Lesson video

Hello, my name's Dr.

Warren.

I'm so pleased that you can join me today for this lesson on mass spectrometry.

It's part of the Chemical Analysis Unit.

I'm here to work with you throughout the lesson and support you all the way, especially through the tricky bits.

Our learning outcome for today's lesson is "I can explain how mass spectrometry is used to identify isotopes and molecular structures and calculate relative atopic masses." And here are our key words for today's lesson.

Isotope, an atom of an element with the same number of proton, but different numbers of neutrons.

Relative atomic mass.

An average value that takes account of the abundance of the isotopes of the element.

Molecular mass.

The sum of the atomic masses of all the atoms in a molecule.

Molecular ion.

The ion produced when the molecule is ionised by loss of an electron from the molecule.

Spectrometer.

An apparatus used for recording and measuring spectra, especially as a method of analysis.

In today's lesson, we have two learning cycles.

The first is isotopes and mass spectrometry.

The second is molecules and mass spectrometry.

So let's get started with our first learning cycle on isotopes and mass spectrometry.

Mass spectrometry is an analytical technique that ionises chemical substances and sorts the ions based on their mass to charge ratio or m/z, where z is the charge.

Here we can see an image of a mass spectrum.

We have on the y-axis, the relative abundance that tells us how much of the isotope is present.

And on the x-axis we have our mass to charge ratio.

It's mainly used to identify isotopes and helps chemists identify unknown organic compounds.

So how does a mass spectrometer work? Well, we can divide it up into different sections in different parts of the process.

First of all, the sample is ionised by knocking off electrons to form positive ions.

Next, the ions are accelerated by an electric field, then they are deflected by a magnetic field, and finally the ions are detected and their abundance is recorded and you end up getting a printout of the mass spectrum that we've already seen.

So quick check for understanding.

True or false.

"In mass spectrometry, the sample is ionised by adding electrons to form negative ions." True or false? Well done if you chose false.

Let's have a look at the reason why.

Is it A, the sample can be ionised either by adding electrons forming negative ions, or removing electrons forming positive ions.

Or B, the sample is ionised to form positive ions only.

Well done if you chose B.

During mass spectrometry, only positive ions are formed.

Well done.

So isotopes are detected as peaks in the mass spectrum at different mass to charge values corresponding to their atomic masses.

So in the diagram graph that we can see here, we have a bromine atom, which has a 79 atomic mass.

And we also have another isotope of the bromine atom, which has a 81 atomic mass.

When analysing a mass spectrum, we're focusing on ions with a single positive charge.

Therefore, the mass to charge ratio corresponds directly to the atomic mass.

So that's Br+ has the atomic mass of 79, and the BR+ 81 has the atomic mass of 81.

Another quick check for understanding.

"What does the mass-to-charge ratio in mass spectrometry help to determine?" Is it A, the colour of each ion? B, the relative abundance of isotopes? C, the identity of the element? Or D, the speed at which the ion travels? So well done if you chose B, the relative abundance of isotopes and also the identity of the element.

It's a well done if you chose C as well.

They are the correct answers.

The relative atomic mass of an element is calculated using the relative abundances, and the atomic mass of the isotopes.

So here, we have our mass spectrum, and to work out the relative atomic mass, what we need to do is take the sum of the isotope mass and multiply it by the relative abundance and divide by 100.

So we have to go through each individual isotope mass that we see on the graph multiplied by the relative abundance and divide by 100.

Since the charge is 1+, the m/z value equals the atomic mass so that keeps it relatively simple when we are talking about isotopes.

Let's have a look at an example.

So here we have the mass spectrum for bromine, we have two bromine+ ions showing one at 79 and one at 81.

So let's see if we can work out the relative atomic mass.

The relative atomic mass is going to equal the sum of the isotope mass times relative abundance divided by 100.

And we do that for each isotope mass showing.

So we have 50.

7 x 79 + 49.

3 x 81 divided by 100.

If we do the math, that works out at the relative atomic mass being 79.

986 or 80.

0 to one decimal place.

Okay, see if you can have a go at working out the relative atomic mass for chlorine.

Look at the diagram and work out what isotope masses are available and their relative abundance.

Let's have a look at the answer.

Well, the relative atomic mass is the sum of the isotope mass times the relative abundance divided by 100.

So that's 75.

8 x 35 + 24.

2 x 37 divided by 100.

We do the maths, that works out at 35.

484 or 35.

5 to one decimal place.

So very well done if you've got that correct.

So this brings us to our first task.

"Some students are discussing mass spectrometry.

Identify who is correct, and update any incorrect statements." So let's have a look and see what they say.

So first of all, Andeep, he says, "In mass spectrometry, the isotopes are identified by their colour changes during ionisation." Lucas says, "The relative atomic mass is calculated using the relative abundances and masses of isotopes." Sofia says, "In a mass spectrometer, the sample is ionised by adding electrons to form negative ions." Laura says, "Mass spectrometry sorts ions by their mass-to-charge ratio and for 1+ ions, this ratio corresponds directly to the atomic mass.

