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PAPER CHROMATOGRAPHY This page is an introduction to paper chromatography - including two way chromatography. Carrying out paper chromatography Background Chromatography is used to separate mixtures of substances into their components. All forms of chromatography work on the same principle. They all have a stationary phase (a solid, or a liquid supported on a solid) and a mobile phase (a liquid or a gas). The mobile phase flows through the stationary phase and carries the components of the mixture with it. Different components travel at different rates. We'll look at the reasons for this further down the page. In paper chromatography, the stationary phase is a very uniform absorbent paper. The mobile phase is a suitable liquid solvent or mixture of solvents. Producing a paper chromatogram You probably used paper chromatography as one of the first things you ever did in chemistry to separate out mixtures of coloured dyes - for example, the dyes which make up a particular ink. That's an easy example to take, so let's start from there. Suppose you have three blue pens and you want to find out which one was used to write a message. Samples of each ink are spotted on to a pencil line drawn on a sheet of chromatography paper. Some of the ink from the message is dissolved in the minimum possible amount of a suitable solvent, and that is also spotted onto the same line. In the diagram, the pens are labelled 1, 2 and 3, and the message ink as M. paper chromatography http://www.chemguide.co.uk/analysis/chromatography/paper... 1 of 10 9/16/17, 3:49 PM

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PAPER CHROMATOGRAPHY

This page is an introduction to paper chromatography -including two way chromatography.

Carrying out paper chromatography

Background

Chromatography is used to separate mixtures ofsubstances into their components. All forms ofchromatography work on the same principle.

They all have a stationary phase (a solid, or a liquidsupported on a solid) and a mobile phase (a liquid or agas). The mobile phase flows through the stationaryphase and carries the components of the mixture with it.Different components travel at different rates. We'll lookat the reasons for this further down the page.

In paper chromatography, the stationary phase is a veryuniform absorbent paper. The mobile phase is a suitableliquid solvent or mixture of solvents.

Producing a paper chromatogram

You probably used paper chromatography as one of thefirst things you ever did in chemistry to separate outmixtures of coloured dyes - for example, the dyes whichmake up a particular ink. That's an easy example to take,so let's start from there.

Suppose you have three blue pens and you want to findout which one was used to write a message. Samples ofeach ink are spotted on to a pencil line drawn on a sheetof chromatography paper. Some of the ink from themessage is dissolved in the minimum possible amount ofa suitable solvent, and that is also spotted onto the sameline. In the diagram, the pens are labelled 1, 2 and 3, andthe message ink as M.

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Note:  The chromatography paper will in fact bepure white - not pale grey. I'm forced to show it asoff-white because of the way I construct thediagrams. Anything I draw as pure white allowsthe background colour of the page to showthrough.

The paper is suspended in a container with a shallowlayer of a suitable solvent or mixture of solvents in it. It isimportant that the solvent level is below the line with thespots on it. The next diagram doesn't show details of howthe paper is suspended because there are too manypossible ways of doing it and it clutters the diagram.Sometimes the paper is just coiled into a loose cylinderand fastened with paper clips top and bottom. Thecylinder then just stands in the bottom of the container.

The reason for covering the container is to make surethat the atmosphere in the beaker is saturated withsolvent vapour. Saturating the atmosphere in the beakerwith vapour stops the solvent from evaporating as it risesup the paper.

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As the solvent slowly travels up the paper, the differentcomponents of the ink mixtures travel at different ratesand the mixtures are separated into different colouredspots.

The diagram shows what the plate might look like afterthe solvent has moved almost to the top.

It is fairly easy to see from the final chromatogram thatthe pen that wrote the message contained the same dyesas pen 2. You can also see that pen 1 contains a mixtureof two different blue dyes - one of which might be thesame as the single dye in pen 3.

Rf values

Some compounds in a mixture travel almost as far as thesolvent does; some stay much closer to the base line.The distance travelled relative to the solvent is a constantfor a particular compound as long as you keep everythingelse constant - the type of paper and the exactcomposition of the solvent, for example.

The distance travelled relative to the solvent is called theRf value. For each compound it can be worked out usingthe formula:

For example, if one component of a mixture travelled 9.6

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cm from the base line while the solvent had travelled 12.0cm, then the Rf value for that component is:

In the example we looked at with the various pens, itwasn't necessary to measure Rf values because you aremaking a direct comparison just by looking at thechromatogram.

You are making the assumption that if you have twospots in the final chromatogram which are the samecolour and have travelled the same distance up thepaper, they are most likely the same compound. It isn'tnecessarily true of course - you could have two similarlycoloured compounds with very similar Rf values. We'lllook at how you can get around that problem furtherdown the page.

What if the substances you are interested in arecolourless?

In some cases, it may be possible to make the spotsvisible by reacting them with something which produces acoloured product. A good example of this is inchromatograms produced from amino acid mixtures.

Suppose you had a mixture of amino acids and wanted tofind out which particular amino acids the mixturecontained. For simplicity we'll assume that you know themixture can only possibly contain five of the commonamino acids.

A small drop of a solution of the mixture is placed on thebase line of the paper, and similar small spots of theknown amino acids are placed alongside it. The paper isthen stood in a suitable solvent and left to develop asbefore. In the diagram, the mixture is M, and the knownamino acids are labelled 1 to 5.

The position of the solvent front is marked in pencil andthe chromatogram is allowed to dry and is then sprayedwith a solution of ninhydrin. Ninhydrin reacts with aminoacids to give coloured compounds, mainly brown or

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purple.

The left-hand diagram shows the paper after the solventfront has almost reached the top. The spots are stillinvisible. The second diagram shows what it might looklike after spraying with ninhydrin.

There is no need to measure the Rf values because youcan easily compare the spots in the mixture with those ofthe known amino acids - both from their positions andtheir colours.

In this example, the mixture contains the amino acidslabelled as 1, 4 and 5.

