let's take a closer look at isotopes isotopes don't always occur in equal amounts in fact fluorine 19 is the only naturally occurring isotope that is 100 abundant this means that all of fluorine is fluorine 19. some other element examples are lithium which is 7.59 percent lithium-6 and 92.41 lithium-7 carbon is a 98.93 percent carbon-12 1.07 percent carbon 13 and approximately 0 carbon 14 it's just really really tiny there is carbon 14 that exists and we use it um for radiological dating but it's a really tiny percentage of what is in existence naturally for carbon so because of this we will look at things like atomic weight and atomic mass in much more depth than you might think is necessary but it helps us understand kind of what these numbers are in the periodic table why they maybe don't correlate to some of the masses you might expect based on isotopes and then what do we do with this information and how do we get this information so the atomic weight that's listed in the periodic table is a weighted average of the masses of the two isotopes according to their natural abundance so calculating a weighted average requires just a little bit of tweaking to the math that would go into calculating an actual average and we'll talk about the differences between those um and i will share an example video that is a really good primer on how to do a weighted average so what this looks like for dealing with atomic masses and calculating it as a weighted average so out of the 100 right all of the naturally occurring lithium we have our percentages of the individual isotopes so calculating a weighted average means taking the percent of lithium-6 and multiplying it by the mass for lithium-6 and then adding to that the percent for lithium-7 multiplied by the mass for lithium-7 this will give you the atomic weight of lithium in that matches what's in the periodic table essentially what this does is it tells you if i were to i wouldn't recommend this with lithium but if i were to scoop up a handful of lithium that's naturally occurring what would its apparent atomic mass be right seven percent almost eight percent of that contribution comes from a lighter isotope and 92 percent comes from a heavier isotope so we would expect that this atomic mass is probably much closer to what we would expect for the lithium-7 but it's going to be a little bit lower because a small portion of that sample is from a lighter isotope the masses of an individual atom are given an atomic mass units this is so that we don't have to deal with those really tiny numbers in grams that you saw in an earlier part of a video so a proton is 1.00 atomic mass units a neutron about the same a little bit heavier which we also saw in grams and then electrons are still super super tiny in relation so what this means is that right electrons are really not contributing to the mass of an atom but we can still talk about what its mass would be in relative terms in atomic mass units and of course let's do an example where we actually plug some numbers in so boron is 19.78 boron 10 and the atomic mass or of boron 10 is 10.0129 amu and it's 80.22 boron 11 which has a mass of 11.00931 amu we are asked to calculate the atomic weight of naturally occurring boron so calculate the atomic weight means do a weighted average of the isotopes that boron exists as so when we set this up we will have the percent for boron 10 times the mass for boron 10 plus the percent for boron 11 times the mass for boron 11. and i've got these highlighted here so the percentages i've highlighted the 19.78 and the 80.22 something to know about this that will be kind of a trick for later problems is that these have to add up to a hundred because it's a percentage of all of the naturally occurring isotopes of that element those two percentages will add up to a hundred so you may find in a later problem that you're only given one of the percentages depending on what it's asking you for you might have to do some other things to solve for that or you might have to do something as simple as 100 minus the given percentage to find the percentage of the other isotope you'll also notice that boron 10 has an atomic mass that's about 10. and this should make sense because the mass number is the number of protons plus the number of neutrons and the electrons are super tiny they don't contribute well each of the protons and neutrons weigh approximately one atomic mass unit so our boron 10 should have an atomic mass around 10 and our boron 11 has an atomic mass around 11. so in this setup right we have 0.1978 now i've taken my percentage and written it as a decimal here you can do 19.78 times 10.0129 but you would then have to divide it by a hundred because the percentage that this represents the 19.78 is really 19.78 over 100. so out of the 100 of the sample out of the whole sample 19.78 of it is boron 10. it's up to you how you set up your calculation i personally think it's a little bit easier to do the decimal but it's completely up to you as long as you don't forget to divide by 100 if you need to so 0.1978 times 10.0129 plus 0.802 times 11.0931 and this would give me 10.812 atomic mass units and if you take a look at the periodic table it's much much closer than either of the individual isotope masses to what the mass the atomic weight is listed in the periodic table now i've also got some markers here tracking sig figs don't forget about sig figs so this is a mixed operations you're going to have to remember your order of operations for tracking sig figs as you go through a problem like this so when you do 0.1978 times 10.0129 atomic mass units you're going to need to write that number down because those two values are going to limit your significant