talking to you about isotopes but i haven't really gone much farther into it than saying that these isotopes exist and that um an isotope just means that it's the same element but the atom just has a different number of neutrons in it how do we know this well what we do now is we use a mass spectrometer a mass spec is how we can identify the isotopes we have present really cool instrument um it works but your sample is injected here and the sample becomes vaporized meaning it's forced into the gas phase the particles or the atoms themselves of your sample become charged meaning we form ions it's ionized and we want them charged because then we're going to send them through a magnetic field and separate them out by their mass to charge ratio now we can set the mass effect so it gives it a very specific charge most often it will give it a charge of plus one so these particles will go through with a plus one charge and we can then identify what they are based on what reaches the detector you don't need to be able to fully explain this instrument it's a senior level chemistry instrument that we use but you do need to understand you know your sample goes in we vaporize it meaning we get into the gas phase we ionize it we run it through some kind of magnetic field that's going to separate out those particles on their mass to charge ratio and then we can detect those particles at the end this helps us identify the different ions we have or the different isotopes we have rather because they're all going to form ions but let's say we looked at you know we looked at carbon 12 and carbon 13. well the carbon-12 ion is going to have a different mass to charge ratio than the carbon-13 ion because carbon-12 has one less neutron in carbon-13 it's going to separate these out based on their individual isotopes we can use this information to better identify what elements we have and how we can use this information to then find what we can what we call the atomic mass which is that number below your elemental symbol on your periodic table so for example the mass spectrum of gallium elemental symbol ga is shown below the atomic mass of gallium is 69.7 amu which of the following statements is correct this atomic mass is found by a weighted average of the um mass and percent of each gallium isotope what do i mean by a weighted average i mean that if i have 100 atoms of gallium 60 of them will have an atomic mass unit or mass of 68.9 and about 40 of them will have an atomic mass unit of 71.9 the atomic mass of gallium will be some kind of average between these two numbers taking into account those percentages so does this mean that all gallium atoms weigh amu well no because 60 percent of them weigh 68.9 71.9 um is the weight of the other roughly 40 of them so no not all gallium atoms actually weigh 69.7 amu even though that is our atomic mass 69.7 isn't actually the mass of any gallium isotope how about number two the atomic mass of gallium is the average of 68.9 and 71.9 no no that's not it either one if i average these two numbers it does not equal 69.7 but two i know that about you know we've got about 60 percent that are due to 68.9 amu whereas only um 40 are due to seventy one point nine eighteen so that's not gonna be true either instead three will be correct the atomic mass of gallium will actually be closer to 69 than 71 because there are more atoms that weigh 68.9 that is in fact correct that's going to be the answer you need to be able to look at these in your textbook or a couple more examples with them as well there's another example in your slides here and then there's examples at the end of the extra problems your textbook has extra problems etc but be cautious when working through the textbook problems because it can often be very difficult to read the graphs they give you with the precision they seem to expect based upon the answers try not to get too frustrated we do want you to be able to read these though and be like okay sixty percent of it's due to sixty eight point nine seventy one um forty percent is due to seventy one point nine it's definitely gonna have a mass closer to this than this but they both do contribute to the overall atomic mass of the um atom itself so how we find this number and we'll talk about how to actually calculate that in just a second so one more example i want you to estimate the atomic mass of each species below go ahead and pause the video and kind of look at these and see if you can have any idea what this means now again the atomic mass is what is found on the periodic table so for example if i was to look up silver on the periodic table i do not have a table in front of me and i do not remember its atomic number at all but the atomic number would be listed here and its atomic mass would be listed here this graph can give me an idea of what this number is that's what i mean by estimate the atomic mass of each species what i see here is i have a silver 107 isotope and a silver 109 isotope now i could go ahead and figure out the number of neutrons of each species but in this case i don't need that i don't need that information i don't need to know how many neutrons and protons there are because they're being treated equally in the weight or mass of the atom itself what i do want to see here though is i look and i'm like okay this is a hundred percent of them are 107 but then i've got peaks at 109.2 how can i have 100 of one in peaks of the other well the way we actually do this works and i cannot draw a straight line um the way it actually works is that we normalize these values so say if 50 atoms go through with a mass of 107 we normalize those 50 items up to 100 percent so that any everything else that's measured is measured in comparison to that so if i had 50 atoms go through at 107 and then say 25 atoms go through 109 this this peak here would be half the height of 107 it would be at 50 so looking at this what i see is that okay 107 is definitely the more abundant isotope but not by much this isotope at 109 is actually pretty close they're not equal so the uh the atomic mass of silver will not be a true average of these two it will not be a hundred and eight it will be a little bit below 108 a little closer to 107. so i'm going to say i'm going to estimate the mass to be 107.9 amu because this looks uh the silver 109 to me looks like it could be present anywhere between 