Isotopes and Mass Spectrometry

[Music] animus Valley we're going to be looking at topic to ear and we're gonna be looking at the first section which is about isotopes and mass spectrometry and this is part of the IES chemistry course of film ed excel so some of us will be revision that we're going to be looking at the structure of the atom and the relative mass and charge of the subatomic particles what we mean by atomic number and mass number how we can use these numbers to determine our subatomic particles what we mean by an isotope and then the new part for a level is about the basic principles of a mass spectrometer and be able to conduct the mass spectra for specific diatomic molecules so mg coz you learned that our atoms contain three subatomic particles so we have protons neutrons and electrons so there should be nothing new to you our mass for a proton and a neutron are each one and our electron is one over one eight four zero sometimes written as one eight three six just depends your charge as your proton as your positive your neutron is neutral and your electron is negative and we also know that we find the protons and neutrons and the nucleus and the electrons are found and what we call quantum shells so in GCSE you probably called them energy levels but we'll talk more about this later and topic to eat when we talk about the electron shells and how we arrange the electrons so we've also discussed back and GCSE atomic number the atomic number as the number of protons that are n an atom we looked at mass number being the number of protons plus the number of neutrons remember is everything that has that mass of one that is in the center of the nucleus we also mentioned isotopes and isotopes are atoms of the same element so they have the same atomic number but a definite mass number so that because they are atoms of the same element they have the same protons but because they have a different number that tells us that they must have a different number of neutrons now be a wiener question could ask you to explain what an isotope is in terms of the subatomic particles you have to mention protons and neutrons or they can simply say in terms of atomic number or mass number you just make sure to meet the question carefully and the most common example that we use as chlorine-35 and chlorine-37 notice that they both have the same number of protons and of course the same number of electrons and these balance out because our atoms are neutral so the positives and negatives are going to balance but we have this different number of protons so coding 35 has so different number of neutrons chlorine 35 has 18 neutrons and coding 37 has 20 now when we're talking about the relative atomic mass you don't meet this last year and we're just going to kind of tighten up the definition that we use we say that this is the weighted mean average of an atom of an element compared to 1/12 of the mass of carbon-12 so when we're talking about relative and it's because all of the masses of these atoms are all measured relative to the mass of carbon-12 so carbon 12 is our standard and then we measure all of the masses of any other atom using that standard and when we have a relative atomic mass of course we're making sure that we're taking into account our average of all of our isotopes and we'll look at the calculation for that understand relative isotopic mass as the mass of a particular isotope relative to 1/12 the mass of carbon-12 so this would be looking at the mass of carbon after coding 35 or a coding 37 as opposed to Cody as a whole and each of these relative isotopic masses and the percentage abundance of the isotopes is used to calculate the relative atomic mass and you that's back and GCSE so this should not be new to you so we have our example here we have lithium that has isotopes of two relative isotopic masses of sex point zero one five and seven point zero one sex notice that and to be a level we have much more accurate values probable back in GCSE you would have just seen sex and seven where is it a level we use much more accurate isotopic values we've got our percentage abundances of seven point five nine percent and ninety two point four one so you calculate the relative atomic mass by doing the mass times the abundance plus the mass tamed the abundance and you do that for as many isotopes as you have and then you divide all of them by 100 here because you're obviously using percentages here so you've got six point zero one five times seven point five nine added to seven point zero one six times ninety two point four one which gives you that number here and we divide it all by a hundred and we get a relative atomic mass of lithium to three significant figures of sex point named for now let's shift to look at mass spectrometry so this is the new part at a level and that's the part that you've most likely not seen before so the first mass spectrometer was built back in 1918 so just over a hundred years ago by Francis Austin and he was a student of GJ Thomson and JD Thomson being famous and winning the Nobel Prize for discovering the electron so what Francis asked him dead is he used the instrument to sure that there were different forms of the same