Learning outcome 11.2(a)This statement is about electrophoresis and the analysis of mixtures of amino acids or proteins.The background to electrophoresisA simple inorganic exampleSuppose you put a single crystal of potassium manganate(VII) (potassium permanganate) onto some damp filter paper supported on a microscope slide. The crystal would dissolve in the water in the filter paper, and the deep purple colour would diffuse out to make a small circle around the original crystal.

Now let's modify this by connecting the filter paper into a simple electric circuit. This time, when you put the potassium manganate(VII) crystal onto the paper, the colour doesn't spread into a circle. Instead, the purple colour starts to move towards the positively charged crocodile clip.

There isn't anything very surprising about this. The colour of the potassium manganate(VII) is due to the manganate(VII) ions present. These are negatively charged, MnO4-, and move towards the positive electrode.This movement of ions in an electric field is the basis of a separation technique known aselectrophoresis. Let's apply this to the more interesting (and more complicated) case of amino acids.

Electrophoresis involving amino acidsI am assuming that you have already done the Chemistry of Life (section 11.1) part of Applications of Chemistry. If you haven't, bits of the following aren't going to make sense to you.

Results of the electrophoresis of a mixture of simple amino acidsIn practice, paper isn't used for serious electrophoresis. Instead, a gel is used. Control of the pH is important, and so the gel is soaked in a buffer solution.Little troughs are made in the gel to hold the solutions being tested. In an exam, you may find that you have to interpret diagrams based on bits of paper or slabs of gel. It makes no difference whatsoever to what you would need to say.Unlike potassium manganate(VII), amino acids are colourless. To find out where they have got to, they are sprayed with ninhydrin which shows up the amino acids as brown or purplish stains.The diagram shows the effect of electrophoresis on a mixture of four amino acids at a known pH. This is just on a strip of paper, and the mixture was originally a single drop on the start line.

We need to explain why the four amino acids have moved in this way.Spot CThe amino acid responsible for spot C hasn't moved. Why not?Remember that an amino acid has both a basic amine group and an acidic carboxylic acid group.

There is an internal transfer of a hydrogen ion from the -COOH group to the -NH2group to leave an ion with both a negative charge and a positive charge.This is called azwitterion.

At a particular pH, known as the isoelectric point, this is how the amino acid exists in solution. The amino acid won't move during electrophoresis, because the two charges cancel each other out, and there won't be any attraction either to the positive or the negative electrode.At any other pH, there will be some movement one way or the other - and we will explore that later.So amino acid C doesn't move because the pH of the solution happens to be exactly the same as the isoelectric point.

Note: For many simple amino acids, the isoelectric point is at a pH of about 6. I have, however, come across the odd exam question which gives a diagram a bit like the one above, but describes the solution as being neutral. This is a bit careless! In fact, an amino acid with an isoelectric point of 6 would in fact move slightly towards the positive electrode at pH 7.This is discussed in detail at the bottom of the page about theacid-base behaviour of amino acids.

Spot DThis amino acid has moved towards the negative electrode, and so must be positively charged.This positive charge results from having an extra -NH2group in the amino acid, like this:

In a buffer solution with a pH around neutral, this will be present as the ion:

The extra -NH2group in the amino acid picks up a hydrogen ion from the water.Amino acids like this carry a net 1+ charge in a solution which is around neutral.The amino acid lysine is an example of this. You don't need to remember this formula (or the formulae of the other named amino acids mentioned below).

During electrophoresis at a pH about neutral, amino acids like this will travel towards the negative electrode.

Spots A and BThese travel towards the positive electrode and so must be negatively charged at this pH. This happens in amino acids which carry an extra -COOH group, for example:

In a buffer solution with a pH around neutral, this will be present as the ion:

The extra -COOH group loses a hydrogen ion to the water, to leave an ion which has an overall charge of 1-, and which therefore moves towards the positive electrode.Two examples of this are aspartic acid and glutamic acid.

The question remains as to why the two amino acids in our electrophoresis example have travelled different distances. One of the factors which affects this is the size of the ions.The ions have to find their way through the fibres in the paper or the pores in the gel. Smaller ions will travel faster than bigger ones. So, for example, the smaller aspartic acid will travel faster than the larger glutamic acid.What do I mean by "smaller" or "larger"? This could be in terms of the masses of the ions, or their shapes. Ions with bulky groups (such as benzene or other rings) will travel more slowly than ones with, say, unbranched chains.Since there are only two naturally occurring amino acids with an extra -COOH group, spot A must be due to aspartic acid and spot B to glutamic acid.

Changing the pH of electrophoresis experiments with amino acidsThis should be fairly obvious if you understand about the acid-base behaviour of amino acids.

What happens if the buffer solution you use has a low pH?In the presence of a sufficiently acidic solution, all the -COOH groups present will exist as -COOH groups, and not as ions.All the -NH2groups will have picked up hydrogen ions to form -NH3+.That means that all of the amino acids will carry a positive charge, and all of them will move towards the negative electrode.The three different sorts of amino acid that we have been looking at will have ions like this at a low enough pH:

All of them will move to the negative electrode, but they will move at different rates because, for example, one of them carries 2+ charges whereas the others carry only 1+. And obviously they will be different sizes as well - both in mass and shape.

What happens if the buffer solution you use has a high pH?In the presence of a sufficiently alkaline solution, all the -COOH groups present will exist as -COO-ions.All the -NH2groups will exist as simple -NH2groups - not ions.That means that all of the amino acids will carry a negative charge, and all of them will move towards the positive electrode.The three different sorts of amino acid that we have been looking at will have ions like this at a high enough pH:

All of them will move to the positive electrode, but they will move at different rates because, for example, one of them carries 2- charges whereas the others carry only 1-. And obviously they will be different sizes as well - both in mass and shape.

Don't waste time learning this. Make sure you understand it, and then work it out if you need to.

Other factors affecting the movement of the amino acidsVoltage matters. The higher the voltage, the greater the speed at which the amino acids move through the paper or gel.Temperature also matters. An increase in temperature can speed up electrophoresis.

Electrophoresis involving proteinsElectrophoresis can be used to estimate the relative molecular mass of a protein, or to sort a mixture of proteins into relative molecular mass order.One technique for doing this is known as SDS-PAGE. SDS-PAGE stands for Sodium Dodecyl Sulfate PolyAcrylamide Gel Electrophoresis.The sodium dodecyl sulfate is a constituent of many detergents and cleaning products. It is also known as sodium lauryl sulfate. If a protein is treated with SDS and then heated, the protein is denatured.That means that its secondary and tertiary structures are lost. It becomes covered in SDS molecules, and this turns the protein molecule into a long tube covered with negative charges on the outside. The negative charges come from the SDS molecules.A protein which has more than one sub-unit is split into these separate bits, each of which is coated in SDS molecules.If these treated protein molecules undergo electrophoresis, all the molecules will move towards the positive electrode, but the smaller molecules will move through the gel faster than the bigger ones.There is a good, simple animation of this on thisYouTubepage.

Note: If this link doesn't work, please let me know via the address on theabout this sitepage.This is only intended to be a quick outline of how this sort of electrophoresis works. You shouldn't need any more than the above for CIE purposes.

Obviously, protein molecules have to be stained in some way in order to find out how far they have travelled. The intensity of the colour is a measure of how much of any particular protein is in the mixture.The apparatus for doing this can be calibrated by using proteins of known relative molecular mass, and finding out how far they travel.This technique is used to detect the presence or absence of proteins which might be an indicator of disease. For example, kidney disease can be recognised by the unusual presence of proteins in urine. Lots of diseases including liver disease and some cancers produce abnormal levels of some proteins in the blood.

Learning outcome 11.2(b)This statement is about the use of electrophoresis in DNA fingerprinting (DNA profiling).DNA fingerprintingThe finished productMost people will have seen pictures like the one on the right showing a finished DNA fingerprint (or profile). This shows a comparison of the DNA of two suspects A and B with DNA in evidence (E) from a crime scene.It is fairly easy to see that A's DNA matches the evidence.This page is designed to give you enough idea of how a DNA profile like this is produced to satisfy the needs of the CIE syllabus.The problem is that there are old and new methods of doing this, and these differ quite a lot in the details. Modern techniques don't actually produce pictures like this either.I am going to concentrate on the older method, because it is a bit easier to understand, and also because it matches the sort of material that CIE are producing.