90 and 95 it actually looks higher than 90 to me when i if i try to break this value up by five it looks like it's a little higher than ninety percent so i'm gonna estimate 107.9 amu if i look at the pr table the actual mass of silver actual atomic mass is 107.87 amu now i actually did guess on that but um i've read a lot more mass effect than you so i don't expect you to have a perfect guess but i expect you to have a value that is above 107 but close to 108 but below 108 because if i was to average these two out i would see it should be below 108 it's not they're not equally um weighted out there's more 107 atoms 108 but it should be near that point as well because again 107 the ice of 107 masking um for the mass number is the lowest we're going for silver next let's look at chlorine chlorine has two naturally occurring isotopes fluorine 35 and chlorine 37. i see that chlorine 35 is the most abundant it's been normalized up to 100 and chlorine 37 is down here this chlorine 37 seems about one third of the height of the chlorine 35 to me so i know that the mass will be closer to chlorine 35 than chlorine 37. 36 would be halfway so if i had 50 this would be chlorine 36 this would have an atomic mass unit of 36 i know to be below 36. i'm going to take this distance here and take it and divide it by 3. so i'm going to estimate that this is close to 35.7 amu saying that that .7 is about one-third of the way between 35 and 37 the difference between those two values if i look up the actual mass it's 35.45 am so i'm a little more off but not far this is what you need to be able to do if i was to give you a graph like this so atomic masses i keep talking to you guys about these atomic masses let's go ahead and actually figure out how the heck we're calculating these things these are again the values straight from the periodic table but how are we calculating that we need to remember some basic definitions first percent percent is your part divided by your whole times 100 percent and possible values here range from 0 to 100 percent assuming we put the smaller number on top and the larger number in the denominator a fraction is your part divided by whole you guys can convert back and forth between fraction and percent hopefully very quickly because you'll need to do it very quickly very intuitively in this unit but the possible values here would be from zero to one if you're unsure remember take your percentage and divide by 100 that brings you back to your fraction so how does this relate to atomic weight though atomic weight is what we call a weighted average of isotopic masses so what do i mean by that well let's look at a weighted average example you most likely have multiple classes that do weighted average to figure out your final course grade we do it for this class so these are not your breakdowns for this class but let's say for a class you had 50 of your grades due to a midterm 25 due to the final exam and 25 due to your lab average and let's say we got an 82 on our midterm a 78 on our final and 97 on our lab average how do we figure out our final grade well our average will be our percent in fractional form in mass we're going to call this percent abundance but we're going to take this percent and bring it back into fractional form so point 5 0 i'm going to write the 0 there because i don't want to lose sig figs times eighty-two point five or fifty percent of my grade is due to that midterm so i wanna see out of a hundred points possible how many points did i earn from my midterm plus 0.25 times 78 25 is due to that final exam plus 0.25 times 97. the last 25 is due to our lab rate when you average this out you'll get a value of 84.75 don't forget to round for sig figs and you should find your answer to be 85 so if this was your class you would have an average of 85. we're going to do the same thing now but for elements to figure out our atomic masses on that table so let me go ahead and erase this for some space hopefully you've paused and recorded what you needed magnesium has three naturally occurring isotopes magnesium 24 magnesium 25 and magnesium 26. magnesium 24 has an atomic mass unit of 23.98 so one atom of magnesium 24 weighs 23.9850 amu that is based on the number of protons and neutrons it has in its nucleus the percent abundance of magnesium 24 is 78.99 which means almost 79 of every magnesium atom analyzed is magnesium 24. magnesium-25 has a mass atomic mass unit of 24.9858 that's 10 of the magnesium atoms that naturally occur at magnesium 26 with a mass of 25.929 has a percent abundance of 11.01 you know these are not whole numbers um the entire periodic table is based on the carbon-12 atom we'll talk about that in just a minute but that is the only one carbon-12 isotope specifically is the only isotope that will have a whole number the rest of these will be in a fractional form because they're based on comparing them um to normalizing them to carbon 12 but we do see that 23.98 is very close to 24 24.98 very close to 25 25.98 close to 26. and that's where we get these numbers here 24 25 and 26 are again the numbers of protons must be transposed electrons but um they're very close to what we're recording for the actual mass so how do i add this up though what do i do to find the atomic weight so we're going to write a t w t a t for atomic wt for weight i'm going to do the same thing i'm going to take the percent abundance but i'm going to convert this to fractional abundance so i'm going to undo the percentage 0.78 0.1000 don't forget the zeros here because these are your precision your measurements and 0.1101 so i've got .7899 times 23.9850 this is amu plus um point zero zero point one zero zero zero times twenty four point nine eight five eight sorry i know some of these numbers are hard to see you guys um i've tried recording on my other devices but they don't want to stay connected for the volume and lastly one 0.1101 is point one one zero one times twenty five point nine eight two nine and that is also atomic mass unit and i apologize running out of space here um you can go ahead and do this in one fell swoop in your calculator but i do want to pay attention to sig figs so if i take 0.7899 times 23.9850 i get an answer of 18.9458 i have four sig figs here and six here so i need to round this number to the fourth decimal place all right sorry fourth stick thing it's a multiplication