ailment and we now call these isotopes so before that's point isotopes were just theorized and potential that we may have different masses but we doesn't necessarily know why so we use the mass spectrometer in order to prove the isotopes exist and then expand our knowledge of atomic theory so remember back in 1918 atomic theory was still being developed and we didn't know the model of the atom that we know now so we need to know what happens in a mass spectrometer and how we then interpret the spectra now what you will find is that all specification papers ask you and get in detail how a mass spectrometer works no the new specification does not require that but it's good for you to know have the process but you're not going to be asked what are the steps and a mass spectrometer you're more likely you're going to be asked about the analysis part so basically what we do in a mass spectrometer is the particles are turned into positive ions so we take our atoms we turn them into ions we accelerate them and then we deflect them using an electric or a magnetic field and I'm gonna show you what that looks like in just a minute okay the path that the ion is going to take when it is deflected is all going to depend on something called its mass to charge ratio and you may see this written as M zette or M II particles with a very large mass to charge value are going to be deflected the least and the mass spectrometer and those with a law ended very are going to be deflected the most and what we have at the end of the mass spectrometer is a detector that allows the machine to detect what mass charge ratios they're going to be and that tells us the mass of each isotope so if we have two isotopes that will each have a different mass to charge ratio they will be deflected differently and then they will be detected at different points so mass spectrometry was initially used to identify isotopes back when they were first being utilized nowadays we use them to calculate molecular masses or if we make a new compound a new synthesis such as an organic substance we will use that just to find some information about the and you will come back and cover this and topics 10 of a level topic 15 and topic 20 of a - so this is what a mass spectrometer looks like okay and as I said and old specification papers you may be given this diagram and asked to explain it you are not going to be asked that new spec but it's good for you to understand how this works so we take our sample that we want to verify the isotopes and work out the abundances and then we vaporize the sample meaning we turn it into a gas it then passes through this point here called the ionization chamber and thus makes the positive ions these positive ions then pass through an electric field and it's at this point that they are accelerated and as they are accelerated they will pass through the tube and you can see here that they pass through a magnetic field and when we have this magnetic field this is where they are deflected and you can see that we now have three different paths and then at the end they are detected so the five steps that we use for a mass spectrometer as we vaporize the sample enter a gas we ionize them and the charge that we get is typically a one plus charge we are looking for the ions to become one plus so that their mass to charge ratio is simply going to be their mass so if you have a mass of let's say 16 and the charge as one my master charge ratio is going to be 16 telling me that that isotope has a mass of 16 and we'll look at what happens if it has a two plus charge later so once they are made into ions they are accelerated by this electric field we are then deflected using the magnetic field and then they are detected using various different methods so either electric or photographic but remember you're not going to be asked which method is being used so the path that we isotope is going to follow is going to all be dependent on this mass to charge ratio or the MZ ratio so because the ions of heavier isotopes are going to have larger MZ values so for example chlorine-37 would have a mass to charge ratio of 37 whereas coding 35 would have a month to charge ratio of 35 so the heavier isotope the 37 is going to have a different path and it's going to have a larger curve as opposed to the coding 35 then as I've said the majority of the ions are one positive charge - the amount of separation is all just due to the mass so you're heavier masses and your lighter masses are going to follow two different pathways actually I am somehow gets a two positive charge it's going to be deflected even more and this is because it's MZ barrier is going to be hashed so if we have 37 over 2 and 35 over 2 or going to get different masses master child ratios as we would if it was a one plus a good thing there's questions about two plus ions don't come up very often and if they do they are simply just asking about the effect of that on the mass-to-charge ratio they're not going to go into a lot more detail about that so you can see here that we've got three different answer talks of neon and the heavier isotopes are going to be deflected less so that's the pathway number one the black one and pathway number two