Background informationYou will know that the structures of all the proteins in the body come from genetic code carried by DNA. In humans, this DNA is found in 46 chromosomes, arranged in 23 pairs. One member of each pair of chromosomes comes from the father, and the other from the mother.Each chromosome has a whole lot of genes strung out along it, and each gene is responsible for the code to make a particular protein.But in between the genes are lengths of DNA which don't code for proteins, and in these lengths you can find repeated patterns of DNA bases.So, for example, you might get short sequences (like this GATA sequence) repeated 5 times:GATAGATAGATAGATAGATA. . . or repeated 10 times:GATAGATAGATAGATAGATAGATAGATAGATAGATAGATA. . . or however many times.The number of repeats will be different from person to person.All the chromosomes will contain repeated sequences of variable lengths. Some sequences are short (as the GATA sequence above); some can be up to 80 "letters" long. And each sequence can be repeated a number of times varying from person to person.

Using teaching material from the Dolan DNA Learning CenterRepeating sequences in DNAI came to this topic as a non-biologist knowing nothing whatsoever about how genetic fingerprinting works. When I was researching it, I found this DNA Learning Center material to be by far the most helpful, and least confusing, of all the sites I visited.So I want to guide you through some of their material. You will need to go to thisDNA Learning Centerpage. I suggest that you open this in a new browser tab (or at least in a new browser window), so that you can flip easily back and forward to this page.

Note: If you try to move back and forwards between this page and the DNA Learning Center page using back and forward buttons on your browser, it won't work! Each time you return to the DNA Learning Center page, it will take you right back to the start of the sequence. The sequence is written in Flash, and you can't break into the middle of it.The simplest way around this is to open the link in a new tab.

Start by choosing the Human Identification module by clicking the box in the bottom left-hand corner.Having read that page, click on the word "profiling" in the top line of the page to the right of the words "Human Identification".Read that page, and then click on the left-hand one of the three big circles above the text, and then, finally, click on the A button to lead you into a simple sequence of animations. Follow that through to the end.If you aren't a biologist, don't worry too much about any long words you come across. CIE will probably expect you to recognise the terms VNTR and STR, though.VNTR are repeated sequences of from about 9 - 90 "letters". STR are repeated sequences of from 2 - 8 "letters". Just recognise that the S in STR stands for short.When you have finished this sequence, you could, if you want to, watch the two short video clips by clicking on the two V buttons just underneath the A button you pressed before.

Cutting the DNA up and producing a DNA fingerprintThis is the core of the DNA fingerprinting process.If you already have the DNA Learning Centre material in a separate browser tab or window, you can go back to this and continue from where you left off. You need to click on the second of the three big circles which explains the first DNA fingerprints.

Note: If you don't still have this available, then go to theDNA Learning Centerpage in a new tab or browser window. Choose the Human Identification module by clicking the box in the bottom left-hand corner. Then choose profiling, and finally click on the second of the three big circles above the text.

Now go into the animation sequence by clicking on the A button. Work through this to the end.

More modern techniquesYoucouldfind out about these from the DNA Learning Center by clicking on the third of the big circles. However, I don't think that this is necessarily a good idea, because it leads you to methods which are quite unlike those asked about by CIE - and you risk getting confused.There is, though, one point which CIE do want you to know about, and which came up as a small part of a major question about DNA fingerprinting in June 2011.The technique you have looked at up to now needs a reasonable amount of DNA to work with. More modern techniques will work with very, very tiny amounts. They do this by multiplying up the DNA using the enzyme DNA polymerase.According to the Chemistry Applications Support Booklet, this is done after the DNA has been chopped up using restriction enzymes. Huge numbers of copies of a particular segment in the resulting mixture which contains the repeating sequence you are interested in can then be produced. This is known as a Polymerisation Chain Reaction (PCR).

Summarising what you are likely to need for CIE exam purposesUp to June 2011, CIE had asked about this just twice - a small bit of a question in November 2008, and a full question in June 2011. Neither of them wanted any real detail. What follows would cope with the process of producing the fingerprint from either question.Extract the DNADNA can be obtained from blood, hair, skin and semen, for example.Chop the DNA into short segmentsA number of different restriction enzymes can cut the DNA into smaller segments. These enzymes can be chosen to target segments containing particular repeating sequences of bases.Multiply up the number of particular segments if your DNA sample is very smallYou can use the polymerisation chain reaction (PCR).Place the samples in troughs on agarose gelThe agarose gel is the support medium for electrophoresis.Carry out the electrophoresisBecause of the phosphate groups in DNA, the segments that the DNA splits into will carry negative charges, and so will all move towards the positive electrode. Segments with a small number of repeats will travel faster, and therefore further, than those longer ones with more repeats.Make the fingerprint visibleThis is done by transferring the fingerprint on the gel to a more stable nylon membrane using a technique called "Southern blotting" (named after the inventor of the technique). This is then treated with radioactively labelled probes.These are short lengths of DNA, radioactively labelled with phosphorus-32 atoms. These are designed to bind with the repeating sequences in the DNA fragments.If a sheet of photographic (or X-ray) film is placed over the membrane, radiation from the radioactive phosphorus fogs the film. That produces the effect that you will have seen in the graphic at the top of this page.

Some uses of DNA fingerprintingForensic scienceWe have already mentioned this at the top of the page. Notice that DNA matching in forensic science doesn't only help to convict guilty people, but it also helps to eliminate innocent people as possible suspects.

Paternity testingYou should remember from the DNA Learning Center material that you have 23 pairs of chromosomes. One of each pair comes from your mother, and one from your father.It is possible to get restriction enzymes which are so selective that they will only cut out a particular sequence of repetitions at one particular site on one particular pair of chromosomes.If you just used that one enzyme, you would get a single band in the fingerprint which was due to your mother and a single band which was due to your father. If you can do this across a number of different chromosome pairs, then half of the bands would come from your mother and half from your father.You will find a short neatpaternity testing animationfrom Sumanas Inc if you follow this link. Use the Continue button to advance through the animation, unless you want to go backwards, in which case use the back arrow.This animation would give you enough understanding to answer a 2 mark part of the June 2011 question.

DNA and archeologyThe same June 2011 question had a 3 mark section about this. You were told about ancient writings on goatskins which had become broken into fragments, with the implication that the fragments might have come from several different goatskins.You were then asked how DNA fingerprinting might be used to identify which fragment came from which skin.Before we look at that, my gut feeling is that you are more likely to be asked a question like this than a question asking you to just quote an example of the use of DNA testing in archeology. For one thing, a question which gives you a context, and asks you to comment on it, is far easier to mark!All they wanted was for you to suggest that you compared the DNA fingerprints of every fragment, and then matched them to find out which came from the same skins. That was worth 2 marks, and is something that you could fairly easily work out even if you hadn't come across the problem before.

DNA fingerprints and medicineThe Chemistry Applications Support Booklet mentions using DNA fingerprinting in tuberculosis (TB) and cancers to see whether the problem is a recurrence of an old disease or is due to a new one.The Chemistry Coursebook has quite a long section about genetic testing (which is much easier to discuss), but I am not convinced that this is the same as DNA fingerprinting. CIE haven't set any questions so far (in the nine exam sessions up to June 2011) which asked for this. It needs a comment from an Examiner's Report to find out what they will or won't accept.I will come back to this if a question comes up in the future. In the meantime, you would probably be safer to stick to what the Support Booklet is saying - they would have to accept that.A simple question asking you for a medical use won't be worth more than 1 mark. Any more complicated question would give you all the background information you need, and then ask you to apply what you already know about DNA fingerprinting to that. So there shouldn't be a problem.

earning outcome 11.2(e)This statement is about the use of NMR and X-ray crystallography in determining the structure of big molecules such as proteins.This statement simply asks you for an "awareness" of the uses of NMR and X-ray crystallography. The second edition of the Applications Support Booklet makes it clear: "Students do not need to know anything about the principles or the methods of X-ray crystallography."What follows is based on the (fairly frequent) exam questions which are asked about this topic.