Who is correct and update any incorrect statements.

So pause the video and have a go at the question and then when you're ready, we'll look at the answer together.

So let's have a look and see who's correct.

Well, Lucas is correct.

The relative atomic mass is calculated using the relative abundance and masses of isotopes.

So well done if you spotted that Lucas was correct, and also Laura.

Mass spectrometry sorts ions by their mass charge ratio, and for 1+ ions, this ratio corresponds directly to the atomic mass.

So very well done if you spotted that both Lucas and Laura were correct.

Now what about Andeep? Well, he said, "In mass spectrometry, isotopes are identified by their colour changes during ionisation," but I'm sorry, that is something that is incorrect.

What Andeep should have said was in mass spectrometry, isotopes are identified by their peaks at different mass to charge values or m/z values corresponding to the atomic mass.

So well done if you spotted he was incorrect, and got that statement updated.

And finally, what about Sofia? Well, she said, "In a mass spectrometer, the sample is ionised by adding electrons to form negative ions." Well, that doesn't happen.

We don't have the negative ions in the mass spectrometer.

What does happen is the sample is ionised by knocking off electrons to form positive ions, typically 1+ ions.

So very well done if you spotted that Sofia was incorrect and you managed to correct the statement.

You're doing really well.

Okay, let's move on to our second question.

"The following samples were analysed using a mass spectrometer.

Calculate the relative of atomic mass for each element based on the given data." And we've got three different samples here.

First of all, copper, where the relative abundances are 69.

17% for Cu+ 63 and 30.

83% for Cu+ at 65.

Part B is magnesium.

And you've got given the relative abundances there for all the Mg+ ions, and then silicon.

Again, you've been given the relative abundances for each of the three Si+ ions present.

So pause the video, and have a go at this question, and then when you're ready, we'll look at the answers together.

All right, so for our first calculation using copper, so what we need to do is work out the sum of the relative abundances times the isotope mass for each of those copper ions and then divide by 100.

So that's 69.

17 x 63 + 30.

83 x 65 divided by 100.

So just work through the different lines of the maths there, and that gives us a final answer of 63.

616, and we round that up to one decimal place.

So the relative atomic mass of copper is 63.

6 to one decimal place.

So very well done if you've got that correct.

B, for magnesium, just do the same thing.

So our relative abundance is 78.

99 x 24 for our first isotope mass plus 10.

00 x 25 for our second isotope plus mass, and then plus 11.

01 x 26 for our third isotope mass, and we divide all that by 100.

So just work through the figures and we end up with a relative atomic mass of 24.

3202.

And again, we are going to round that up to one decimal place, so it is gonna be 24.

3 to one decimal place.

So well done if you've got that correct.

2c, silicon.

Just going to go through the same calculation again.

So to work out the relative atomic mass of silicon, well we have again three isotopes present, first one 92.

23%, and that's gonna be x 28 + 4.

67 x 29 + 3.

10 x 30, and all of that will be divided by 100.

Again, work through the maths, and you end up with a relative atomic mass of 28.

1087.

And so to one decimal place, the relative atomic mass of silicon is 28.

1.

So really well done if you've got all of those calculations correct.

And it's really important to show the work as you go through because if you make a mistake, you'll still get credit.

If you don't show you working, then you won't get any credit at all.

So that brings us to the end of our first learning cycle and we move on to our second learning cycle where we are going to look at molecules and mass spectrometry.

In organic chemistry, mass spectrometry is crucial for determining molecular masses, understanding molecular structures, and identifying unknown compounds.

So we've got an example here of propanone and you have got the a molecular model of propanone which shows three carbon atoms, one oxygen atom, and three hydrogen atoms. Now mass spectrometry comes into its own where we can use it to understand the molecular structure.

We can use it to work out how all of those atoms, the three carbons, the six hydrogens, and the one oxygen is put together by looking at the molecular ions fragments and the mass-to-charge ratio.

We have a suite of tools including infrared spectrometry and NMR, which is Nuclear Magnetic Resonance, something you'll learn about if you carry on Chemistry post-16.

They are instrumental techniques used in organic chemistry and all of them have a slightly different role to play.

So let's have a little look at this mass spectrum a bit closer.

So mass spectrometry of molecules generates a mass spectrum with multiple peaks.

As you can see here, we've got a peak at 15, one at 43, and one at 58.

And each peak refers to a slightly different part of the molecule.

So for example, we have the CH3, which we have done in the green circle.

That refers to the CH3 part or the methyl group of the molecule, and it gives a peak at 15 because carbon is 12, and three hydrogens are one.

So these peak correspond to different fragments of the molecule, and you can show the bits that have been ionised.

So this purple peak at 43, well, we've got there an ion with the methyl group and the carbon and the oxygen.

And if you work out that, its mass-to-charge ratio is 43.

And then finally, we find that the most significant peak is often the one that represents the molecular ion, and in this case it is at 58.

This is the ion formed when the entire molecule just loses one electron resulting in a positively charged ion.

And the m/z ratio value of this peak indicates the molecular mass of the compound.