the blue one is your medium isotope so it's deflected more than the neon 22 but it's deflected less than the neon 20 because that is the lace latest isotope so undergoes more deflection and then we get a spectrum that looks like this so at the detector it will check how much of each of these ions so we are measuring the abundance of each of these and that is our spectrum that has been formed there so a mass spectrum is going to give the following information we are going to see the possessions of the peaks to give us the mass of substances we're going to see the peak intensity to give us the abundance so the peak intensity is otherwise known as the peak height and the possessions gives us the atomic mass and then the highest abundance is scaled to a hundred percent and then the other values are skilled as well so we the mass spectrometer does all of this automatically and we get a spectrum that has got the different peaks what their abundances already labeled so you may get something that looks like this where we've got a sample that has been passed through a mass spectrometer and we get this mass spectrum so you can see that we have our master charge ratio and we have our height so we have master charge ratio of 79 is at 50.5% and 81 as a 49.5 and you can then determine the relative atomic mass so you would have fifty point five times seventy nine plus forty-nine point five times 81 all divided by 100 and you would get a relative atomic mass of seventy nine point nine nine so Battin IGCSE you were given the abundances and you were asked anything about where they come from this is nowhere the abundances come from now we can focus on the mass spectrum of diatomic molecules at this point you will leave as a polyatomic mass spectrum later on and that's course but we're just gonna focus on diatomic molecules because these are makes and simple so diatomic molecules are going to contain only two atoms and we can analyze them by mass spectrometry and we can determine our molec relative molecular mass of the element and we look at our Peaks and our abundances in order to do this so if we have something that looks like this we can see that we have a mass spectrum of chlorine which is going to exist as two diatomic molecules and we have a number of different Peaks here so this one here is a peak at 35 this one here is a peak at 37 and then you can see that we have three other Peaks so we have peaks at 70 72 and 74 at a level you are expected to be able to explain where these Peaks come from and it's actually a lot more simple than you think so we have two main Peaks for at 35 and 37 and if we measure the abundance we will get bundles of 75 and 25% now the other Peaks come from the fact that we have a diatomic molecule and all these Peaks are is just different isotopes joining together so if I have two chlorine-35 joining together I'm going to have a mass of 70 however if I get according thirty-five joining with the codeine 37 I'm going to get a different mass that's a matt72 and if I have according 37 bonding with the codeine 37 then I'm going to get a mass of 74 now this is the part that is a little bit more difficult is understanding the peak Heights so let's take this nice and slow so we said that chlorine 35 has an abundance of 75% and chlorine 37 has an abundance of 25 so we had that from the previous slide this means there is a three and four chance of selecting according thirty-five atom from a sample of codeine atoms so if I have a chance of selecting two chlorine thirty fives I have to multiply 3 over 4 times 3 over 4 which gives me a chance of 9 over 16 of me selecting 2 chlorine 35 atoms now we'll look at the chlorine 37 that is a 1 in 4 chance of me selecting a Corinne 37 so fi the chance of me combining about 37 with a 35 as 1 over 4 times 3 over 4 giving me 3 over 16 now I can have a 35 and a 37 so I could select the 35 first followed by the 37 or I could collect the 37 first followed by the 35 now both of these combinations have the same chance of being selected they both have the chance of 3 over 16 but I have to take into account the fact that I could pick the 35 first or the 37 first so I have to do is multiply it by 2 telling me that the chances of me getting a combination with a mass spectrum I saw a mass of 72 is sex over 16 whereas the 17 was 9 over 16 the chances of me selecting too cold in 37 s as 1 over 4 times 1 over 4 which is a 1 and 16 chance that my mass is going to be 74 so this now gives me a ratio of 9 to sex to 1 and less corresponds to the peak Heights so if I go back to my spectrum here the 70 is going to have a ratio of 9 to sex to 1 for the 70 to 72 to 74 and you can see that and the spectrum you can see the different heights relative to each other so what we can now do is we can use that ratio in order to determine the relative molecular mass of coding so we do 9 our ratio times 70 Plus sex times 72 plus 1 times 74 and divide at all that's 10 not by 100 but by the total value of 16 because remember all of these are 9 over 16 sex over 16 and 1 over 16 and that gives us relative molecular mass of 71 for coding now because this is a diatomic co2 that tail and that 71 that tells me when I take my individual atoms I get 