X-ray crystallographyX-ray crystallography works by shining X-rays onto a very pure crystal of the substance you are interested in. The X-rays are diffracted (bent from their original path) when they interact with regions of high electron density.The result is a diffraction pattern made up of lots of spots of varying intensities.You will find an example of adiffraction patternby following this link.Measurements of patterns like this lead toelectron density mapslike the one you will find from the same site by following this link.Interpretation of these enables you to see where the main atoms are to be found in the molecule, and to measure bond lengths and angles. This is useful in looking at the geometry of active sites in enzyme molecules, for example.

Note: Using material from other sites is always a bit risky, because they can change. If you have trouble loading either of the last two links, please let me know via the address on theabout this sitepage.

What a map like this doesn't show are the positions of hydrogen atoms. There isn't enough electron density around a hydrogen atom for it to have any effect on the X-rays.Key things to remember X-ray crystallography needs very pure crystals to work with. X-ray crystallography detects positions of high electron density in a structure. Electrons cause diffraction of the X-rays which results in a pattern of spots. The pattern of spots can be used to produce an electron density map which allows you to work out the positions of various atoms, and the shapes of important features like active sites in enzymes. X-ray crystallography won't detect the positions of hydrogen atoms.

NMRAll you need to realise here is that: NMR is carried out in solution. NMR tells you about the position of hydrogen atoms because it affects the protons in the nuclei of hydrogen atoms. NMR works because of the two possible magnetic orientations of the protons - aligned with or against an applied magnetic field. The energy gap between these corresponds with energies in radio frequencies.

Note: You may come across the expression that a proton in the nucleus of a hydrogen can have two possible "spin states". These two different spin states correspond to the two different magnetic orientations of the proton. You don't need to understand what exactly spin state refers to.

MRI and X-rays in medicineThis is another example of CIE at its irritating worst. There is no mention of either MRI scans or the use of medical X-rays in the syllabus, and a literal reading of syllabus statement 11.2(e) couldn't possibly be taken to imply that either are wanted, even if you trace the links back to the other statements mentioned.There is a brief mention of MRI scans in the Applications Support Booklet. This seems, therefore, to have been arbitrarily introduced by an author of the Support Booklet - bearing no relationship to what the syllabus is asking.CIE have asked about this twice in the nine exam sessions up to June 2011, each time implying in the Examiner's Reports that students didn't know enough about this. Why should they?The following is written to give you enough to be able answer a question on this non-syllabus material.First, you must realise that medical use of X-rays has nothing whatsoever to do with X-ray crystallography. Medical X-rays show up bones because bones are more opaque to X-rays than soft tissue, but are not so good at identifying soft tissue. Medical X-rays are harmful (causing tissue damage) if over-used.Secondly, although MRI (Magnetic Resonance Imaging) scans rely on the same sort of energy gaps between the two magnetic states of protons in hydrogen nuclei, the way this is used, and what the results show, bears no resemblance to NMR as used in chemistry or biochemistry.You need to know that: Unlike X-rays, MRI doesn't cause any tissue damage. Unlike X-rays, MRI isn't obscured by bones. MRI picks up particular concentrations of hydrogen nuclei in different tissues - for example, in water or fat molecules. MRI can be fine-tuned to pick out particular tissue types which can be shown on a scan as different shades of grey. This includes the ability to distinguish between cancerous or normal tissue.

Learning outcome 11.2(f)This statement is about partition coefficents.PartitionIf you have two immiscible liquids like ether and water, and shake them up in a separating funnel, they obviously form two layers. The ether is less dense than water, and so forms the top layer.Now suppose you shake up a mixture of ether and water containing a substance which is soluble in both of them. Let's suppose that the substance, X, is more soluble in ether than it is in water.

Particles of X will cross the boundary between the two liquid layers, and you will soon get a dynamic equilibrium set up. For every particle which moves into the top layer, one will move back down into the bottom one.You could write an equation for this:

. . . and like any other equilibrium, you can find an equilibrium constant:

This equilibrium constant is called thepartition coefficient, and is often given the symbol Kpc.Like other equilibrium constants, partition coefficients are constant at a constant temperature, but they have some other restrictions as well. They only work properly for fairy dilute solutions, and the solute must be in the same chemical form in both solvents. It mustn't react, or ionise or associate (join together in dimers, for example).

Partition coefficient calculationsA note on unitsNotice that the partition coefficient is a simple ratio of two concentrations. It doesn't matter what concentration units you use - as long as you use the same ones top and bottom.You could use mol dm-3, but more often you use g cm-3- grams per cubic centimetre.Technically, the square brackets can only be used for concentration in mol dm-3, but the Application Support Booklet and CIE's mark schemes both use it for other units as well.I'm not prepared to do that, and so shall use the term "concentration of X" rather than [X] where non-standard concentration units are used.

Calculating a partition coefficientWhen a solution of 1.00 g of X in 100 cm3of water was shaken with 10 cm3of ether, 0.80 g of X was transferred to the ether layer. Calculate the partition coefficient of X between ether and water.If you are asked to calculate a partition coefficient between two solvents, the concentration of the first solvent mentioned goes on top of the Kpcexpression. So in this case:

You have enough information to calculate both concentrations in g cm-3.concentration of X in ether = 0.80/10 g cm-3If 0.80 g were transferred to the ether, 1.00 - 0.80 g = 0.20 g were left in the water.concentration of X in water = 0.20/100 g cm-3So:

Obviously, you could work out the concentrations in ether and in water as actual numbers before you put them into the expression. Do it however you feel most comfortable.Partition coefficients like this don't have units - the units cancel out because they are the same top and bottom.

Calculations involving partition coefficientsThe CIE syllabus says specifically that you should be able to calculate a partition coefficient (in other words, what we have just done). It says nothing whatsoever about using them to calculate other things. These extra calculations are discussed in the Application Support Booklet and have been asked twice in exams up to June 2011.The Coursebook doesn't include them (quite properly!) because the syllabus doesn't mention them, and this book has been endorsed by CIE examiners as a "complete and precise coverage" of the syllabus. So does that mean that CIE won't ask any more questions in the future? Don't count on it!A basic exampleWe will use the same case as before - the same solvents, the same X and the same partition coefficient we have just calculated.This time we will work out how much would have been extracted into the ether layer if we had shaken the original solution of 1.00 g of X in 100 cm3of water with just 5 cm3of ether.We are trying to work out the mass of X extracted. Let's call that m.Now work out an expression for the concentration of the solution of X in ether.concentration of X in ether = m/5 g cm-3What about the water? There will be (1.00 - m) g of X left in the water. So:concentration of X in water = (1.00 - m)/100 g cm-3Now you can put all this into the partition coefficient expression. Remember that we have already calculated the partition coefficient of X between ether and water as 40.

You are then just faced with a simple, but slightly tedious, bit of algebra:

Making it more tedious!In the original calculation to find the partition coefficient, you were told that if you shook the original solution of 1.00 g of X in 100 cm3of water with 10 cm3of ether, you extracted 0.80 g of X.Shaking it with 5 cm3of ether, we have just worked out that you would extract 0.67 g of X.That would leave 0.33 g of X behind in the 100 cm3of water. Suppose you had carefully kept this solution, and then shook it with a second fresh 5 cm3of ether.How much of X would you extractin totalby using the ether as two separate lots of 5 cm3instead of the 10 cm3in one go?Let's call the mass of X extracted by the second lot of ether n - so that we don't get confused.Work out an expression for the concentration of the solution of X in ether.concentration of X in ether = n/5 g cm-3What about the water? There will be (0.33 - n) g of X left in the water after the second extraction. So:concentration of X in water = (0.33 - n)/100 g cm-3Now you can put all this into the partition coefficient expression for X between ether and water as before.