So it's often the highest peak that you see on the graph and it's an important one because once you have the molecular compound, you can start to work out what the structure actually is.

So let's have a quick check for understanding.

"What do the peaks in mass spectrum typically represent?" Is it, A, different fragments of the ionise molecule? B, negatively charged ions? C, ions with different mass-to-charge ratios? Or D, the absolute number of neutrons in the molecule? So well done if you chose A and C.

The peaks in the mass spectrum show the different fragments of the ionised molecule, and the ions with different mass-to-charge ratios.

So very, very well done if you've got both of those answers correct.

"What does the M+ peak in a mass spectrum signify?" Is it A, the molecular ion of the compound? B, the neutral molecule? C, the gain of a proton in the molecule? And D, the molecular mass of the molecule? Well done if you chose both A and D.

It is the molecular ion of a compound.

So it's that compound that's just lost or had knocked out one electron and that represents the molecular mass of the molecule.

So very well done if you got that right as well.

So often, when you look at a mass spectrum, you'll see that there are some smaller peaks as well as these major peaks.

And this is shown clearly in the mass spectrum of pentane C5 H12, which is now shown.

And these smaller peaks result from fragmentation of the molecular ion.

So that means a molecular ion just loses little parts of it here and there because chemical bonds within the molecule break producing smaller ionised fragments.

So this is quite a normal thing that happens during mass spectrometry.

The fragmentation patterns are unique to each molecule and it that can be really useful because it can provide detailed structural information that helps the chemists actually work out what the structure is.

So by analysing the m/z values of the fragments, chemists can reduce the arrangement of atoms within the molecule and identify the compound.

So it's a bit of detective work really putting together all these different parts of the jigsaw until we come up with that actual structural arrangement of the atoms. So we have got a mass spectrum here for you to have a look at.

Four peaks have been highlighted on the mass spectrum of pentane.

So what we want you to do is match each peak to the responsible ion.

So we've got four responsible ions, and we've got four peaks that have been shown, and we've also given you the displayed formula of pentane to help.

So our possible ions are C4H9+, C2H5+, C3H12+, and C3H7+.

So can we match those to the peaks? Okay, so our first peak at 29, so the mass-to-charge ratio at 29 comes from C2H5 because 12 is carbon and hydrogen is a value of one.

So that adds up to 24 + 5 is 29.

Our next one at 43, that's gonna be our next biggest fragment, C3H7.

Then at 57, we've got C4H9, and finally at 72, we've got C5H12.

So very well done if you've got those right.

Excellent work.

All right.

So that brings us to our second task.

"A sample of ethanol C2H5OH is analysed using a mass spectrometer and five peaks have been highlighted, and you can see those peaks on the mass spectrum.

They're the colours in green.

So first of all, what we'd like you to do is to identify the molecular ion, the M+ peak, and then explain using all five highlighted peaks, how this mass spectrum shows the sample contains ethanol.

And again, to make it a little bit easier, we've actually given you the displayed formula of ethanol.

So pause the video and have a go at this question and then when you're ready, we'll look at the answer together.

Okay, so part one, identify the molecular iron M+ peak for this corresponds to the molecular mass of ethanol.

The peak is at m/z 46, and this is the molecular ion peak as the molecular formula of ethanol is C2H5OH, and it's relative molecular mass is 46.

So carbon is 12 x 2 is 24, hydrogen, 1 x 6 is 6, and oxygen 16.

If you add those up together, you get 46.

So remember when you're working out that molecular ion M+ peak, it's usually the highest peak that you are going to see on the mass spectrum.

And then if you go and work out its relative molecular mass, you can check that through.

So very well done if you've got that answer correct.

So question two, explain using all five highlighted peaks how this mass spectrum shows the sample contains ethanol.

So again, the best way to to do this is to look at your mass spectrum and look at the displayed formula, and see if we can match up the different fragmentations.

So at 15, we have a CH3+ fragment, and if we look at our diagram, that looks pretty good because we can just have that C to C bond breaking.

So carbon is 12 + 3 H's gives 15, so that's CH3+ fragment.

At 29, we have the fragment CH3CH2+ and again, we can see from our displayed formula that that CO bond could just break easily to form that fragment.

At 31, we have CH2OH fragments.

And again, you can see from the diagram that that CC bond will break easily to form that fragment.

At 45, we have CH3CH2O+ fragment, so that hydrogen OH bond has broken there.

And then we have our molecular ion of ethanol at m/z 46.

So very well done If you've got all of those right.

Remember, working out the relative molecular formulas for each bit is the way to make sure and double check that it all works well.

So excellent work if you've got that done.

We now come to our key learning points for today's lesson.

Isotopes are atoms of the same element with different numbers of neutrons.

The relative abundance of isotopes can be used to calculate the relative atomic mass or the RAM of an element.

Mass spectrometry ionises samples to create positive ions for analysis.

The M+ peak represents the molecular ion, indicating the molecular mass.

Fragmentation in mass spectrometry helps identify molecular structure by breaking molecules into smaller ions.

I hope that you have enjoyed today's lesson and look forward to learning with you again very soon.