35.5 and we already know that the relative atomic mass of coding is thirty five point five because we can get that from the abundance so this is just further proof that isotopes exist and how we can use them to determine the relative molecular mass as well as confirm what we already know about the relative atomic mass so we're going to do some questions that are gonna give us some information about why we have this particular spectrum on the next slide so we're gonna be looking at bromine Mustang so we have what are a relative isotopic masses identifying the particles response both the peaks and then deducing the relative abundance and explained the heights of the three peaks so we've got these here so our first question was what are our isotopic masses and our isotopic masses are these two here so our isotopic masses are 79 and 81 the second one the second question was looking at what are the particles responsible for the three peaks so for 158 that must be 279 bromine atoms combining 460 it must be an 81 and a 79 or it could be a 79 and an 81 remember the order is important here and for the hundred and sixty-two it could be an 81 and then 81 okay so now we want to look at the relative Heights and Tim see how these Taylor's the why we get there the peak heights for the combinations so we can see that the abundance for 79 and 81 are going to be exactly the same so that tells us that they are 50/50 so you get a 50% chance of finding one of these two so that in other words has a one in two chance so if we want to figure it out well for 158 the chance of 270 names is going to be 1 over 2 times 1 over 2 which gives me 1 over 4 the chance of me getting a 79 and an 81 is 1 over 2 times 1 over 2 which is 1 over 4 again but remember that's gonna be equal to one sex day but so as this one the 81 first followed by the 79 so again 1 over 2 times 1 over 2 is 1 over 4 but I need to add these two together because both of these two are going to have a mass of one sex day so this becomes a total chance of failing I also talked that is that one sex day as 1 over 2 and then the chance of failing 280 once again is 1 over 2 times 1 over 2 giving us 1 over 4 so you can see that the 150 in 162 I've exactly half because my ratio should be a 1 to 2 to 1 because we have 1/4 to 1/2 to 1/4 so this is a passed paper question and it's using very much the same spectrum that we just use I just want to show you in terms of the marks how you get these marks so we want to complete the mass spectrum for a sample bromine gas that tin contains approximately half of the 79 and a half of the 81 isotopes now the intensity that we use doesn't matter as long as we get the correct ratios so if I have 79 and 81 as long as they are identical height then we would get a mark here so let's make them there is 79 and there is 81 at the same height of course you would have a really when you're drawing these and that would give you your first mark then we have how did he combined so if we have 79 times 2 we know that that's 150 yet if we have 79 plus 81 we know that is 160 and 81 times 2 is 162 so our combinations should be our other peaks at 1 5 8 1 sex date and one sex - so we have to then put that onto our graph here our spectrum so we have the one sex day that's going to be the most common one remember these are going to be in a ratio of 1 to 2 to 1 then we have 1 5 8 and 1 6 2 and these should be half the height of one sex day so let's have it here again of course you're gonna have a ruler when you draw your's so we get something like that so you get a mark point 2 4 having the peaks and the right place for 1 5 8 and one sex - mark point 3 4 having the peak at once XD a mark point 4 as for having them and the cadet ratio so first Klee drawing five legs using a ruler and just making sure that the heights are the same you can get four marks on an exam these should be some of the easiest marks that you will pick up and less paper so then we want to calculate the relative atomic mass for a sample that was found to contain forty-seven percent seventy nine and fifty three percent eighty one so this is exactly what you learned back in GCSE so we have 47 times 79 plus 53 times 81 all divided by 100 that gets you your first mark and the second mark is for the answer of 80 point one of course putting all of that into your calculator and then what effect FME on the M Z value of the peak if the iron detected had lost two electrons well what we would see as our MZ or any value is going to be haft because if you have let's imagine 79 over 1 that's going to give you a ratio of 79 but if you have 79 over 2 you're going to get thirty nine point five so you're gonna have that value and that's for your last March so that's everything for topic 2 e isotopes and mass spectral mass spectrometry I would strongly recommend that you do practice how to draw the mass spectra and you try the questions on the textbook if you have any questions on the topic please feel free to leave a comment below and we hope to see you back soon [Music]