And what if the mixture contained amino acids other thanthe ones we have used for comparison? There would bespots in the mixture which didn't match those from theknown amino acids. You would have to re-run theexperiment using other amino acids for comparison.

Two way paper chromatography

Two way paper chromatography gets around the problemof separating out substances which have very similar Rfvalues.

I'm going to go back to talking about colouredcompounds because it is much easier to see what ishappening. You can perfectly well do this with colourlesscompounds - but you have to use quite a lot ofimagination in the explanation of what is going on!

This time a chromatogram is made starting from a single

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spot of mixture placed towards one end of the base line.It is stood in a solvent as before and left until the solventfront gets close to the top of the paper.

In the diagram, the position of the solvent front is markedin pencil before the paper dries out. This is labelled asSF1 - the solvent front for the first solvent. We shall beusing two different solvents.

If you look closely, you may be able to see that the largecentral spot in the chromatogram is partly blue and partlygreen. Two dyes in the mixture have almost the same Rfvalues. They could equally well, of course, both havebeen the same colour - in which case you couldn't tellwhether there was one or more dye present in that spot.

What you do now is to wait for the paper to dry outcompletely, and then rotate it through 90°, and developthe chromatogram again in a different solvent.

It is very unlikely that the two confusing spots will havethe same Rf values in the second solvent as well as thefirst, and so the spots will move by a different amount.

The next diagram shows what might happen to thevarious spots on the original chromatogram. The positionof the second solvent front is also marked.

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You wouldn't, of course, see these spots in both theiroriginal and final positions - they have moved! The finalchromatogram would look like this:

Two way chromatography has completely separated outthe mixture into four distinct spots.

If you want to identify the spots in the mixture, youobviously can't do it with comparison substances on thesame chromatogram as we looked at earlier with thepens or amino acids examples. You would end up with ameaningless mess of spots.

You can, though, work out the Rf values for each of thespots in both solvents, and then compare these withvalues that you have measured for known compoundsunder exactly the same conditions.

How does paper chromatography work?

Although paper chromatography is simple to do, it is quitedifficult to explain compared with thin layerchromatography. The explanation depends to someextent on what sort of solvent you are using, and manysources gloss over the problem completely. If you haven'talready done so, it would be helpful if you could read theexplanation for how thin layer chromatography works(link below). That will save me a lot of repetition, and Ican concentrate on the problems.

Note:  You will find the explanation for how thin

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layer chromatography works by following this link.

Use the BACK button on your browser to returnquickly to this page when yhou have read it.

The essential structure of paper

Paper is made of cellulose fibres, and cellulose is apolymer of the simple sugar, glucose.

The key point about cellulose is that the polymer chainshave -OH groups sticking out all around them. To thatextent, it presents the same sort of surface as silica gelor alumina in thin layer chromatography.

It would be tempting to try to explain paperchromatography in terms of the way that differentcompounds are adsorbed to different extents on to thepaper surface. In other words, it would be nice to be ableto use the same explanation for both thin layer and paperchromatography. Unfortunately, it is more complicatedthan that!

The complication arises because the cellulose fibresattract water vapour from the atmosphere as well as anywater that was present when the paper was made. Youcan therefore think of paper as being cellulose fibres witha very thin layer of water molecules bound to the surface.

It is the interaction with this water which is the mostimportant effect during paper chromatography.

Paper chromatography using a non-polar solvent

Suppose you use a non-polar solvent such as hexane todevelop your chromatogram.

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Non-polar molecules in the mixture that you are trying toseparate will have little attraction for the water moleculesattached to the cellulose, and so will spend most of theirtime dissolved in the moving solvent. Molecules like thiswill therefore travel a long way up the paper carried bythe solvent. They will have relatively high Rf values.

On the other hand, polar molecules will have a highattraction for the water molecules and much less for thenon-polar solvent. They will therefore tend to dissolve inthe thin layer of water around the cellulose fibres muchmore than in the moving solvent.

Because they spend more time dissolved in thestationary phase and less time in the mobile phase, theyaren't going to travel very fast up the paper.

The tendency for a compound to divide its time betweentwo immiscible solvents (solvents such as hexane andwater which won't mix) is known as partition. Paperchromatography using a non-polar solvent is therefore atype of partition chromatography.

Paper chromatography using a water and other polarsolvents

A moment's thought will tell you that partition can't be theexplanation if you are using water as the solvent for yourmixture. If you have water as the mobile phase and thewater bound on to the cellulose as the stationary phase,there can't be any meaningful difference between theamount of time a substance spends in solution in eitherof them. All substances should be equally soluble (orequally insoluble) in both.

And yet the first chromatograms that you made wereprobably of inks using water as your solvent.

If water works as the mobile phase as well being thestationary phase, there has to be some quite differentmechanism at work - and that must be equally true forother polar solvents like the alcohols, for example.Partition only happens between solvents which don't mixwith each other. Polar solvents like the small alcohols domix with water.

In researching this topic, I haven't found any easyexplanation for what happens in these cases. Most

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sources ignore the problem altogether and just quote thepartition explanation without making any allowance forthe type of solvent you are using. Other sources quotemechanisms which have so many strands to them thatthey are far too complicated for this introductory level. I'mtherefore not taking this any further - you shouldn't needto worry about this at UK A level, or its variousequivalents.

Note:  If I have missed something obvious in myresearch and you know of a straightforwardexplanation (worth about 1 or 2 marks in anexam) for what happens with water and otherpolar solvents, could you contact me via theaddress on the about this site page.

Questions to test your understanding

If this is the first set of questions you have done, please readthe introductory page before you start. You will need to usethe BACK BUTTON on your browser to come back hereafterwards.

questions on paper chromatography

answers

Where would you like to go now?

To the chromatography menu . . .

To the analysis menu . . .

To Main Menu . . .