That means that if you were to combine the two 5 cm3lots of ether, you would have extracted a total of 0.67 + 0.22 g of X. That is 0.89 g.You were originally told that if you had only done this once, using the ether as a single lot of 10 cm3, you would only have extracted 0.80 g.You get a more efficient extraction by splitting your solvent up into smaller volumes as above.You use this sort of technique during the preparation of some organic compounds. You extract what you are trying to make from some messy solution in water so that it ends up in an organic solvent. You then remove the solvent by careful distillation.

CIE asked a question similar to this in June 2009 paper 4 Q8. You had to calculate a value for the partition coefficient, and then use it in a two step extraction exactly as above. The calculation was worth 4 marks. This may be tedious and (apart from calculating the partition coefficient) not on the syllabus, but you can't afford not to be able to do it.Learning outcome 11.2(g)This statement is about various sorts of chromatography.You will find everything you need if you work through all the pages from thechromatography menu.It is important that you work through these in the order listed in the menu, andnotin the order given in the syllabus statement. You might think that it is easier to start with something simple like paper chromatography, but there are problems with explaining paper chromatography.

Thin layer chromatography: TLCSo start with thin layer chromatography. This is just like the paper chromatography you may be familiar with, except that you use something other than paper. Make sure that you fully understand why some things move further than others during the chromatography.

Column chromatographyThe next page from the menu is about column chromatography. This isn't mentioned by the syllabus, but is a useful bridge between thin layer and high performance liquid chromatography. It is easy to understand, and will make HPLC look less scary.You don't need to remember any details about this, of course.

High performance liquid chromatography: HPLCThere are two types of HPLC. The Applications Support Booklet mentions this in passing, and the Chemistry Coursebook seems to be describing reversed phase HPLC, but without actually calling it that.I suggest that you concentrate on reversed phase HPLC, and ignore the other form. If you try to learn both forms, you are only going to get confused. Make sure that you understand what is going on in the column, because that is what is most likely to be tested.

Gas-liquid chromatographyIn gas-liquid chromatography, you should again concentrate on what is happening in the column so that you can explain why some things pass through faster than others. Again, that is the most likely thing for you to get tested on.A past question (November 2010 paper 43 Q10) gave you an output chart with two triangular shaped peaks labelled X and Y, and asked you to work out the percentage of each of X and Y in the mixture.What you needed to do was to find the areas under the two peaks, add them together, and then work out the percentages. To do that, you need to remember a simple bit of geometry - the formula for the area of a triangle. That's 1/2 x base x height.You needed to take measurements from the exam paper. In fact, in this case, the bases of the two triangles were identical, and so you could take a short-cut, and just work from the heights if you had enough confidence.

Paper chromatographyFinally, you can return to the more familiar paper chromatography. You will find that much of this is a repeat of the thin layer chromatography you started with.CIE count the mechanism for paper chromatography as partition of the solutes between the solvent and water trapped in the cellulose fibres. Although I think you should read about the problem with this explanation in the final section of the page, don't try to remember it. If you are ever asked for an example of paper chromatography, just avoid the use of water (or any water-soluble solvent such as alcohol) as a solvent.Be sure that you understand two-way paper chromatography. This was asked three times in the first nine exam sessions of the current syllabus. You must be able to interpret a 2-way chromatogram.For example, suppose you were given a 2-way chromatogram with a set of seven spots (labelled A to F) like this:

M represents the position of the original drop of mixture.Can you answer questions like these?How many spots would there have been in the chromatogram after you had used solvent 1 and before you used solvent 2?4. B and C wouldn't have separated out, and neither would D and E.Which spot is due to a substance almost insoluble in solvent 2?D - because it has hardly moved at all when solvent 2 was used?Which spot is due to a substance almost insoluble in solvent 1, but very soluble in solvent 2?F - because it only travelled a little way up the paper when solvent 1 was used, but was carried a long way by solvent 2.

Partition or adsorption?You must be sure that you can define partition and adsorption in the context of chromatography.Partition chromatographyis a separation due to the different solubilities of a solute in two different solvents - one the mobile phase and the other the stationary phase.Paper chromatography and gas-liquid chromatography both involve partition. If you aren't sure about this, go and read those pages again.Adsorption chromatographyis a separation due to molecules having different attractions for the solid stationary phase and the liquid mobile phase. If there are strong attractions between the solute and the stationary phase, the solute won't move very far - it will spend most of its time sticking to the stationary phase.On the other hand, if there are weak attractions between the solute and the stationary phase, but strong attractions between the solute and the solvent, it will spend most of its time in solution, and so travel much further.Thin layer chromatography involves adsorption.

What about HPLC? This depends on which method you are talking about. If you are talking about the most commonly used form (reversed phase HPLC), it is debatable whether it involves partition or adsorption. CIE have sensibly avoided asking you which this is up to now (August 2011), because there is no obviously right answer.THIN LAYER CHROMATOGRAPHY

This page is an introduction to chromatography using thin layer chromatography as an example. Although if you are a beginner you may be more familiar with paper chromatography, thin layer chromatography is equally easy to describe and more straightforward to explain.

Note: I'm taking a simple view of the way that thin layer chromatography works in terms of adsorption (see below) which should be adequate for students doing courses for 16 - 18 year olds. The reality is more complicated and the explanation will vary depending on what sort of solvent or solvent mixture you are using. Some similar problems are discussed on the page about paper chromatography, but I am unwilling to do the same thing on this page which is intended as a fairly gentle introduction to chromatography.

Carrying out thin layer chromatographyBackgroundChromatography is used to separate mixtures of substances into their components. All forms of chromatography work on the same principle.They all have astationary phase(a solid, or a liquid supported on a solid) and amobile 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.Thin layer chromatography is done exactly as it says - using a thin, uniform layer of silica gel or alumina coated onto a piece of glass, metal or rigid plastic.The silica gel (or the alumina) is the stationary phase. The stationary phase for thin layer chromatography also often contains a substance which fluoresces in UV light - for reasons you will see later. The mobile phase is a suitable liquid solvent or mixture of solvents.

Producing the chromatogramWe'll start with a very simple case - just trying to show that a particular dye is in fact a mixture of simpler dyes.

Note: The chromatography plate will in fact be pure white - not pale grey. I'm forced to show it as off-white because of the way I construct the diagrams. Anything I draw as pure white allows the background colour of the page to show through.

A pencil line is drawn near the bottom of the plate and a small drop of a solution of the dye mixture is placed on it. Any labelling on the plate to show the original position of the drop must also be in pencil. If any of this was done in ink, dyes from the ink would also move as the chromatogram developed.When the spot of mixture is dry, the plate is stood in a shallow layer of solvent in a covered beaker. It is important that the solvent level is below the line with the spot on it.The reason for covering the beaker is to make sure that the atmosphere in the beaker is saturated with solvent vapour. To help this, the beaker is often lined with some filter paper soaked in solvent. Saturating the atmosphere in the beaker with vapour stops the solvent from evaporating as it rises up the plate.As the solvent slowly travels up the plate, the different components of the dye mixture travel at different rates and the mixture is separated into different coloured spots.

The diagram shows the plate after the solvent has moved about half way up it.The solvent is allowed to rise until it almost reaches the top of the plate. That will give the maximum separation of the dye components for this particular combination of solvent and stationary phase.

Measuring RfvaluesIf all you wanted to know is how many different dyes made up the mixture, you could just stop there. However, measurements are often taken from the plate in order to help identify the compounds present. These measurements are the distance travelled by the solvent, and the distance travelled by individual spots.When the solvent front gets close to the top of the plate, the plate is removed from the beaker and the position of the solvent is marked with another line before it has a chance to evaporate.These measurements are then taken:

The Rfvalue for each dye is then worked out using the formula:

For example, if the red component travelled 1.7 cm from the base line while the solvent had travelled 5.0 cm, then the Rfvalue for the red dye is:

If you could repeat this experiment underexactlythe same conditions, then the Rfvalues for each dye would always be the same. For example, the Rfvalue for the red dye would always be 0.34. However, if anything changes (the temperature, the exact composition of the solvent, and so on), that is no longer true. You have to bear this in mind if you want to use this technique to identify a particular dye. We'll look at how you can use thin layer chromatography for analysis further down the page.