[Music] animus Valley we&#39;re going to be looking at topic to ear and we&#39;re gonna be looking at the first section which is about isotopes and mass spectrometry and this is part of the IES chemistry course of film ed excel so some of us will be revision that we&#39;re going to be looking at the structure of the atom and the relative mass and charge of the subatomic particles what we mean by atomic number and mass number how we can use these numbers to determine our subatomic particles what we mean by an isotope and then the new part for a level is about the basic principles of a mass spectrometer and be able to conduct the mass spectra for specific diatomic molecules so mg coz you learned that our atoms contain three subatomic particles so we have protons neutrons and electrons so there should be nothing new to you our mass for a proton and a neutron are each one and our electron is one over one eight four zero sometimes written as one eight three six just depends your charge as your proton as your positive your neutron is neutral and your electron is negative and we also know that we find the protons and neutrons and the nucleus and the electrons are found and what we call quantum shells so in GCSE you probably called them energy levels but we&#39;ll talk more about this later and topic to eat when we talk about the electron shells and how we arrange the electrons so we&#39;ve also discussed back and GCSE atomic number the atomic number as the number of protons that are n an atom we looked at mass number being the number of protons plus the number of neutrons remember is everything that has that mass of one that is in the center of the nucleus we also mentioned isotopes and isotopes are atoms of the same element so they have the same atomic number but a definite mass number so that because they are atoms of the same element they have the same protons but because they have a different number that tells us that they must have a different number of neutrons now be a wiener question could ask you to explain what an isotope is in terms of the subatomic particles you have to mention protons and neutrons or they can simply say in terms of atomic number or mass number you just make sure to meet the question carefully and the most common example that we use as chlorine-35 and chlorine-37 notice that they both have the same number of protons and of course the same number of electrons and these balance out because our atoms are neutral so the positives and negatives are going to balance but we have this different number of protons so coding 35 has so different number of neutrons chlorine 35 has 18 neutrons and coding 37 has 20 now when we&#39;re talking about the relative atomic mass you don&#39;t meet this last year and we&#39;re just going to kind of tighten up the definition that we use we say that this is the weighted mean average of an atom of an element compared to 1/12 of the mass of carbon-12 so when we&#39;re talking about relative and it&#39;s because all of the masses of these atoms are all measured relative to the mass of carbon-12 so carbon 12 is our standard and then we measure all of the masses of any other atom using that standard and when we have a relative atomic mass of course we&#39;re making sure that we&#39;re taking into account our average of all of our isotopes and we&#39;ll look at the calculation for that understand relative isotopic mass as the mass of a particular isotope relative to 1/12 the mass of carbon-12 so this would be looking at the mass of carbon after coding 35 or a coding 37 as opposed to Cody as a whole and each of these relative isotopic masses and the percentage abundance of the isotopes is used to calculate the relative atomic mass and you that&#39;s back and GCSE so this should not be new to you so we have our example here we have lithium that has isotopes of two relative isotopic masses of sex point zero one five and seven point zero one sex notice that and to be a level we have much more accurate values probable back in GCSE you would have just seen sex and seven where is it a level we use much more accurate isotopic values we&#39;ve got our percentage abundances of seven point five nine percent and ninety two point four one so you calculate the relative atomic mass by doing the mass times the abundance plus the mass tamed the abundance and you do that for as many isotopes as you have and then you divide all of them by 100 here because you&#39;re obviously using percentages here so you&#39;ve got six point zero one five times seven point five nine added to seven point zero one six times ninety two point four one which gives you that number here and we divide it all by a hundred and we get a relative atomic mass of lithium to three significant figures of sex point named for now let&#39;s shift to look at mass spectrometry so this is the new part at a level and that&#39;s the part that you&#39;ve most likely not seen before so the first mass spectrometer was built back in 1918 so just over a hundred years ago by Francis Austin and he was a student of GJ Thomson and JD Thomson being famous and winning the Nobel Prize for discovering the electron so what Francis asked him dead is he used the instrument to sure that there were different forms of the same ailment and we now call these isotopes so before that&#39;s point isotopes were just theorized and potential that we may have different masses