© Jim Clark 2007 (modified July 2016)

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THE MASS SPECTROMETER

This page describes how a mass spectrum is producedusing a mass spectrometer.

How a mass spectrometer works

The basic principle

If something is moving and you subject it to a sidewaysforce, instead of moving in a straight line, it will move in acurve - deflected out of its original path by the sidewaysforce.

Suppose you had a cannonball travelling past you andyou wanted to deflect it as it went by you. All you've got isa jet of water from a hose-pipe that you can squirt at it.Frankly, its not going to make a lot of difference! Becausethe cannonball is so heavy, it will hardly be deflected atall from its original course.

But suppose instead, you tried to deflect a table tennisball travelling at the same speed as the cannonball usingthe same jet of water. Because this ball is so light, youwill get a huge deflection.

The amount of deflection you will get for a givensideways force depends on the mass of the ball. If youknew the speed of the ball and the size of the force, youcould calculate the mass of the ball if you knew what sortof curved path it was deflected through. The less thedeflection, the heavier the ball.

Note:  I'm not suggesting that you personallywould have to do the calculation, although themaths isn't actually very difficult - certainly nomore than A'level standard!

You can apply exactly the same principle to atomic sized

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particles.

An outline of what happens in a mass spectrometer

Atoms and molecules can be deflected by magnetic fields- provided the atom or molecule is first turned into an ion.Electrically charged particles are affected by a magneticfield although electrically neutral ones aren't.

The sequence is :

Stage 1: Ionisation

The atom or molecule is ionised by knocking one or moreelectrons off to give a positive ion. This is true even forthings which you would normally expect to form negativeions (chlorine, for example) or never form ions at all(argon, for example). Most mass spectrometers work withpositive ions.

Note:  All mass spectrometers that you will comeacross if you are doing a course for 16 - 18 yearolds work with positive ions. Even if a few atomsin a sample of chlorine, for example, captured anelectron instead of losing one, the negative ionsformed wouldn't get all the way through theordinary mass spectrometer. But it has beenpointed out to me that there is work being done onnegative ion mass spectrometers, although theyuse a different ionisation technique.

My thanks to Professor John Todd of theUniversity of Kent for drawing this to my attention.

Stage 2: Acceleration

The ions are accelerated so that they all have the samekinetic energy.

Stage 3: Deflection

The ions are then deflected by a magnetic field accordingto their masses. The lighter they are, the more they aredeflected.

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The amount of deflection also depends on the number ofpositive charges on the ion - in other words, on howmany electrons were knocked off in the first stage. Themore the ion is charged, the more it gets deflected.

Stage 4: Detection

The beam of ions passing through the machine isdetected electrically.

A full diagram of a mass spectrometer

Understanding what's going on

The need for a vacuum

It's important that the ions produced in the ionisationchamber have a free run through the machine withouthitting air molecules.

Ionisation

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The vaporised sample passes into the ionisationchamber. The electrically heated metal coil gives offelectrons which are attracted to the electron trap which isa positively charged plate.

The particles in the sample (atoms or molecules) aretherefore bombarded with a stream of electrons, andsome of the collisions are energetic enough to knock oneor more electrons out of the sample particles to makepositive ions.

Most of the positive ions formed will carry a charge of +1because it is much more difficult to remove furtherelectrons from an already positive ion.

These positive ions are persuaded out into the rest of themachine by the ion repeller which is another metal platecarrying a slight positive charge.

Note:  As you will see in a moment, the wholeionisation chamber is held at a positive voltage ofabout 10,000 volts. Where we are talking aboutthe two plates having positive charges, thesecharges are in addition to that 10,000 volts.

Acceleration

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The positive ions are repelled away from the very positiveionisation chamber and pass through three slits, the finalone of which is at 0 volts. The middle slit carries someintermediate voltage. All the ions are accelerated into afinely focused beam.

Deflection

Different ions are deflected by the magnetic field bydifferent amounts. The amount of deflection depends on:

the mass of the ion. Lighter ions are deflected morethan heavier ones.

the charge on the ion. Ions with 2 (or more) positivecharges are deflected more than ones with only 1positive charge.

These two factors are combined into the mass/chargeratio. Mass/charge ratio is given the symbol m/z (orsometimes m/e).

For example, if an ion had a mass of 28 and a charge of1+, its mass/charge ratio would be 28. An ion with amass of 56 and a charge of 2+ would also have amass/charge ratio of 28.

In the last diagram, ion stream A is most deflected - it willcontain ions with the smallest mass/charge ratio. Ion

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stream C is the least deflected - it contains ions with thegreatest mass/charge ratio.

It makes it simpler to talk about this if we assume that thecharge on all the ions is 1+. Most of the ions passingthrough the mass spectrometer will have a charge of 1+,so that the mass/charge ratio will be the same as themass of the ion.

Note:  You must be aware of the possibility of 2+(etc) ions, but the vast majority of A'levelquestions will give you mass spectra which onlyinvolve 1+ ions. Unless there is some hint in thequestion, you can reasonably assume that theions you are talking about will have a charge of1+.

Assuming 1+ ions, stream A has the lightest ions, streamB the next lightest and stream C the heaviest. Lighterions are going to be more deflected than heavy ones.

Detection

Only ion stream B makes it right through the machine tothe ion detector. The other ions collide with the wallswhere they will pick up electrons and be neutralised.Eventually, they get removed from the massspectrometer by the vacuum pump.

When an ion hits the metal box, its charge is neutralisedby an electron jumping from the metal on to the ion (righthand diagram). That leaves a space amongst theelectrons in the metal, and the electrons in the wireshuffle along to fill it.

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A flow of electrons in the wire is detected as an electriccurrent which can be amplified and recorded. The moreions arriving, the greater the current.

Detecting the other ions

How might the other ions be detected - those in streamsA and C which have been lost in the machine?