What if the substances you are interested in are colourless?There are two simple ways of getting around this problem.Using fluorescenceYou may remember that I mentioned that the stationary phase on a thin layer plate often has a substance added to it which will fluoresce when exposed to UV light. That means that if you shine UV light on it, it will glow.That glow is masked at the position where the spots are on the final chromatogram - even if those spots are invisible to the eye. That means that if you shine UV light on the plate, it will all glow apart from where the spots are. The spots show up as darker patches.

While the UV is still shining on the plate, you obviously have to mark the positions of the spots by drawing a pencil circle around them. As soon as you switch off the UV source, the spots will disappear again.

Showing the spots up chemicallyIn some cases, it may be possible to make the spots visible by reacting them with something which produces a coloured product. A good example of this is in chromatograms produced from amino acid mixtures.The chromatogram is allowed to dry and is then sprayed with a solution ofninhydrin. Ninhydrin reacts with amino acids to give coloured compounds, mainly brown or purple.

In another method, the chromatogram is again allowed to dry and then placed in an enclosed container (such as another beaker covered with a watch glass) along with a fewiodine crystals.The iodine vapour in the container may either react with the spots on the chromatogram, or simply stick more to the spots than to the rest of the plate. Either way, the substances you are interested in may show up as brownish spots.

Using thin layer chromatography to identify compoundsSuppose you had a mixture of amino acids and wanted to find out which particular amino acids the mixture contained. For simplicity we'll assume that you know the mixture can only possibly contain five of the common amino acids.A small drop of the mixture is placed on the base line of the thin layer plate, and similar small spots of the known amino acids are placed alongside it. The plate is then stood in a suitable solvent and left to develop as before. In the diagram, the mixture is M, and the known amino acids are labelled 1 to 5.The left-hand diagram shows the plate after the solvent front has almost reached the top. The spots are still invisible. The second diagram shows what it might look like after spraying with ninhydrin.

There is no need to measure the Rfvalues because you can easily compare the spots in the mixture with those of the known amino acids - both from their positions and their colours.In this example, the mixture contains the amino acids labelled as 1, 4 and 5.And what if the mixture contained amino acids other than the ones we have used for comparison? There would be spots in the mixture which didn't match those from the known amino acids. You would have to re-run the experiment using other amino acids for comparison.

How does thin layer chromatography work?The stationary phase - silica gelSilica gel is a form of silicon dioxide (silica). The silicon atoms are joined via oxygen atoms in a giant covalent structure. However, at the surface of the silica gel, the silicon atoms are attached to -OH groups.

So, at the surface of the silica gel you have Si-O-H bonds instead of Si-O-Si bonds. The diagram shows a small part of the silica surface.

The surface of the silica gel is very polar and, because of the -OH groups, can form hydrogen bonds with suitable compounds around it as well as van der Waals dispersion forces and dipole-dipole attractions.

The other commonly used stationary phase is alumina - aluminium oxide. The aluminium atoms on the surface of this also have -OH groups attached. Anything we say about silica gel therefore applies equally to alumina.What separates the compounds as a chromatogram develops?As the solvent begins to soak up the plate, it first dissolves the compounds in the spot that you have put on the base line. The compounds present will then tend to get carried up the chromatography plate as the solvent continues to move upwards.How fast the compounds get carried up the plate depends on two things: How soluble the compound is in the solvent. This will depend on how much attraction there is between the molecules of the compound and those of the solvent. How much the compound sticks to the stationary phase - the silica gel, for example. This will depend on how much attraction there is between the molecules of the compound and the silica gel.Suppose the original spot contained two compounds - one of which can form hydrogen bonds, and one of which can only take part in weaker van der Waals interactions.The one which can hydrogen bond will stick to the surface of the silica gel more firmly than the other one. We say that one isadsorbedmore strongly than the other. Adsorption is the name given to one substance forming some sort of bonds to the surface of another one.Adsorption isn't permanent - there is a constant movement of a molecule between being adsorbed onto the silica gel surface and going back into solution in the solvent.Obviously the compound can only travel up the plate during the time that it is dissolved in the solvent. While it is adsorbed on the silica gel, it is temporarily stopped - the solvent is moving on without it. That means that the more strongly a compound is adsorbed, the less distance it can travel up the plate.In the example we started with, the compound which can hydrogen bond will adsorb more strongly than the one dependent on van der Waals interactions, and so won't travel so far up the plate.What if both components of the mixture can hydrogen bond?It is very unlikely that both will hydrogen bond to exactly the same extent, and be soluble in the solvent to exactly the same extent. It isn't just the attraction of the compound for the silica gel which matters. Attractions between the compound and the solvent are also important - they will affect how easily the compound is pulled back into solution away from the surface of the silica.However, it may be that the compounds don't separate out very well when you make the chromatogram. In that case, changing the solvent may well help - including perhaps changing the pH of the solvent.This is to some extent just a matter of trial and error - if one solvent or solvent mixture doesn't work very well, you try another one. (Or, more likely, given the level you are probably working at, someone else has already done all the hard work for you, and you just use the solvent mixture you are given and everything will work perfectly!)

Note: You will find detailed descriptions with photographs of how to carry out thin layer chromatography by going to theColorado University site.

COLUMN CHROMATOGRAPHY

This page shows how the same principles used in thin layer chromatography can be applied on a larger scale to separate mixtures in column chromatography. Column chromatography is often used to purify compounds made in the lab.

Note: It is important to read the introductory page aboutthin layer chromatographybefore you continue with this one - particularly the part about how thin layer chromatography works, although you will also need some idea about how to make a thin layer chromatogram.

Carrying out column chromatographyThe columnIn thin layer chromatography, the stationary phase is a thin layer of silica gel or alumina on a glass, metal or plastic plate. Column chromatography works on a much larger scale by packing the same materials into a vertical glass column.Various sizes of chromatography columns are used, and if you follow a link at the bottom of the page to the Organic Chemistry section of the Colorado University site, you will find photographs of various columns. In a school lab, it is often convenient to use an ordinary burette as a chromatography column.

Using the columnSuppose you wanted to separate a mixture of two coloured compounds - one yellow, one blue. The mixture looks green.You would make a concentrated solution of the mixture preferably in the solvent used in the column.First you open the tap to allow the solvent already in the column to drain so that it is level with the top of the packing material, and then add the solution carefully to the top of the column. Then you open the tap again so that the coloured mixture is all absorbed into the top of the packing material, so that it might look like this:

Next you add fresh solvent to the top of the column, trying to disturb the packing material as little as possible. Then you open the tap so that the solvent can flow down through the column, collecting it in a beaker or flask at the bottom. As the solvent runs through, you keep adding fresh solvent to the top so that the column never dries out.The next set of diagrams shows what might happen over time.

Note: These diagrams are very simplified in order to make them easier to draw. In reality, the colours won't separate out into these neat blocks, but will probably be much more spread out - more so the further down the column they get.

Explaining what is happeningThis assumes that you have read the explanation for what happens during thin layer chromatography. If you haven't, follow the very first link at the top of the page and come back to this point afterwards.The blue compound is obviously more polar than the yellow one - it perhaps even has the ability to hydrogen bond. You can tell this because the blue compound doesn't travel through the column very quickly. That means that it must adsorb more strongly to the silica gel or alumina than the yellow one. The less polar yellow one spends more of its time in the solvent and therefore washes through the column much faster.The process of washing a compound through a column using a solvent is known aselution. The solvent is sometimes known as theeluent.