but we doesn&#39;t necessarily know why so we use the mass spectrometer in order to prove the isotopes exist and then expand our knowledge of atomic theory so remember back in 1918 atomic theory was still being developed and we didn&#39;t know the model of the atom that we know now so we need to know what happens in a mass spectrometer and how we then interpret the spectra now what you will find is that all specification papers ask you and get in detail how a mass spectrometer works no the new specification does not require that but it&#39;s good for you to know have the process but you&#39;re not going to be asked what are the steps and a mass spectrometer you&#39;re more likely you&#39;re going to be asked about the analysis part so basically what we do in a mass spectrometer is the particles are turned into positive ions so we take our atoms we turn them into ions we accelerate them and then we deflect them using an electric or a magnetic field and I&#39;m gonna show you what that looks like in just a minute okay the path that the ion is going to take when it is deflected is all going to depend on something called its mass to charge ratio and you may see this written as M zette or M II particles with a very large mass to charge value are going to be deflected the least and the mass spectrometer and those with a law ended very are going to be deflected the most and what we have at the end of the mass spectrometer is a detector that allows the machine to detect what mass charge ratios they&#39;re going to be and that tells us the mass of each isotope so if we have two isotopes that will each have a different mass to charge ratio they will be deflected differently and then they will be detected at different points so mass spectrometry was initially used to identify isotopes back when they were first being utilized nowadays we use them to calculate molecular masses or if we make a new compound a new synthesis such as an organic substance we will use that just to find some information about the and you will come back and cover this and topics 10 of a level topic 15 and topic 20 of a - so this is what a mass spectrometer looks like okay and as I said and old specification papers you may be given this diagram and asked to explain it you are not going to be asked that new spec but it&#39;s good for you to understand how this works so we take our sample that we want to verify the isotopes and work out the abundances and then we vaporize the sample meaning we turn it into a gas it then passes through this point here called the ionization chamber and thus makes the positive ions these positive ions then pass through an electric field and it&#39;s at this point that they are accelerated and as they are accelerated they will pass through the tube and you can see here that they pass through a magnetic field and when we have this magnetic field this is where they are deflected and you can see that we now have three different paths and then at the end they are detected so the five steps that we use for a mass spectrometer as we vaporize the sample enter a gas we ionize them and the charge that we get is typically a one plus charge we are looking for the ions to become one plus so that their mass to charge ratio is simply going to be their mass so if you have a mass of let&#39;s say 16 and the charge as one my master charge ratio is going to be 16 telling me that that isotope has a mass of 16 and we&#39;ll look at what happens if it has a two plus charge later so once they are made into ions they are accelerated by this electric field we are then deflected using the magnetic field and then they are detected using various different methods so either electric or photographic but remember you&#39;re not going to be asked which method is being used so the path that we isotope is going to follow is going to all be dependent on this mass to charge ratio or the MZ ratio so because the ions of heavier isotopes are going to have larger MZ values so for example chlorine-37 would have a mass to charge ratio of 37 whereas coding 35 would have a month to charge ratio of 35 so the heavier isotope the 37 is going to have a different path and it&#39;s going to have a larger curve as opposed to the coding 35 then as I&#39;ve said the majority of the ions are one positive charge - the amount of separation is all just due to the mass so you&#39;re heavier masses and your lighter masses are going to follow two different pathways actually I am somehow gets a two positive charge it&#39;s going to be deflected even more and this is because it&#39;s MZ barrier is going to be hashed so if we have 37 over 2 and 35 over 2 or going to get different masses master child ratios as we would if it was a one plus a good thing there&#39;s questions about two plus ions don&#39;t come up very often and if they do they are simply just asking about the effect of that on the mass-to-charge ratio they&#39;re not going to go into a lot more detail about that so you can see here that we&#39;ve got three different answer talks of neon and the heavier isotopes are going to be deflected less so that&#39;s the pathway number one the black one and pathway number two the blue one is your medium isotope so