Remember that stream A was most deflected - it has thesmallest value of m/z (the lightest ions if the charge is1+). To bring them on to the detector, you would need todeflect them less - by using a smaller magnetic field (asmaller sideways force).

To bring those with a larger m/z value (the heavier ions ifthe charge is +1) on to the detector you would have todeflect them more by using a larger magnetic field.

If you vary the magnetic field, you can bring each ionstream in turn on to the detector to produce a currentwhich is proportional to the number of ions arriving. Themass of each ion being detected is related to the size ofthe magnetic field used to bring it on to the detector. Themachine can be calibrated to record current (which is ameasure of the number of ions) against m/z directly. Themass is measured on the 12C scale.

Note:  The 12C scale is a scale on which the 12Cisotope weighs exactly 12 units.

What the mass spectrometer output looks like

The output from the chart recorder is usually simplifiedinto a "stick diagram". This shows the relative currentproduced by ions of varying mass/charge ratio.

The stick diagram for molybdenum looks lilke this:

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You may find diagrams in which the vertical axis islabelled as either "relative abundance" or "relativeintensity". Whichever is used, it means the same thing.The vertical scale is related to the current received by thechart recorder - and so to the number of ions arriving atthe detector: the greater the current, the more abundantthe ion.

As you will see from the diagram, the commonest ion hasa mass/charge ratio of 98. Other ions have mass/chargeratios of 92, 94, 95, 96, 97 and 100.

That means that molybdenum consists of 7 differentisotopes. Assuming that the ions all have a charge of 1+,that means that the masses of the 7 isotopes on thecarbon-12 scale are 92, 94, 95, 96, 97, 98 and 100.

Note:  If there were also 2+ ions present, youwould know because every one of the lines in thestick diagram would have another line at exactlyhalf its m/z value (because, for example, 98/2 =49). Those lines would be much less tall than the1+ ion lines because the chances of forming 2+ions are much less than forming 1+ ions.

If you want to go straight on to how you use thesemass spectra to calculate relative atomic massesyou can jump straight to that page by followingthis link rather than going via the menus below.

Questions to test your understanding

If this is the first set of questions you have done, please readthe introductory page before you start. You will need to usethe BACK BUTTON on your browser to come back here

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afterwards.

questions on how a mass spectrometer works

answers

Where would you like to go now?

To the mass spectrometry menu . . .

To the instrumental analysis menu . . .

To Main Menu . . .

© Jim Clark 2000 (last modified February 2015)

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THE MASS SPECTRA OFELEMENTS

This page looks at the information you can get from themass spectrum of an element. It shows how you can findout the masses and relative abundances of the variousisotopes of the element and use that information tocalculate the relative atomic mass of the element.

It also looks at the problems thrown up by elements withdiatomic molecules - like chlorine, Cl2.

The mass spectrum of monatomic elements

Monatomic elements include all those except for thingslike chlorine, Cl2, with molecules containing more thanone atom.

The mass spectrum for boron

Note:  If you need to know how this diagram isobtained, you should read the page describinghow a mass spectrometer works.

The number of isotopes

The two peaks in the mass spectrum shows that thereare 2 isotopes of boron - with relative isotopic masses of

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10 and 11 on the 12C scale.

Notes:  Isotopes are atoms of the same element(and so with the same number of protons), butwith different masses due to having differentnumbers of neutrons.

We are assuming (and shall do all through thispage) that all the ions recorded have a charge of1+. That means that the mass/charge ratio (m/z)gives you the mass of the isotope directly.

The carbon-12 scale is a scale on which themass of the 12C isotope weighs exactly 12 units.

The abundance of the isotopes

The relative sizes of the peaks gives you a directmeasure of the relative abundances of the isotopes. Thetallest peak is often given an arbitrary height of 100 - butyou may find all sorts of other scales used. It doesn'tmatter in the least.

You can find the relative abundances by measuring thelines on the stick diagram.

In this case, the two isotopes (with their relativeabundances) are:

boron-10 23boron-11 100

Working out the relative atomic mass

The relative atomic mass (RAM) of an element is giventhe symbol Ar and is defined as:

The relative atomic mass of an element is theweighted average of the masses of the isotopeson a scale on which a carbon-12 atom has amass of exactly 12 units.

A "weighted average" allows for the fact that there won'tbe equal amounts of the various isotopes. The examplecoming up should make that clear.

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Suppose you had 123 typical atoms of boron. 23 of thesewould be 10B and 100 would be 11B.

The total mass of these would be (23 x 10) + (100 x 11) =1330

The average mass of these 123 atoms would be 1330 /123 = 10.8 (to 3 significant figures).

10.8 is the relative atomic mass of boron.

Notice the effect of the "weighted" average. A simpleaverage of 10 and 11 is, of course, 10.5. Our answer of10.8 allows for the fact that there are a lot more of theheavier isotope of boron - and so the "weighted" averageought to be closer to that.

The mass spectrum for zirconium

The number of isotopes

The 5 peaks in the mass spectrum shows that there are 5isotopes of zirconium - with relative isotopic masses of90, 91, 92, 94 and 96 on the 12C scale.

The abundance of the isotopes

This time, the relative abundances are given aspercentages. Again you can find these relativeabundances by measuring the lines on the stick diagram.

In this case, the 5 isotopes (with their relative percentageabundances) are:

zirconium-90 51.5zirconium-91 11.2zirconium-92 17.1

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zirconium-94 17.4zirconium-96 2.8

Note:  You almost certainly wouldn't be able tomeasure these peaks to this degree of accuracy,but your examiners may well give you the data innumber form anyway. We'll do the sum with themore accurate figures.

Working out the relative atomic mass

Suppose you had 100 typical atoms of zirconium. 51.5 ofthese would be 90Zr, 11.2 would be 91Zr and so on.

Note:  If you object to the idea of having 51.5atoms or 11.2 atoms and so on, just assumeyou've got 1000 atoms instead of 100. That wayyou will have 515 atoms, 112 atoms, etc. Mostpeople don't get in a sweat over this, and just usethe numbers as they are!