What if you want to collect the blue compound as well?It is going to take ages to wash the blue compound through at the rate it is travelling at the moment! However, there is no reason why you can't change the solvent during elution.Suppose you replace the solvent you have been using by a more polar solvent once the yellow has all been collected. That will have two effects, both of which will speed the blue compound through the column. The polar solvent will compete for space on the silica gel or alumina with the blue compound. Any space temporarily occupied by solvent molecules on the surface of the stationary phase isn't available for blue molecules to stick to and this will tend to keep them moving along in the solvent. There will be a greater attraction between the polar solvent molecules and the polar blue molecules. This will tend to attract any blue molecules sticking to the stationary phase back into solution.The net effect is that with a more polar solvent, the blue compound spends more time in solution, and so moves faster.So why not use this alternative solvent in the first place? The answer is that if both of the compounds in the mixture travel quickly through the column right from the beginning, you probably won't get such a good separation.

What if everything in your mixture is colourless?If you were going to use column chromatography to purify the product of an organic preparation, it is quite likely that the product that you want will be colourless even if one or more of the impurities is coloured. Let's assume the worst case that everything is colourless.How do you know when the substance you want has reached the bottom of the column?There is no quick and easy way of doing this! What you do is collect what comes out of the bottom of the column in a whole series of labelled tubes. How big each sample is will obviously depend on how big the column is - you might collect 1 cm3samples or 5 cm3samples or whatever is appropriate.You can then take a drop from each solution and make a thin layer chromatogram from it. You would place the drop on the base line alongside a drop from a pure sample of the compound that you are making. By doing this repeatedly, you can identify which of your samples collected at the bottom of the column contain the desired product, and only the desired product.Once you know this, you can combine all of the samples which contain your pure product, and then remove the solvent. (How you would separate the solvent from the product isn't directly relevant to this topic and would vary depending on their exact nature - so I'm not even going to attempt a generalisation.)

HIGH PERFORMANCE LIQUID CHROMATOGRAPHY - HPLC

High performance liquid chromatography is a powerful tool in analysis. This page looks at how it is carried out and shows how it uses the same principles as in thin layer chromatography and column chromatography.

Note: It is important to read the introductory page aboutthin layer chromatographybefore you continue with this one - particularly the part about how thin layer chromatography works. High performance liquid chromatography works on the same basic principle. HPLC is essentially an adaptation ofcolumn chromatography- so it might be a good idea to have a (very quick) look at that as well.

Carrying out HPLCIntroductionHigh performance liquid chromatography is basically a highly improved form of column chromatography. Instead of a solvent being allowed to drip through a column under gravity, it is forced through under high pressures of up to 400 atmospheres. That makes it much faster.It also allows you to use a very much smaller particle size for the column packing material which gives a much greater surface area for interactions between the stationary phase and the molecules flowing past it. This allows a much better separation of the components of the mixture.The other major improvement over column chromatography concerns the detection methods which can be used. These methods are highly automated and extremely sensitive.

The column and the solventConfusingly, there are two variants in use in HPLC depending on the relative polarity of the solvent and the stationary phase.Normal phase HPLCThis is essentially just the same as you will already have read about in thin layer chromatography or column chromatography. Although it is described as "normal", it isn't the most commonly used form of HPLC.The column is filled with tiny silica particles, and the solvent is non-polar - hexane, for example. A typical column has an internal diameter of 4.6 mm (and may be less than that), and a length of 150 to 250 mm.Polar compounds in the mixture being passed through the column will stick longer to the polar silica than non-polar compounds will. The non-polar ones will therefore pass more quickly through the column.Reversed phase HPLCIn this case, the column size is the same, but the silica is modified to make it non-polar by attaching long hydrocarbon chains to its surface - typically with either 8 or 18 carbon atoms in them. A polar solvent is used - for example, a mixture of water and an alcohol such as methanol.In this case, there will be a strong attraction between the polar solvent and polar molecules in the mixture being passed through the column. There won't be as much attraction between the hydrocarbon chains attached to the silica (the stationary phase) and the polar molecules in the solution. Polar molecules in the mixture will therefore spend most of their time moving with the solvent.Non-polar compounds in the mixture will tend to form attractions with the hydrocarbon groups because of van der Waals dispersion forces. They will also be less soluble in the solvent because of the need to break hydrogen bonds as they squeeze in between the water or methanol molecules, for example. They therefore spend less time in solution in the solvent and this will slow them down on their way through the column.That means that now it is the polar molecules that will travel through the column more quickly.Reversed phase HPLC is the most commonly used form of HPLC.

Note: I have been a bit careful about how I have described the attractions of the non-polar molecules to the surface of the stationary phase. In particular, I have avoided the use of the word "adsorpion". Adsorption is when a molecule sticks to the surface of a solid. Especially if you had small molecules in your mixture, some could get in between the long C18chains to give what is essentially a solution.You could therefore say that non-polar molecules were more soluble in the hydrocarbon on the surface of the silica than they are in the polar solvent - and so spend more time in this alternative "solvent". Where a solute divides itself between two different solvents because it is more soluble in one than the other, we call itpartition.So is this adsorption or partition? You could argue it both ways! Be prepared to find it described as either.

Looking at the whole processA flow scheme for HPLC

Injection of the sampleInjection of the sample is entirely automated, and you wouldn't be expected to know how this is done at this introductory level. Because of the pressures involved, it isnotthe same as in gas chromatography (if you have already studied that).

Retention timeThe time taken for a particular compound to travel through the column to the detector is known as itsretention time. This time is measured from the time at which the sample is injected to the point at which the display shows a maximum peak height for that compound.Different compounds have different retention times. For a particular compound, the retention time will vary depending on: the pressure used (because that affects the flow rate of the solvent) the nature of the stationary phase (not only what material it is made of, but also particle size) the exact composition of the solvent the temperature of the columnThat means that conditions have to be carefully controlled if you are using retention times as a way of identifying compounds.

The detectorThere are several ways of detecting when a substance has passed through the column. A common method which is easy to explain uses ultra-violet absorption.Many organic compounds absorb UV light of various wavelengths. If you have a beam of UV light shining through the stream of liquid coming out of the column, and a UV detector on the opposite side of the stream, you can get a direct reading of how much of the light is absorbed.The amount of light absorbed will depend on the amount of a particular compound that is passing through the beam at the time.

You might wonder why the solvents used don't absorb UV light. They do! But different compounds absorb most strongly in different parts of the UV spectrum.Methanol, for example, absorbs at wavelengths below 205 nm, and water below 190 nm. If you were using a methanol-water mixture as the solvent, you would therefore have to use a wavelength greater than 205 nm to avoid false readings from the solvent.

Note: If you are interested, there is a whole section aboutUV-visible spectroscopyon the site. This explores the question of the absorption of UV and visible light by organic compounds in some detail.

Interpreting the output from the detectorThe output will be recorded as a series of peaks - each one representing a compound in the mixture passing through the detector and absorbing UV light. As long as you were careful to control the conditions on the column, you could use the retention times to help to identify the compounds present - provided, of course, that you (or somebody else) had already measured them for pure samples of the various compounds under those identical conditions.But you can also use the peaks as a way of measuring the quantities of the compounds present. Let's suppose that you are interested in a particular compound, X.If you injected a solution containing a known amount of pure X into the machine, not only could you record its retention time, but you could also relate the amount of X to the peak that was formed.The area under the peak is proportional to the amount of X which has passed the detector, and this area can be calculated automatically by the computer linked to the display. The area it would measure is shown in green in the (very simplified) diagram.

If the solution of X was less concentrated, the area under the peak would be less - although the retention time will still be the same. For example:

This means that it is possible to calibrate the machine so that it can be used to find how much of a substance is present - even in very small quantities.Be careful, though! If you had two different substances in the mixture (X and Y) could you say anything about their relative amounts?Not if you were using UV absorption as your detection method.

In the diagram, the area under the peak for Y is less than that for X. That may be because there is less Y than X, but it could equally well be because Y absorbs UV light at the wavelength you are using less than X does. There might be large quantities of Y present, but if it only absorbed weakly, it would only give a small peak.

Note: If you want lots more detail about HPLC you could explore the site operated byWaters Corporation- a supplier of HPLC equipment.You will also find a usefulindustry training videowhich talks through the whole process by following this link.

Coupling HPLC to a mass spectrometerThis is where it gets really clever! When the detector is showing a peak, some of what is passing through the detector at that time can be diverted to a mass spectrometer. There it will give a fragmentation pattern which can be compared against a computer database of known patterns. That means that the identity of a huge range of compounds can be found without having to know their retention times.

