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AP®︎/College Chemistry
Course: AP®︎/College Chemistry > Unit 1
Lesson 2: Mass spectrometry of elementsWorked example: Identifying an element from its mass spectrum
Using mass spectrometry data, we can estimate an element's average atomic mass and determine its identity. Created by Sal Khan.
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1:57When looking for the mass spectrum of the isotopes of a pure element, why is the "weighted average" calculated like this instead of averaged out like a mean in statistics?
For example, if I had the numbers 5, 3, and 6, the way to find their average would be (5 + 3 + 6) / 3.(5 votes)
The kind of mean you're describing is called an arithmetic mean. The issue with using an arithmetic mean for calculating something like average atomic mass is that it assumes that all the masses of the isotopes are present in nature in equally abundant amounts. And of course they are not, some isotopes are more stable than others leading to wildly different abundances often.
And the whole point of an average atomic mass to determine how much mass there would be in a pure sample of a single element. If I use your example and assign random abundances to the numbers, say 90% for 6, 9% for 5, and 1% for 3; we can see the differences in using arithmetic compared to a weighted average. An athematic average would be 4.67, but a weighted average would be 5.88. The weighted average is much closer to 6 because there so much of 6 compared to the other numbers. So if I had an element with isotopes of those masses and abundances, in real life we'll see the 5.88 number since most of the mass of the sample is coming from the 6 and 5 isotopes and relatively little from the 3 isotope.
So arithmetic averages are good for finding averages of things which are equally possible like dice rolls or coin tosses. But weighted averages are necessary for things which do have unequal representations like atomic mass or GPA.
Hope that helps.(5 votes)
Does not knowing the precise percentage of the isotopes affect the calculation of average atomic mass drastically? This is if one is using a graph, like in the video.(1 vote)- Drastically? Not really. With Sal's example we were only trying to estimate the atomic mass to determine what element we had. And using the numbers that he did with numbers with a low amount of sig figs accomplished that.
But just in case we cared what the numbers were, using Sal's numbers gives you an average atomic mass of 87.69 u for strontium, while the official atomic mass of strontium using more precise numbers with more sig figs is listed as 87.62 u. So there is a difference, but not a very big one.
Hope that helps.(4 votes)
so this works with a pure element, such as in this example. but say your sample was not pure - would you still be able to infer with the mass spectrometer what (multiple) elements comprised the sample?(2 votes)- If the two elements has similar isotope masses, then no. We wouldn’t be able to distinguish the signals between the two elements. Essentially they would overlap with each other. Ideally we do some separation process on samples to separate the components of the mixture before they enter the MS so that we only analyze one component at a time.
Hope that helps.(2 votes)
why did you put (.) in every number like 82 to 0.82?(1 vote)- The numbers on the y-axis are percentages, so to use them in a calculation you have to convert them to decimals first. This is because 82 ≠ 82%. 82% is actually 82/100, which as a decimal is 0.82 and this explains the decimalized percentages in the calculations.
Hope that helps.(4 votes)
- @22:2 I am a little confused, in the notes before this it was stated that isotopes that were lighter, were deflected more therefore making them the most abundant; but in this video the heaviest isotope seems to be more abundant(1 vote)
- An isotope’s mass determines how much it will be deflected in the magnetic field of a mass spectrometer. So less massive, or lighter, isotopes will be deflected more than more massive isotopes.
How the mass spectrometer determines abundance though is not by the trajectory path of the isotopes, but rather how many isotopes of a certain mass make contact with a certain spot on the detector in the instrument.
In this video, the most massive isotope of the element will be deflected the least in the mass spectrometer, but collide most with the detector in a certain spot to show that it is the most abundant.
Hope that helps.(3 votes)
- I got it.
Here my Q. How I can figure out how many protons, neutrons & electron in the nucleus.