The total mass of these 100 typical atoms would be

(51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + (2.8x 96) = 9131.8

The average mass of these 100 atoms would be 9131.8 /100 = 91.3 (to 3 significant figures).

91.3 is the relative atomic mass of zirconium.

Note:  If you want further examples of calculatingrelative atomic masses from mass spectra, youmight like to refer to my book, Calculations in Alevel Chemistry.

The mass spectrum of chlorine

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Chlorine is taken as typical of elements with more thanone atom per molecule. We'll look at its mass spectrumto show the sort of problems involved.

Chlorine has two isotopes, 35Cl and 37Cl, in theapproximate ratio of 3 atoms of 35Cl to 1 atom of 37Cl.You might suppose that the mass spectrum would looklike this:

You would be wrong!

The problem is that chlorine consists of molecules, notindividual atoms. When chlorine is passed into theionisation chamber, an electron is knocked off themolecule to give a molecular ion, Cl2+. These ions won'tbe particularly stable, and some will fall apart to give achlorine atom and a Cl+ ion. The term for this isfragmentation.

If the Cl atom formed isn't then ionised in the ionisationchamber, it simply gets lost in the machine - neitheraccelerated nor deflected.

The Cl+ ions will pass through the machine and will givelines at 35 and 37, depending on the isotope and youwould get exactly the pattern in the last diagram. Theproblem is that you will also record lines for theunfragmented Cl2+ ions.

Think about the possible combinations of chlorine-35 andchlorine-37 atoms in a Cl2+ ion.

Both atoms could be 35Cl, both atoms could be 37Cl, oryou could have one of each sort. That would give youtotal masses of the Cl2+ ion of:

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35 + 35 = 70

35 + 37 = 72

37 + 37 = 74

That means that you would get a set of lines in the m/z =70 region looking like this:

These lines would be in addition to the lines at 35 and 37.

The relative heights of the 70, 72 and 74 lines are in theratio 9:6:1. If you know the right bit of maths, it's veryeasy to show this. If not, don't worry. Just remember thatthe ratio is 9:6:1.

What you can't do is make any predictions about therelative heights of the lines at 35/37 compared with thoseat 70/72/74. That depends on what proportion of themolecular ions break up into fragments. That's whyyou've got the chlorine mass spectrum in two separatebits so far. You must realise that the vertical scale in thediagrams of the two parts of the spectrum isn't the same.

The overall mass spectrum looks like this:

Note:  This is based on information from the NISTChemistry WebBook. NIST is the US National

the mass spectra of elements http://www.chemguide.co.uk/analysis/masspec/elements.htm...

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Institute of Standards and Technology.

Questions to test your understanding

If this is the first set of questions you have done, please readthe introductory page before you start. You will need to usethe BACK BUTTON on your browser to come back hereafterwards.

questions on the mass spectra of elements

answers

Where would you like to go now?

To the mass spectrometry menu . . .

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To Main Menu . . .

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FRAGMENTATION PATTERNS INTHE MASS SPECTRA OF ORGANICCOMPOUNDS

This page looks at how fragmentation patterns areformed when organic molecules are fed into a massspectrometer, and how you can get information from themass spectrum.

The origin of fragmentation patterns

The formation of molecular ions

When the vaporised organic sample passes into theionisation chamber of a mass spectrometer, it isbombarded by a stream of electrons. These electronshave a high enough energy to knock an electron off anorganic molecule to form a positive ion. This ion is calledthe molecular ion - or sometimes the parent ion.

Note:  If you aren't sure about how a massspectrum is produced, it might be worth taking aquick look at the page describing how a massspectrometer works.

The molecular ion is often given the symbol M+ or -the dot in this second version represents the fact thatsomewhere in the ion there will be a single unpairedelectron. That's one half of what was originally a pair ofelectrons - the other half is the electron which wasremoved in the ionisation process.

Fragmentation

The molecular ions are energetically unstable, and someof them will break up into smaller pieces. The simplestcase is that a molecular ion breaks into two parts - one ofwhich is another positive ion, and the other is an

mass spectra - fragmentation patterns http://www.chemguide.co.uk/analysis/masspec/fragment.htm...

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uncharged free radical.

Note:  A free radical is an atom or group of atomswhich contains a single unpaired electron.

More complicated break-ups are beyond thescope of A'level syllabuses.

The uncharged free radical won't produce a line on themass spectrum. Only charged particles will beaccelerated, deflected and detected by the massspectrometer. These uncharged particles will simply getlost in the machine - eventually, they get removed by thevacuum pump.

The ion, X+, will travel through the mass spectrometerjust like any other positive ion - and will produce a line onthe stick diagram.

All sorts of fragmentations of the original molecular ionare possible - and that means that you will get a wholehost of lines in the mass spectrum. For example, themass spectrum of pentane looks like this:

Note:  All the mass spectra on this page havebeen drawn using data from the Spectral DataBase System for Organic Compounds (SDBS) atthe National Institute of Materials and ChemicalResearch in Japan.

mass spectra - fragmentation patterns http://www.chemguide.co.uk/analysis/masspec/fragment.htm...

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They have been simplified by omitting all theminor lines with peak heights of 2% or less of thebase peak (the tallest peak).

It's important to realise that the pattern of lines in themass spectrum of an organic compound tells yousomething quite different from the pattern of lines in themass spectrum of an element. With an element, each linerepresents a different isotope of that element. With acompound, each line represents a different fragmentproduced when the molecular ion breaks up.

Note:  If you are interested in the mass spectra ofelements, you could follow this link.

The molecular ion peak and the base peak

In the stick diagram showing the mass spectrum ofpentane, the line produced by the heaviest ion passingthrough the machine (at m/z = 72) is due to the molecularion.