GAS-LIQUID CHROMATOGRAPHY

Gas-liquid chromatography (often just called gas chromatography) is a powerful tool in analysis. It has all sorts of variations in the way it is done - if you want full details, a Google search on gas chromatography will give you scary amounts of information if you need it! This page just looks in a simple introductory way at how it can be carried out.

Carrying out gas-liquid chromatographyIntroductionAll forms of chromatography involve astationary phaseand amobile phase. In all the other forms of chromatography you will meet at this level, the mobile phase is a liquid. In gas-liquid chromatography, the mobile phase is a gas such as helium and the stationary phase is a high boiling point liquid adsorbed onto a solid.How fast a particular compound travels through the machine will depend on how much of its time is spent moving with the gas as opposed to being attached to the liquid in some way.

A flow scheme for gas-liquid chromatography

Note: You will have to imagine the coiled column in its oven. Drawing a convincing and tidy coil defeated me completely!

Injection of the sampleVery small quantities of the sample that you are trying to analyse are injected into the machine using a small syringe. The syringe needle passes through a thick rubber disc (known as a septum) which reseals itself again when the syringe is pulled out.The injector is contained in an oven whose temperature can be controlled. It is hot enough so that all the sample boils and is carried into the column as a gas by the helium (or other carrier gas).

How the column worksThe packing materialThere are two main types of column in gas-liquid chromatography. One of these is a long thin tube packed with the stationary phase; the other is even thinner and has the stationary phase bonded to its inner surface.To keep things simple, we are just going to look at the packed column.The column is typically made of stainless steel and is between 1 and 4 metres long with an internal diameter of up to 4 mm. It is coiled up so that it will fit into a thermostatically controlled oven.The column is packed with finely grounddiatomaceous earth, which is a very porous rock. This is coated with a high boiling liquid - typically a waxy polymer.The column temperatureThe temperature of the column can be varied from about 50C to 250C. It is cooler than the injector oven, so that some components of the mixture may condense at the beginning of the column.In some cases, as you will see below, the column starts off at a low temperature and then is made steadily hotter under computer control as the analysis proceeds.How separation works on the columnOne of three things might happen to a particular molecule in the mixture injected into the column: It may condense on the stationary phase. It may dissolve in the liquid on the surface of the stationary phase. It may remain in the gas phase.None of these things is necessarily permanent.A compound with a boiling point higher than the temperature of the column will obviously tend to condense at the start of the column. However, some of it will evaporate again in the same way that water evaporates on a warm day - even though the temperature is well below 100C. The chances are that it will then condense again a little further along the column.Similarly, some molecules may dissolve in the liquid stationary phase. Some compounds will be more soluble in the liquid than others. The more soluble ones will spend more of their time absorbed into the stationary phase; the less soluble ones will spend more of their time in the gas.The process where a substance divides itself between two immiscible solvents because it is more soluble in one than the other is known aspartition. Now, you might reasonably argue that a gas such as helium can't really be described as a "solvent". But the termpartitionis still used in gas-liquid chromatography.You can say that a substance partitions itself between the liquid stationary phase and the gas. Any molecule in the substance spends some of its time dissolved in the liquid and some of its time carried along with the gas.Retention timeThe time taken for a particular compound to travel through the column to the detector is known as itsretention time. This time is measured from the time at which the sample is injected to the point at which the display shows a maximum peak height for that compound.Different compounds have different retention times. For a particular compound, the retention time will vary depending on: the boiling point of the compound. A compound which boils at a temperature higher than the column temperature is going to spend nearly all of its time condensed as a liquid at the beginning of the column. So high boiling point means a long retention time. the solubility in the liquid phase. The more soluble a compound is in the liquid phase, the less time it will spend being carried along by the gas. High solubility in the liquid phase means a high retention time. the temperature of the column. A higher temperature will tend to excite molecules into the gas phase - either because they evaporate more readily, or because they are so energetic that the attractions of the liquid no longer hold them. A high column temperature shortens retention times for everything in the column.For a given sample and column, there isn't much you can do about the boiling points of the compounds or their solubility in the liquid phase - but you do have control over the temperature.The lower the temperature of the column, the better the separation you will get - but it could take averylong time to get the compounds through which are condensing at the beginning of the column!On the other hand, using a high temperature, everything will pass through the column much more quickly - but less well separated out. If everything passed through in a very short time, there isn't going to be much space between their peaks on the chromatogram.The answer is to start with the column relatively cool, and then gradually and very regularly increase the temperature.At the beginning, compounds which spend most of their time in the gas phase will pass quickly through the column and be detected. Increasing the temperature a bit will encourage the slightly "stickier" compounds through. Increasing the temperature still more will force the very "sticky" molecules off the stationary phase and through the column.

The detectorThere are several different types of detector in use. The flame ionisation detector described below is commonly used and is easier to describe and explain than the alternatives.A flame ionisation detectorIn terms of reaction mechanisms, the burning of an organic compound is very complicated. During the process, small amounts of ions and electrons are produced in the flame. The presence of these can be detected.The whole detector is enclosed in its own oven which is hotter than the column temperature. That stops anything condensing in the detector.

Note: This is simplified for clarity. There obviously has to be some way of lighting the flame. This is done with an electrically heated coil, but including it clutters the diagram.

If there is nothing organic coming through from the column, you just have a flame of hydrogen burning in air. Now suppose that one of the compounds in the mixture you are analysing starts to come through.As it burns, it will produce small amounts of ions and electrons in the flame. The positive ions will be attracted to the cylindrical cathode. Negative ions and electrons will be attracted towards the jet itself which is the anode.This is much the same as what happens during normal electrolysis.At the cathode, the positive ions will pick up electrons from the cathode and be neutralised. At the anode, any electrons in the flame will transfer to the positive electrode; and negative ions will give their electrons to the electrode and be neutralised.This loss of electrons from one electrode and gain at the other will result in a flow of electrons in the external circuit from the anode to the cathode. In other words, you get an electric current.The current won't be very big, but it can be amplified. The more of the organic compound there is in the flame, the more ions will be produced, and so the higher the current will be. As a reasonable approximation, especially if you are talking about similar compounds, the current you measure is proportional to the amount of compound in the flame.Disadvantages of the flame ionisation detectorThe main disadvantage is that it destroys everything coming out of the column as it detects it. If you wanted to send the product to a mass spectrometer, for example, for further analysis, you couldn't use a flame ionisation detector.

Interpreting the output from the detectorThe output will be recorded as a series of peaks - each one representing a compound in the mixture passing through the detector. As long as you were careful to control the conditions on the column, you could use the retention times to help to identify the compounds present - provided, of course, that you (or somebody else) had already measured them for pure samples of the various compounds under those identical conditions.But you can also use the peaks as a way of measuring the relative quantities of the compounds present. This is only accurate if you are analysing mixtures of similar compounds - for example, of similar hydrocarbons.The areas under the peaks are proportional to the amount of each compound which has passed the detector, and these areas can be calculated automatically by the computer linked to the display. The areas it would measure are shown in green in the (very simplified) diagram.

Note that it isn't the peak height that matters, but the total area under the peak. In this particular example, the left-hand peak is both tallest and has the greatest area. That isn't necessarily always so.There might be a lot of one compound present, but it might emerge from the column in relatively small amounts over quite a long time. Measuring the area rather than the peak height allows for this.

Coupling a gas chromatogram to a mass spectrometerThis can't be done with a flame ionisation detector which destroys everything passing through it. Assuming you are using a non-destructive detector . . .When the detector is showing a peak, some of what is passing through the detector at that time can be diverted to a mass spectrometer. There it will give a fragmentation pattern which can be compared against a computer database of known patterns. That means that the identity of a huge range of compounds can be found without having to know their retention times.

PAPER CHROMATOGRAPHY

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

Carrying out paper chromatographyBackgroundChromatography is used to separate mixtures of substances into their components. All forms of chromatography work on the same principle.They all have astationary phase(a solid, or a liquid supported on a solid) and amobile 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 chromatogramYou 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.