Thnks.(1 vote)
If you have a periodic table available, the number of protons is equal to the atomic number (the little number usually on the top left of the element symbol. The number of neutrons is equal to Mass of your isotope minus atomic number. Remember that there are different isotopes of the element, so there may be many different values. The mass on the periodic table is the average of all of the different isotopes. For example, let's say you have a carbon-13. Carbon, by definition, has 6 protons, so 13-6, you get 7 neutrons. On the other hand, you might have a carbon-12, the more common isotope of Carbon, 12-6, 6 neutrons. The number of electrons (which are not in the nucleus) is equal to the number of protons in an atom if the atom has a neutral charge. if the atom has a +x charge, take away x number of electrons. if the atom has a -x charge add x number of electrons. Thanks to richard for the Suggestions.
Hope this helps(2 votes)
Since the real relative abundance of each isotope on earth is obtained via carrying a number samples around the world and analyzing their mass spectrum.
Does this mean that the average atomic mass can be calculated and the element known after the examination of all those samples?(1 vote)- If you are able purify your samples so that only a single element is being analyzed in the mass spectrometer, then yes you would probably be able to determine what element it is using Sal method. It's important for the sample to be as pure as possible and free of contaminants because you can imagine if your sample of scandium had significant amounts of rubidium or yttrium, if would make your analysis problematic by distorting the calculations.
Hope that helps.(1 vote)
How did scientists determine the actual values of the periodic table to begin with? That is, how did they know they had strontium isotopes to measure its average atomic mass?(1 vote)
they discovered the different atoms, saw that there were atoms with different proton amounts, and decided to make a chart.(1 vote)
- im self-studying for ap chemistry like this: 1.watch khan academy videos. 2.make study sheets abt the topic from diff sources such as khan academy articles, byjus,and libretexts.
3.practicing on solving problems found in khan academy and libretexts.
is that enough?(1 vote) - 1:57. What are you adding/multiplying?(1 vote)
- Sal is calculating the average atomic mass of the element. We use a type of average called a weighted average which takes into account the abundances of the different isotopes. It involves multiplying the mass of each individual isotope by its abundance (as a decimal) and adding these terms for all the isotopes.
Hope that helps.(1 vote)
1:57Video transcript
- [Instructor] So let's say that we have some mystery substance here, and we know that it's a pure element, and we need to figure out what it is. Well, scientists have a method, and we go into the details, or more details, in other videos, called mass, sometimes it's
known as mass spectrometry or mass spectroscopy. It's a technique where you can
take a sample of a substance and think about the various atomic masses of the different isotopes
in that substance. And that's what we have right over here. They tell us the mass
spectrum for an average sample of a pure element is shown below. So let's say it's this pure element. So what this is telling us is, this looks like maybe, I don't know, let's call this 82% of our sample has an atomic mass of 88
universal atomic mass units. About, this looks like
about 7% of our sample has an atomic mass of 87
universal atomic mass units. It looks like 10% has an atomic mass of 86 universal atomic mass units, and it looks like about 1% of our sample has an atomic mass of 84
universal atomic mass units. And so from this information, we can try to estimate what
the average atomic mass of this mystery element is. We could calculate it as 0.82 times 88, plus, let's call this 7%, so 0.07 times 87, plus 10%, 0.1, times 86, plus, let's see, it should add up to 100%. This is 89, and then this gets us to 99, so then another 1%, 0.01 times 84. And so if we were to do this calculation, this is our estimate of the average atomic
mass of this element. We could type this into a calculator and get some number and then look that up on a
periodic table of elements, or we could just try to estimate it. We can see that it's
going to be close to 88 because that's where the
highest percentage is. When we're taking the weighted average, we have the highest
weight right over there. But these other isotopes, these other versions of the element that have a different number of neutrons, which changes its atomic mass, they're going to bring the average down. So our average atomic mass is going to be a little bit less than 88. So let's look up a
periodic table of elements. What element here has an atomic mass a little bit less than 88? Well, yttrium is 88.91, but we know it can't be that
because none of the isotopes have an atomic mass above 88. So we can rule out yttrium. Strontium is looking pretty good. It's exactly what we predicted, a little bit less than 88, and rubidium is a lot less than 88. So our, even if we were
to do the calculation, we could feel confident we're not going to be as low as rubidium. So I'm feeling very
confident just eyeballing it, just estimating, this is going to be a little bit, have an average atomic mass
a little bit less than 88, which tells me that this is strontium.