Note:  You have to be a bit careful about this,because in some cases, the molecular ion is sounstable that every single one of them splits up,and none gets through the machine to register inthe mass spectrum. You are very unlikely to comeacross such a case at A'level.

The tallest line in the stick diagram (in this case at m/z =43) is called the base peak. This is usually given anarbitrary height of 100, and the height of everything elseis measured relative to this. The base peak is the tallestpeak because it represents the commonest fragment ionto be formed - either because there are several ways inwhich it could be produced during fragmentation of theparent ion, or because it is a particularly stable ion.

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Using fragmentation patterns

This section will ignore the information you can get fromthe molecular ion (or ions). That is covered in three otherpages which you can get at via the mass spectrometrymenu. You will find a link at the bottom of the page.

Working out which ion produces which line

This is generally the simplest thing you can be asked todo.

The mass spectrum of pentane

Let's have another look at the mass spectrum forpentane:

What causes the line at m/z = 57?

How many carbon atoms are there in this ion? Therecan't be 5 because 5 x 12 = 60. What about 4? 4 x 12 =48. That leaves 9 to make up a total of 57. How aboutC4H9+ then?

C4H9+ would be [CH3CH2CH2CH2]+, and this would beproduced by the following fragmentation:

The methyl radical produced will simply get lost in themachine.

The line at m/z = 43 can be worked out similarly. If youplay around with the numbers, you will find that thiscorresponds to a break producing a 3-carbon ion:

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The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:

The other lines in the mass spectrum are more difficult toexplain. For example, lines with m/z values 1 or 2 lessthan one of the easy lines are often due to loss of one ormore hydrogen atoms during the fragmentation process.You are very unlikely to have to explain any but the mostobvious cases in an A'level exam.

The mass spectrum of pentan-3-one

This time the base peak (the tallest peak - and so thecommonest fragment ion) is at m/z = 57. But this isn'tproduced by the same ion as the same m/z value peak inpentane.

If you remember, the m/z = 57 peak in pentane wasproduced by [CH3CH2CH2CH2]+. If you look at thestructure of pentan-3-one, it's impossible to get thatparticular fragment from it.

Work along the molecule mentally chopping bits off untilyou come up with something that adds up to 57. With asmall amount of patience, you'll eventually find[CH3CH2CO]+ - which is produced by this fragmentation:

You would get exactly the same products whichever sideof the CO group you split the molecular ion.

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The m/z = 29 peak is produced by the ethyl ion - whichonce again could be formed by splitting the molecular ioneither side of the CO group.

Peak heights and the stability of ions

The more stable an ion is, the more likely it is to form.The more of a particular sort of ion that's formed, thehigher its peak height will be. We'll look at two commonexamples of this.

Examples involving carbocations (carbonium ions)

Important!  If you don't know what a carbocation(or carbonium ion) is, or why the various sortsvary in stability, it's essential that you follow thislink before you go on.

Use the BACK button on your browser to returnquickly to this page.

Summarizing the most important conclusion from thepage on carbocations:

Order of stability of carbocations

primary < secondary < tertiary

Note:  The symbol "<" means "is less than". Sowhat this is saying is that primary ions are lessstable than secondary ones which in turn are lessstable than tertiary ones.

Applying the logic of this to fragmentation patterns, itmeans that a split which produces a secondary

mass spectra - fragmentation patterns http://www.chemguide.co.uk/analysis/masspec/fragment.htm...

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carbocation is going to be more successful than oneproducing a primary one. A split producing a tertiarycarbocation will be more successful still.

Let's look at the mass spectrum of 2-methylbutane.2-methylbutane is an isomer of pentane - isomers aremolecules with the same molecular formula, but adifferent spatial arrangement of the atoms.

Look first at the very strong peak at m/z = 43. This iscaused by a different ion than the corresponding peak inthe pentane mass spectrum. This peak in2-methylbutane is caused by:

The ion formed is a secondary carbocation - it has twoalkyl groups attached to the carbon with the positivecharge. As such, it is relatively stable.

The peak at m/z = 57 is much taller than thecorresponding line in pentane. Again a secondarycarbocation is formed - this time, by:

You would get the same ion, of course, if the left-handCH3 group broke off instead of the bottom one as we'vedrawn it.

In these two spectra, this is probably the most dramatic

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example of the extra stability of a secondary carbocation.

Examples involving acylium ions, [RCO]+

Ions with the positive charge on the carbon of a carbonylgroup, C=O, are also relatively stable. This is fairlyclearly seen in the mass spectra of ketones like pentan-3-one.

The base peak, at m/z=57, is due to the [CH3CH2CO]+

ion. We've already discussed the fragmentation thatproduces this.

Note:  There are lots of other examples ofpositive ions with extra stability and which areproduced in large numbers in a massspectrometer as a result. Without making thisarticle even longer than it already is, it'simpossible to cover every possible case.

Check past exam papers to find out whether youare likely to need to know about otherpossibilities. If you haven't got past papers, followthe link on the syllabuses page to find out how toget hold of them.

Using mass spectra to distinguish betweencompounds

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Suppose you had to suggest a way of distinguishingbetween pentan-2-one and pentan-3-one using theirmass spectra.

pentan-2-one CH3COCH2CH2CH3pentan-3-one CH3CH2COCH2CH3

Each of these is likely to split to produce ions with apositive charge on the CO group.

In the pentan-2-one case, there are two different ions likethis:

[CH3CO]+

[COCH2CH2CH3]+

That would give you strong lines at m/z = 43 and 71.

With pentan-3-one, you would only get one ion of thiskind:

[CH3CH2CO]+

In that case, you would get a strong line at 57.

You don't need to worry about the other lines in thespectra - the 43, 57 and 71 lines give you plenty ofdifference between the two. The 43 and 71 lines aremissing from the pentan-3-one spectrum, and the 57 lineis missing from the pentan-2-one one.