Note: The chromatography paper will in fact be pure white - not pale grey. I'm forced to show it as off-white because of the way I construct the diagrams. Anything I draw as pure white allows the background colour of the page to show through.

The paper is suspended in a container with a shallow layer of a suitable solvent or mixture of solvents in it. It is important that the solvent level is below the line with the spots on it. The next diagram doesn't show details of how the paper is suspended because there are too many possible ways of doing it and it clutters the diagram. Sometimes the paper is just coiled into a loose cylinder and fastened with paper clips top and bottom. The cylinder then just stands in the bottom of the container.The reason for covering the container is to make sure that the atmosphere in the beaker is saturated with solvent vapour. Saturating the atmosphere in the beaker with vapour stops the solvent from evaporating as it rises up the paper.

As the solvent slowly travels up the paper, the different components of the ink mixtures travel at different rates and the mixtures are separated into different coloured spots.The diagram shows what the plate might look like after the solvent has moved almost to the top.

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

RfvaluesSome compounds in a mixture travel almost as far as the solvent does; some stay much closer to the base line. The distance travelled relative to the solvent is a constant for a particular compound as long as you keep everything else constant - the type of paper and the exact composition of the solvent, for example.The distance travelled relative to the solvent is called the Rfvalue. For each compound it can be worked out using the formula:

For example, if one component of a mixture travelled 9.6 cm from the base line while the solvent had travelled 12.0 cm, then the Rfvalue for that component is:

In the example we looked at with the various pens, it wasn't necessary to measure Rfvalues because you are making a direct comparison just by looking at the chromatogram.You are making the assumption that if you have two spots in the final chromatogram which are the same colour and have travelled the same distance up the paper, they are most likely the same compound. It isn't necessarily true of course - you could have two similarly coloured compounds with very similar Rfvalues. We'll look at how you can get around that problem further down the page.

What if the substances you are interested in are colourless?In some cases, it may be possible to make the spots visible by reacting them with something which produces a coloured product. A good example of this is in chromatograms produced from amino acid mixtures.Suppose you had a mixture of amino acids and wanted to find out which particular amino acids the mixture contained. For simplicity we'll assume that you know the mixture can only possibly contain five of the common amino acids.A small drop of a solution of the mixture is placed on the base line of the paper, and similar small spots of the known amino acids are placed alongside it. The paper is then stood in a suitable solvent and left to develop as before. In the diagram, the mixture is M, and the known amino acids are labelled 1 to 5.The position of the solvent front is marked in pencil and the chromatogram is allowed to dry and is then sprayed with a solution ofninhydrin. Ninhydrin reacts with amino acids to give coloured compounds, mainly brown or purple.The left-hand diagram shows the paper after the solvent front has almost reached the top. The spots are still invisible. The second diagram shows what it might look like after spraying with ninhydrin.

There is no need to measure the Rfvalues because you can easily compare the spots in the mixture with those of the known amino acids - both from their positions and their colours.In this example, the mixture contains the amino acids labelled as 1, 4 and 5.And what if the mixture contained amino acids other than the ones we have used for comparison? There would be spots in the mixture which didn't match those from the known amino acids. You would have to re-run the experiment using other amino acids for comparison.

Two way paper chromatographyTwo way paper chromatography gets around the problem of separating out substances which have very similar Rfvalues.I'm going to go back to talking about coloured compounds because it is much easier to see what is happening. You can perfectly well do this with colourless compounds - but you have to use quite a lot of imagination in the explanation of what is going on!This time a chromatogram is made starting from a single spot of mixture placed towards one end of the base line. It is stood in a solvent as before and left until the solvent front gets close to the top of the paper.In the diagram, the position of the solvent front is marked in pencil before the paper dries out. This is labelled as SF1 - the solvent front for the first solvent. We shall be using two different solvents.

If you look closely, you may be able to see that the large central spot in the chromatogram is partly blue and partly green. Two dyes in the mixture have almost the same Rfvalues. They could equally well, of course, both have been the same colour - in which case you couldn't tell whether there was one or more dye present in that spot.What you do now is to wait for the paper to dry out completely, and then rotate it through 90, and develop the chromatogram again in a different solvent.It is very unlikely that the two confusing spots will have the same Rfvalues in the second solvent as well as the first, and so the spots will move by a different amount.The next diagram shows what might happen to the various spots on the original chromatogram. The position of the second solvent front is also marked.

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

Two way chromatography has completely separated out the mixture into four distinct spots.If you want to identify the spots in the mixture, you obviously can't do it with comparison substances on the same chromatogram as we looked at earlier with the pens or amino acids examples. You would end up with a meaningless mess of spots.You can, though, work out the Rfvalues for each of the spots in both solvents, and then compare these with values that you have measured for known compounds under exactly the same conditions.

How does paper chromatography work?Although paper chromatography is simple to do, it is quite difficult to explain compared with thin layer chromatography. The explanation depends to some extent on what sort of solvent you are using, and many sources gloss over the problem completely. If you haven't already done so, it would be helpful if you could read the explanation for how thin layer chromatography works (link below). That will save me a lot of repetition, and I can concentrate on the problems.

Note: You will find the explanation forhow thin layer chromatography worksby following this link.

The essential structure of paperPaper is made of cellulose fibres, and cellulose is a polymer of the simple sugar, glucose.

The key point about cellulose is that the polymer chains have -OH groups sticking out all around them. To that extent, it presents the same sort of surface as silica gel or alumina in thin layer chromatography.It would be tempting to try to explain paper chromatography in terms of the way that different compounds are adsorbed to different extents on to the paper surface. In other words, it would be nice to be able to use the same explanation for both thin layer and paper chromatography. Unfortunately, it is more complicated than that!The complication arises because the cellulose fibres attract water vapour from the atmosphere as well as any water that was present when the paper was made. You can therefore think of paper as being cellulose fibres with a very thin layer of water molecules bound to the surface.It is the interaction with this water which is the most important effect during paper chromatography.

Paper chromatography using a non-polar solventSuppose you use a non-polar solvent such as hexane to develop your chromatogram.Non-polar molecules in the mixture that you are trying to separate will have little attraction for the water molecules attached to the cellulose, and so will spend most of their time dissolved in the moving solvent. Molecules like this will therefore travel a long way up the paper carried by the solvent. They will have relatively high Rfvalues.On the other hand, polar moleculeswillhave a high attraction for the water molecules and much less for the non-polar solvent. They will therefore tend to dissolve in the thin layer of water around the cellulose fibres much more than in the moving solvent.Because they spend more time dissolved in the stationary phase and less time in the mobile phase, they aren't going to travel very fast up the paper.The tendency for a compound to divide its time between two immiscible solvents (solvents such as hexane and water which won't mix) is known aspartition. Paper chromatography using a non-polar solvent is therefore a type ofpartition chromatography.

Paper chromatography using a water and other polar solventsA moment's thought will tell you that partition can't be the explanation if you are using water as the solvent for your mixture. If you have water as the mobile phase and the water bound on to the cellulose as the stationary phase, there can't be any meaningful difference between the amount of time a substance spends in solution in either of them. All substances should be equally soluble (or equally insoluble) in both.And yet the first chromatograms that you made were probably of inks using water as your solvent.If water works as the mobile phase as well being the stationary phase, there has to be some quite different mechanism at work - and that must be equally true for other polar solvents like the alcohols, for example. Partition only happens between solvents which don't mix with each other. Polar solvents like the small alcohols do mix with water.In researching this topic, I haven't found any easy explanation for what happens in these cases. Most sources ignore the problem altogether and just quote the partition explanation without making any allowance for the type of solvent you are using. Other sources quote mechanisms which have so many strands to them that they are far too complicated for this introductory level. I'm therefore not taking this any further - you shouldn't need to worry about this at UK A level, or its various equivalents.

Note: If I have missed something obvious in my research and you know of a straightforward explanation (worth about 1 or 2 marks in an exam) for what happens with water and other polar solvents,