it&#39;s deflected more than the neon 22 but it&#39;s deflected less than the neon 20 because that is the lace latest isotope so undergoes more deflection and then we get a spectrum that looks like this so at the detector it will check how much of each of these ions so we are measuring the abundance of each of these and that is our spectrum that has been formed there so a mass spectrum is going to give the following information we are going to see the possessions of the peaks to give us the mass of substances we&#39;re going to see the peak intensity to give us the abundance so the peak intensity is otherwise known as the peak height and the possessions gives us the atomic mass and then the highest abundance is scaled to a hundred percent and then the other values are skilled as well so we the mass spectrometer does all of this automatically and we get a spectrum that has got the different peaks what their abundances already labeled so you may get something that looks like this where we&#39;ve got a sample that has been passed through a mass spectrometer and we get this mass spectrum so you can see that we have our master charge ratio and we have our height so we have master charge ratio of 79 is at 50.5% and 81 as a 49.5 and you can then determine the relative atomic mass so you would have fifty point five times seventy nine plus forty-nine point five times 81 all divided by 100 and you would get a relative atomic mass of seventy nine point nine nine so Battin IGCSE you were given the abundances and you were asked anything about where they come from this is nowhere the abundances come from now we can focus on the mass spectrum of diatomic molecules at this point you will leave as a polyatomic mass spectrum later on and that&#39;s course but we&#39;re just gonna focus on diatomic molecules because these are makes and simple so diatomic molecules are going to contain only two atoms and we can analyze them by mass spectrometry and we can determine our molec relative molecular mass of the element and we look at our Peaks and our abundances in order to do this so if we have something that looks like this we can see that we have a mass spectrum of chlorine which is going to exist as two diatomic molecules and we have a number of different Peaks here so this one here is a peak at 35 this one here is a peak at 37 and then you can see that we have three other Peaks so we have peaks at 70 72 and 74 at a level you are expected to be able to explain where these Peaks come from and it&#39;s actually a lot more simple than you think so we have two main Peaks for at 35 and 37 and if we measure the abundance we will get bundles of 75 and 25% now the other Peaks come from the fact that we have a diatomic molecule and all these Peaks are is just different isotopes joining together so if I have two chlorine-35 joining together I&#39;m going to have a mass of 70 however if I get according thirty-five joining with the codeine 37 I&#39;m going to get a different mass that&#39;s a matt72 and if I have according 37 bonding with the codeine 37 then I&#39;m going to get a mass of 74 now this is the part that is a little bit more difficult is understanding the peak Heights so let&#39;s take this nice and slow so we said that chlorine 35 has an abundance of 75% and chlorine 37 has an abundance of 25 so we had that from the previous slide this means there is a three and four chance of selecting according thirty-five atom from a sample of codeine atoms so if I have a chance of selecting two chlorine thirty fives I have to multiply 3 over 4 times 3 over 4 which gives me a chance of 9 over 16 of me selecting 2 chlorine 35 atoms now we&#39;ll look at the chlorine 37 that is a 1 in 4 chance of me selecting a Corinne 37 so fi the chance of me combining about 37 with a 35 as 1 over 4 times 3 over 4 giving me 3 over 16 now I can have a 35 and a 37 so I could select the 35 first followed by the 37 or I could collect the 37 first followed by the 35 now both of these combinations have the same chance of being selected they both have the chance of 3 over 16 but I have to take into account the fact that I could pick the 35 first or the 37 first so I have to do is multiply it by 2 telling me that the chances of me getting a combination with a mass spectrum I saw a mass of 72 is sex over 16 whereas the 17 was 9 over 16 the chances of me selecting too cold in 37 s as 1 over 4 times 1 over 4 which is a 1 and 16 chance that my mass is going to be 74 so this now gives me a ratio of 9 to sex to 1 and less corresponds to the peak Heights so if I go back to my spectrum here the 70 is going to have a ratio of 9 to sex to 1 for the 70 to 72 to 74 and you can see that and the spectrum you can see the different heights relative to each other so what we can now do is we can use that ratio in order to determine the relative molecular mass of coding so we do 9 our ratio times 70 Plus sex times 72 plus 1 times 74 and divide at all that&#39;s 10 not by 100 but by the total value of 16 because remember all of these are 9 over 16 sex over 16 and 1 over 16 and that gives us relative molecular mass of 71 for coding now because this is a diatomic co2 that tail and that 71 that tells me when I take my individual