Note:  Don't confuse the line at m/z = 58 in thepentan-2-one spectrum. That's due to acomplicated rearrangement which you couldn'tpossibly predict at A'level.

The two spectra look like this:

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Computer matching of mass spectra

As you've seen, the mass spectrum of even very similarorganic compounds will be quite different because of thedifferent fragmentations that can occur. Provided youhave a computer data base of mass spectra, any unkownspectrum can be computer analysed and simply matchedagainst the data base.

Questions to test your understanding

If this is the first set of questions you have done, please readthe introductory page before you start. You will need to usethe BACK BUTTON on your browser to come back hereafterwards.

questions on fragmentation patterns

answers

mass spectra - fragmentation patterns http://www.chemguide.co.uk/analysis/masspec/fragment.htm...

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Where would you like to go now?

To the mass spectrometry menu . . .

To the instrumental analysis menu . . .

To Main Menu . . .

© Jim Clark 2000 (modified February 2014)

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11 of 11 9/16/17, 3:52 PM

MASS SPECTRA - THEMOLECULAR ION (M+) PEAK

This page explains how to find the relative formula mass(relative molecular mass) of an organic compound fromits mass spectrum. It also shows how high resolutionmass spectra can be used to find the molecular formulafor a compound.

Using a mass spectrum to find relativeformula mass

The formation of molecular ions

When the vaporised organic sample passes into theionisation chamber of a mass spectrometer, it isbombarded by a stream of electrons. These electronshave a high enough energy to knock an electron off anorganic molecule to form a positive ion. This ion is calledthe molecular ion.

Note:  If you aren't sure about how a massspectrum is produced, it might be worth taking aquick look at the page describing how a massspectrometer works.

The molecular ion is often given the symbol M+ or -the dot in this second version represents the fact thatsomewhere in the ion there will be a single unpairedelectron. That's one half of what was originally a pair ofelectrons - the other half is the electron which wasremoved in the ionisation process.

The molecular ions tend to be unstable and some ofthem break into smaller fragments. These fragmentsproduce the familiar stick diagram. Fragmentation isirrelevant to what we are talking about on this page - allwe're interested in is the molecular ion.

mass spectra - the molecular ion (M+) peak http://www.chemguide.co.uk/analysis/masspec/mplus.html#top

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MASS SPECTRA - THE M+1 PEAK

This page explains how the M+1 peak in a massspectrum can be used to estimate the number of carbonatoms in an organic compound.

Note:  This is a small corner of massspectrometry. It would be a good idea not toattack this page unless you have a reasonableidea about how a mass spectrum is produced andthe sort of information you can get from it. If youhaven't already done so, explore the massspectrometry menu before you go on.

What causes the M+1 peak?

What is an M+1 peak?

If you had a complete (rather than a simplified) massspectrum, you will find a small line 1 m/z unit to the rightof the main molecular ion peak. This small peak is calledthe M+1 peak.

In questions at this level (UK A level or its equivalent),the M+1 peak is often left out to avoid confusion -particularly if you were being asked to find the relativeformula mass of the compound from the molecular ionpeak.

mass spectra - the M+1 peak http://www.chemguide.co.uk/analysis/masspec/mplus1.html#top

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The carbon-13 isotope

The M+1 peak is caused by the presence of the 13Cisotope in the molecule. 13C is a stable isotope of carbon- don't confuse it with the 14C isotope which isradioactive. Carbon-13 makes up 1.11% of all carbonatoms.

If you had a simple compound like methane, CH4,approximately 1 in every 100 of these molecules willcontain carbon-13 rather than the more commoncarbon-12. That means that 1 in every 100 of themolecules will have a mass of 17 (13 + 4) rather than 16(12 + 4).

The mass spectrum will therefore have a linecorresponding to the molecular ion [13CH4]+ as well as[12CH4]+.

The line at m/z = 17 will be much smaller than the line atm/z = 16 because the carbon-13 isotope is much lesscommon. Statistically you will have a ratio ofapproximately 1 of the heavier ions to every 99 of thelighter ones. That's why the M+1 peak is much smallerthan the M+ peak.

Using the M+1 peak

What happens when there is more than 1 carbonatom in the compound?

Imagine a compound containing 2 carbon atoms. Eitherof them has an approximately 1 in 100 chance of being13C.

There's therefore a 2 in 100 chance of the molecule as awhole containing one 13C atom rather than a 12C atom -which leaves a 98 in 100 chance of both atoms being

mass spectra - the M+1 peak http://www.chemguide.co.uk/analysis/masspec/mplus1.html#top

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12C.

That means that the ratio of the height of the M+1 peakto the M+ peak will be approximately 2 : 98. That's prettyclose to having an M+1 peak approximately 2% of theheight of the M+ peak.

Note:  You might wonder why both atoms can't becarbon-13, giving you an M+2 peak. They can -and do! But statistically the chance of bothcarbons being 13C is approximately 1 in 10,000.The M+2 peak will be so small that you couldn'tobserve it.

Using the relative peak heights to predict the numberof carbon atoms

If there are small numbers of carbon atoms

If you measure the peak height of the M+1 peak as apercentage of the peak height of the M+ peak, that givesyou the number of carbon atoms in the compound.

We've just seen that a compound with 2 carbons willhave an M+1 peak approximately 2% of the height of theM+ peak.

Similarly, you could show that a compound with 3carbons will have the M+1 peak at about 3% of the heightof the M+ peak.

With larger numbers of carbon atoms

The approximations we are making won't hold with morethan 2 or 3 carbons. The proportion of carbon atomswhich are 13C isn't 1% - it's 1.11%. And the appoximationthat a ratio of 2 : 98 is about 2% doesn't hold as the smallnumber increases.