so not just one place four sig fig second decimal place i'm gonna mark it right now but i'm not gonna round so i don't want to create round off error the point one zero zero zero times twenty four point nine eight five eight i calculate this out and get two point four nine eight five eight again four sig figs versus six i count to the fourth sixth big and in this case it's actually the third decimal place and point one one zero one times twenty five point nine eight two nine gives me two point eight zero fourth sig fig is the third decimal place now i'm going to add these up when i add these up in my calculator i get zero 24.30501 five seven six but i need to pay attention to my sig figs before i was doing uh multiplication so i was looking at overall number of sig figs now however i need to pay attention to my number of decimal places so i have because i'm doing addition i have two decimal places here three here and three here i need to limit my final answer to the second decimal place so i'm going to round this to 24.31 amu and if you look up magnesium on the periodic table it is a group 2 metal alkaline earth metal i do not remember its atomic number but you will find the weight to be 24.31 now you may have a table that has more precision in your specimen it may be out for decimal places but if you run two decimal places you'll find it to be 24.3 so again we get these atomic masses off the product table some of these numbers may seem a little off to you though but that's um because the entire table again is based on carbon 12. everything has been normalized to a carbon-12 atom there's only one mass you need to memorize the mass of one carbon-12 atom is exactly 12 amu meaning it does not limited sig figs it is exactly 12. 12.000 repeating forever amu and that is because the definition of the attempt of attack the atomic mass unit scale or the amu scale is 1 amu equals 1 12 based on the carbon-12 atom so 1amu equals 112 based on carbon-12 atom therefore one carbon-12 atom equals 12 amu why is that information useful because of examples such as this the atomic weight of carbon is 12.01 that is the atomic weight you would get the average atomic weight off the periodic table i usually write it down as 12.01 such 24 sig figs but you will find it written down as 12.011 given that the naturally occurring carbon isotopes there's only two of them there's carbon 12 and carbon 13. if we're told the mass of carbon 13 is 13.0035 and u what is the percent abundance of each isotope again why is the atomic weight of carbon not exactly 12 well because i have carbon 13 present two making up some of those carbon atoms how do i figure this out though well i know that my atomic weight will be equal to the fractional abundance of carbon-12 times the mass of carbon-12 plus the fractional abundance of carbon 13 times the mass of carbon 13. if i can find the fractional abundance i can then put that into percent form and tell you the percentage of each species okay well the question gave me the mass of carbon 13 i know the mass of carbon 12 it's the only one i have to memorize carbon 12 is equal to 12 amu i know the atomic weight because the question gave me it but i have two unknowns fractional abundance of carbon 12 and fractional abundance of carbon 13. i am going to let the fractional abundance of carbon 12 equal x and the fractional abundance of carbon 13 equal y i know that x plus y needs to equal one why well because i know that these are the two naturally occurring isotopes so i know that their percentages have to add up to 100 percent which in fractional form is one so therefore i can substitute in and say y is equal to x or to 1 minus x and substitute that into my overall equation i've got 12.011 and these are amu i'm not going to put units in here right now because it's just going to make things very messy so the units would cancel out to get to fractional form so i'll deal with that um i'll show you where that occurs at but again i'm not going to write units and write this in this question specifically because it's just going to make it messy to read it on a slide um 12.011 equals x times the mass of a carbon-12 atom which is 12amu plus the fractional abundance of carbon 13 which i have just said where i'm going to represent with y and i'm going to substitute in and say 1 minus x equals y times the mass of carbon 13 which is 13.0035 amu so i'm going to do some algebraic manipulation 12.0 is equal to 12x plus remember your distribution rules you're going to distribute this through 13.0035 minus thirteen point zero zero three five x go ahead and get your variables on one side and your numbers on the other it doesn't matter what side you put them on i personally always put my variables on the left side it's just what i was taught when i was learning algebra and the way i continue to do so again i don't care which side you put them on it doesn't matter i also need you to be very aware remember this is an exact number so it does not limit sig figs i go ahead and combine all my variables on one side of my numbers on the other that would give me minus 12x plus 13.0035 amu ah keep writing amu that should be x thirteen point zero zero three five x again i just like my x's on the left side you can do one either sides doesn't matter and this will equal 13.0035 minus 12.011 simplify this out if i do the math here this gives me 1.0035 x equal to 0.99235 and solve for x so divide both sides by 1.0035 0.99235 divided by let me 1.0035 that in really quick this gives me a calculated answer of 0.98888 repeating and it's going to keep repeating look at your sig figs let me find a color we have not used okay so looking at our sig figs we know that this is an exact number so it doesn't matter this has five sig figs in it here i need to keep this to three decimal places so my decimal place will be here because addition subtraction is number of decimal places this limits me to three decimal places which means that this number has three sig figs in it this has three sig figs so i need to limit my overall answer to three sig figs so i'm going to round this to 0.989 which represents x remember x equals the fractional abundance of carbon 12. 0.989 is the same as 98 so carbon 12 is 98.9 for its fraction or its percent abundance what does that leave for carbon 13 well the rest to get up to 100 1.1 percent of the carbon atoms will be carbon 13.