-two point nine two nine eight by X so that'll give us 60 three point five five equals sixty two point nine two nine eight X plus sixty four point nine two seven eight minus sixty four point nine two seven eight X so we'll group Arnon X terms together and our X terms together on different sides of the equal sign so I'll subtract sixty four point nine two seven eight from both sides then I'll add these X terms together so that gives me negative one point three seven seven eight and this is two decimals for decimals I'm going to mark that as two decimals just to remind myself and when I add the X terms together I got negative one point nine nine eight zero X don't forget that zero by the way because it is for decimals - for decimals should still have for decimals when I want to isolate the X so divide both sides by negative one point nine nine eight zero one point nine nine eight zero and X came out to be point six eight nine five and it's three sigfigs remember I marked you here divided by five is going to be three sig figs so X came out to be basically point six nine zero or sixty nine point zero cinge well which abundance is this well X we made copper 63 right this was all copper 63 we also label to appear to remind ourselves and this one was all copper 65 so how do we find the abundance of the copper 65 because it asks for the abundances of both well again if one of them 69.0 the other one has to be a hundred minus sixty nine point zero or one minus 0.6 nine zero so we could do it either way so one minus x would give you one minus point six nine zero basically point three one zero or thirty one point zero percent because again these two have to add up to be a hundred percent so it's 69 percent copper 63 and it's 31 percent copper 65 well last question as always does this answer seem reasonable so if you look at the information that went in the calculation would you've expected it to be basically 70 percent 63 and thirty percent 65 so once again let's look at the data we know the masculine would be somewhere between 63 and 65 so somewhere in this range approximately and the average was 63 point five five so 64 is right here in the middle our mass is kind of right in here so it's definitely closer to 63 than it is 65 so you would expect there to be a larger amount of copper 63 than copper 65 and in fact there is