Consider a molecule with 5 carbons in it. You could workout that 5.55 (5 x 1.11) molecules will contain 1 13C toevery 94.45 (100 - 5.55) which contain only 12C atoms. Ifyou convert that to how tall the M+1 peak is as apercentage of the M+ peak, you get an answer of 5.9%(5.55/94.45 x 100). That's close enough to 6% that you

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might assume wrongly that there are 6 carbon atoms.

Above 3 carbon atoms, then, you shouldn't really bemaking the approximation that the height of the M+1peak as a percentage of the height of the M+ peak tellsyou the number of carbons - you will need to do somefiddly sums!

Important!  Most syllabuses at this level (UK Alevel or its equivalent) don't ask for this. Beforeyou get too bogged down in all this, check yoursyllabus and recent past papers to see whetheryou need to bother. This is a case where youreally need to know what line your examiners arecurrently taking - for example, how far are theypushing the approximation.

As far as I am aware, the only UK based syllabusto want this is Cambridge International (CIE). Ihave covered this specifically on this page in mynotes for the CIE syllabus

But to be sure, if you are doing a UK-based exam,and haven't got a syllabus and past papers, followthis link to find out how to get them.

Questions to test your understanding

I am not asking any questions about this page because I can'tfind any peak height data for the small simple molecules forwhich the approximation works. Without that, I can onlyproduce completely trivial questions. If you are doing CIEexams, you will find plenty of examples from their past papersusing their slightly more complicated formula.

Otherwise, just be aware that the M+1 peak is due to thepresence of C-13 atoms.

Where would you like to go now?

To the mass spectrometry menu . . .

To the instrumental analysis menu . . .

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To Main Menu . . .

© Jim Clark 2000 (updated August 2014)

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MASS SPECTRA - THE M+2 PEAK

This page explains how the M+2 peak in a massspectrum arises from the presence of chlorine or bromineatoms in an organic compound. It also deals briefly withthe origin of the M+4 peak in compounds containing twochlorine atoms.

Note:  Before you start this page, it would be agood idea to have a reasonable understandingabout how a mass spectrum is produced and thesort of information you can get from it. If youhaven't already done so, explore the massspectrometry menu before you go on.

The effect of chlorine or bromine atoms onthe mass spectrum of an organic compound

Compounds containing chlorine atoms

One chlorine atom in a compound

Note:  All the mass spectra on this page havebeen drawn using data from the Spectral DataBase System for Organic Compounds (SDBS) atthe National Institute of Materials and Chemical

mass spectra - the M+2 peak http://www.chemguide.co.uk/analysis/masspec/mplus2.html#top

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Research in Japan.

With one exception, they have been simplified byomitting all the minor lines with peak heights of2% or less of the base peak (the tallest peak).

The molecular ion peaks (M+ and M+2) each contain onechlorine atom - but the chlorine can be either of the twochlorine isotopes, 35Cl and 37Cl.

The molecular ion containing the 35Cl isotope has arelative formula mass of 78. The one containing 37Cl hasa relative formula mass of 80 - hence the two lines at m/z= 78 and m/z = 80.

Notice that the peak heights are in the ratio of 3 : 1. Thatreflects the fact that chlorine contains 3 times as much ofthe 35Cl isotope as the 37Cl one. That means that therewill be 3 times more molecules containing the lighterisotope than the heavier one.

So . . . if you look at the molecular ion region, and findtwo peaks separated by 2 m/z units and with a ratio of 3 :1 in the peak heights, that tells you that the moleculecontains 1 chlorine atom.

You might also have noticed the same pattern at m/z =63 and m/z = 65 in the mass spectrum above. Thatpattern is due to fragment ions also containing onechlorine atom - which could either be 35Cl or 37Cl. Thefragmentation that produced those ions was:

Note:  If you aren't sure about fragmentation youmight like to have a look at this link.

Two chlorine atoms in a compound

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Note:  This spectrum has been simplified byomitting all the minor lines with peak heights ofless than 1% of the base peak (the tallest peak).This contains more minor lines than other massspectra in this section. It was necessary becauseotherwise an important line in the molecular ionregion would have been missing.

The lines in the molecular ion region (at m/z values of 98,100 ands 102) arise because of the various combinationsof chlorine isotopes that are possible. The carbons andhydrogens add up to 28 - so the various possiblemolecular ions could be:

28 + 35 + 35 = 98

28 + 35 + 37 = 100

28 + 37 + 37 = 102

If you have the necessary maths, you could show that thechances of these arrangements occurring are in the ratioof 9:6:1 - and this is the ratio of the peak heights. If youdon't know the right bit of maths, just learn this ratio!

So . . . if you have 3 lines in the molecular ion region(M+, M+2 and M+4) with gaps of 2 m/z units betweenthem, and with peak heights in the ratio of 9:6:1, thecompound contains 2 chlorine atoms.

Compounds containing bromine atoms

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Bromine has two isotopes, 79Br and 81Br in anapproximately 1:1 ratio (50.5 : 49.5 if you want to befussy!). That means that a compound containing 1bromine atom will have two peaks in the molecular ionregion, depending on which bromine isotope themolecular ion contains.

Unlike compounds containing chlorine, though, the twopeaks will be very similar in height.

The carbons and hydrogens add up to 29. The M+ andM+2 peaks are therefore at m/z values given by:

29 + 79 = 108

29 + 81 = 110

So . . . if you have two lines in the molecular ion regionwith a gap of 2 m/z units between them and with almostequal heights, this shows the presence of a bromineatom in the molecule.

Questions to test your understanding

If this is the first set of questions you have done, please readthe introductory page before you start. You will need to usethe BACK BUTTON on your browser to come back hereafterwards.

questions on the M+2 peak

answers

mass spectra - the M+2 peak http://www.chemguide.co.uk/analysis/masspec/mplus2.html#top

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Where would you like to go now?

To the mass spectrometry menu . . .

To the instrumental analysis menu . . .

To Main Menu . . .

© Jim Clark 2000 (modified August 2014)

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