atoms I get 35.5 and we already know that the relative atomic mass of coding is thirty five point five because we can get that from the abundance so this is just further proof that isotopes exist and how we can use them to determine the relative molecular mass as well as confirm what we already know about the relative atomic mass so we&#39;re going to do some questions that are gonna give us some information about why we have this particular spectrum on the next slide so we&#39;re gonna be looking at bromine Mustang so we have what are a relative isotopic masses identifying the particles response both the peaks and then deducing the relative abundance and explained the heights of the three peaks so we&#39;ve got these here so our first question was what are our isotopic masses and our isotopic masses are these two here so our isotopic masses are 79 and 81 the second one the second question was looking at what are the particles responsible for the three peaks so for 158 that must be 279 bromine atoms combining 460 it must be an 81 and a 79 or it could be a 79 and an 81 remember the order is important here and for the hundred and sixty-two it could be an 81 and then 81 okay so now we want to look at the relative Heights and Tim see how these Taylor&#39;s the why we get there the peak heights for the combinations so we can see that the abundance for 79 and 81 are going to be exactly the same so that tells us that they are 50/50 so you get a 50% chance of finding one of these two so that in other words has a one in two chance so if we want to figure it out well for 158 the chance of 270 names is going to be 1 over 2 times 1 over 2 which gives me 1 over 4 the chance of me getting a 79 and an 81 is 1 over 2 times 1 over 2 which is 1 over 4 again but remember that&#39;s gonna be equal to one sex day but so as this one the 81 first followed by the 79 so again 1 over 2 times 1 over 2 is 1 over 4 but I need to add these two together because both of these two are going to have a mass of one sex day so this becomes a total chance of failing I also talked that is that one sex day as 1 over 2 and then the chance of failing 280 once again is 1 over 2 times 1 over 2 giving us 1 over 4 so you can see that the 150 in 162 I&#39;ve exactly half because my ratio should be a 1 to 2 to 1 because we have 1/4 to 1/2 to 1/4 so this is a passed paper question and it&#39;s using very much the same spectrum that we just use I just want to show you in terms of the marks how you get these marks so we want to complete the mass spectrum for a sample bromine gas that tin contains approximately half of the 79 and a half of the 81 isotopes now the intensity that we use doesn&#39;t matter as long as we get the correct ratios so if I have 79 and 81 as long as they are identical height then we would get a mark here so let&#39;s make them there is 79 and there is 81 at the same height of course you would have a really when you&#39;re drawing these and that would give you your first mark then we have how did he combined so if we have 79 times 2 we know that that&#39;s 150 yet if we have 79 plus 81 we know that is 160 and 81 times 2 is 162 so our combinations should be our other peaks at 1 5 8 1 sex date and one sex - so we have to then put that onto our graph here our spectrum so we have the one sex day that&#39;s going to be the most common one remember these are going to be in a ratio of 1 to 2 to 1 then we have 1 5 8 and 1 6 2 and these should be half the height of one sex day so let&#39;s have it here again of course you&#39;re gonna have a ruler when you draw your&#39;s so we get something like that so you get a mark point 2 4 having the peaks and the right place for 1 5 8 and one sex - mark point 3 4 having the peak at once XD a mark point 4 as for having them and the cadet ratio so first Klee drawing five legs using a ruler and just making sure that the heights are the same you can get four marks on an exam these should be some of the easiest marks that you will pick up and less paper so then we want to calculate the relative atomic mass for a sample that was found to contain forty-seven percent seventy nine and fifty three percent eighty one so this is exactly what you learned back in GCSE so we have 47 times 79 plus 53 times 81 all divided by 100 that gets you your first mark and the second mark is for the answer of 80 point one of course putting all of that into your calculator and then what effect FME on the M Z value of the peak if the iron detected had lost two electrons well what we would see as our MZ or any value is going to be haft because if you have let&#39;s imagine 79 over 1 that&#39;s going to give you a ratio of 79 but if you have 79 over 2 you&#39;re going to get thirty nine point five so you&#39;re gonna have that value and that&#39;s for your last March so that&#39;s everything for topic 2 e isotopes and mass spectral mass spectrometry I would strongly recommend that you do practice how to draw the mass spectra and you try the questions on the textbook if you have any questions on the topic please feel free to leave a comment below and we hope to see you back soon [Music]

Transcript for:Isotopes and Mass Spectrometry

Transcript for:
Isotopes and Mass Spectrometry