figures for the value that you get from that multiplication and you're going to need to write down the value 4.8022 times 11.00931 because that those two numbers are going to limit the sig figs that you get from that operation but then you add them together so the next step requires a different sig fig propagation rule before assigning sig figs to your final answer so don't forget about sig figs they're still here they're still important they're still adding additional complications um but right just just a reminder i will also post the video on weighted averages so that you can take a closer look at that our next topic is mass spectrometry and this is a technique used to identify molecules it's a huge hugely important technique we actually at washington university have some pretty famous mass spectrometrists i guess spectroscopist spectrometrists i think is the right word there so this is a technique that's used mostly in organic chemistry and higher and it's cool because it separates the molecules by mass and it indicates an abundance and it kind of sneakily looks a little bit like a cathode ray tube i don't have a super fancy picture but we take atoms and we shoot them through a vacuum and put them between magnetic poles and they are bent because they are charged molecules when they go through it's remarkably similar to a cathode ray tube so here i have an example of what you might see when you get some data that's generated from a mass spectrometer this is super simplified it's showing example data for if we were to only put silver into a mass spec what would the data look like and it's computer generated data so on our plot what we have listed is an intensity on the y-axis that goes up to a hundred percent and one of my peaks goes up to a hundred percent now that one isotope right because we're going to take a look at that x-axis in a second which has masses listed that one isotope is not 100 abundant what the computer has done is just taken the most abundant peak and set it to a hundred percent everything else is listed relative to that peak so on the x-axis i have mass here's an atomic mass units so i have an isotope that's at 106.9 and an isotope that's at 108.9 so this is silver 107 and silver 109. this and that's this is it for data from a mass spectrometer the intensity tells you about the natural abundance and it is listed relative to each other so the highest peak is set to 100 and then the position on the x-axis gives you the mass of the isotope we can use this data to calculate the atomic weight of silver so you could put a sample where you don't know the different abundances into a mass spectrometer get this output and then use it to determine the atomic weight of silver now if you are given an intensity where it's relative and the tallest one is 100 what we need are the actual percent abundance values because the total percent abundance needs to add up to 100 so i have one way of solving this on here there are many ways and you are welcome to use any other way that makes more sense to you but one of the ways that you can do this is to take each isotope so silver 107 is listed as 100 abundant we want to know that 100 relative to the entire sample what does that mean so if i do 100 divided by 100 plus 92.9 which is the abundance of the other isotope that's the silver 107 sample on the top the total sample on the bottom and that gives me 0.5184 which is 51.84 now for the other isotope i could set up the same calculation i could do silver 109 over the total sample or this is where i could do maybe a simpler calculation and bring in the fact that both of these if this is the entire isotope lineup for silver they have to add up to a hundred percent so i could do one minus 0.5184 to get the percent abundance for silver 109 and you should try this out you should get the same answer both ways now that i have my percent abundances i can calculate the weighted average to find the atomic weight of silver this is the same setup that we had for boron and it will be the percent abundance for silver 107 times the atomic mass for silver 107 plus the percent abundance for silver 109 times the atomic mass for silver 109 now let's do this with sig figs so from my previous example i don't have in my percentages i don't have sig figs really listed i only take my 100 as 100 which is nice and ambiguous it doesn't have any decimal places so for this example i have set it with four sig figs for the percents and then the atomic masses are given and each of those has six sig figs so when i do this my percent times atomic mass for silver 107 gives me 55.4195 now i do need to limit my sig figs here because i have four sig figs for one and six sig figs for the other this means that 55.4195 would be cut off at the one in the decimal places however i am not going to round i'm just going to put an underline on this and keep track of it in my next step when i do silver 109 times its percent abundance i also end up with four sig figs so 52.4482 limited at the four but i am not going to cut that number off and round it i'm going to leave it but underline the four so i know where to cut it off if i add those two numbers together i get 107.8677 and i have to use the addition subtraction rule to limit my sig figs in this answer that means i'm basing it on the number of decimal places and my answer should go to two decimal places so 107.87 would be my final answer and this is one of the many examples where addition and subtraction gives you what feels like a very non-intuitive value for your answer because we took two numbers with four sig figs each we added them together and we got an answer with five sig figs but that is correct it's the correct way to do it it's totally fine