Chromatography

Transcription

Chromatography

A Science Enrichment Programme for Secondary 3-4 Students : Teaching and Learning Resources
and its Application
in Chemical Analysis and Separation
by Dr. Wing-Fat CHAN,
Dr. Kin-Wah MAK
Chemistry
Chromatography
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Chromatography and its Application in Chemical Analysis and Separation
I. Summary Notes
Subject Area
Chemistry
Level
SS1 to SS2
Keywords
Chromatography, chemical analysis, sample separation and
purification
Prerequisites
You should be familiar with some basic chemistry concepts
such as the properties of matter, chemical bonding, various
types of intermolecular forces, and the chemical structures
of compounds. You should also have some basic knowledge
of organic chemistry, the characteristic properties of some
simple organic compounds and some experience in carrying
out practical chemical work.
Introduction
We introduce the basic concepts of chromatography and its application in chemical
analysis and sample separation. Chromatography is a very useful and versatile
practical technique for analysing and separating chemical mixtures. It is useful
for finding out the identities and the amounts of the components that present in a
mixture, and for isolating the components from the mixture in pure forms. It has very
wide application and has been developed into many different forms, some of which
are rather simple and can easily be carried out in most secondary school science
laboratories with very simple apparatus. Others are more sophisticated and require
expensive and delicate modern instruments. As this technique has many different
forms, it is usually covered in several university courses at both elementary and
advanced levels. We focus on the basic principles of chromatography and introduce
some simple laboratory activities that can be conveniently carried out with commonly
used apparatus. Brief introductions to some modern instrumental techniques are
included to allow you to appreciate how science and technology are applied to areas
closely connected to modern life.
Chemistry
Learning Outcomes
Through laboratory activities, you will:
•
•
•
•
•
Acquire a basic understanding of the principles of chromatography;
Recognise the applications of chromatographic techniques in chemical analysis
and the separation of compound mixtures;
Acquire practical experience in performing simple chromatographic analysis and
separation;
Recognise the concepts of qualitative analysis and quantitative analysis; and
Appreciate the applications of modern chromatographic techniques in monitoring
the quality and safety of consumer products.
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II. Learning Materials
A. Principles of Chromatography
Before discussing the principles of chromatography in detail, the following laboratory
activities can be conducted as an introduction.
Activity 1: Analysing the composition of dyes in pen inks by paper chromatography
This is a very simple experiment that you may have already carried out in science
lessons. You may skip this part if you have recently conducted a very similar
experiment, but should do so again now if that experiment was a long time ago.
Procedure
1. Cut a few rectangular pieces of filter paper (2.5 cm x 5 cm).
2. Take a 100 mL beaker and add 2-3 mL of ethyl acetate (ethyl ethanoate) to it.
Cover the beaker with a watch glass.
3. Take a piece of the prepared rectangular filter paper and use a pencil to draw
a faint straight line about 5 mm away from and parallel to one of the narrower
edges.
4. Collect several fine felt-tip pens of different colours from your classmates. Lightly
spot the different colour inks along the faint pencil line, one spot for each colour.
Each spot should be about 4-5 mm apart.
5. Put the piece of filter paper spotted with coloured inks vertically into the beaker,
with the line of ink spots as the lower edge. Cover the beaker with the watch
glass.
6. The solvent (ethyl acetate) will move slowly upward along the filter paper. Take
the filter paper out of the beaker when the solvent front line reaches the top of
the filter paper.
7. Place the filter paper on the bench and allow it to dry. Describe what you see on
the filter paper.
The simplest form of chromatography – paper chromatography
What you have done in this activity is a very simple form of chromatography – paper
chromatography. It allows you to analyse what colour dyes were used to make the
various colour pen inks. Some pen inks may contain only a single dye while some
may be made by mixing several dyes of different colours. The filter paper served
as the medium for separating the dyes. This technique allows you to indentify the
components that present in a mixture, which is a simple example of qualitative
analysis.
Chemistry
You will see that the coloured ink spots have moved up the filter paper, carried by
the solvent, and now appear in different positions. Some spots may have moved
further up than others. Some coloured inks may have remained as single spots,
while others may have separated into several coloured spots at different positions.
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Principles of paper chromatography
In chromatography two phases are involved – the stationary phase and the mobile
phase. In the previous example the filter paper was the stationary phase and the
solvent (ethyl acetate) was the mobile phase. In a separation, the sample mixture
(ink samples) is applied (spotted) onto the stationary phase (the filter paper). The
mobile phase (solvent - ethyl acetate) is allowed to move along the stationary phase,
carrying the components in the sample mixture along with it. Hence, the coloured
dye spots moved upward along the filter paper as the solvent moved up.
Chemistry
Different components in a mixture have different chemical structures and different
properties. They thus have different degrees of interaction (attraction) with the
mobile and stationary phases. Those components that are attracted more strongly
to the stationary phase (or less strongly to the mobile phase) move slowly along the
stationary phase, and appear at lower positions on the paper. Components that are
attracted less strongly to the stationary phase (or more strongly to the mobile phase)
move more quickly, and appear at higher positions. Put simply, in chromatography
the different components in a mixture are separated by the differences of their
interactions with the mobile and stationary phases.
Figure 1 Chromatography
Stationary phase
Paper is made mainly of cellulose, a natural polymer comprising a larger number
of glucose molecules linked together. It is a very polar substance because, as
you can see in its chemical structure shown below, it has many hydroxyl (─OH)
groups attached along the polymer chain. These groups can interact strongly with
the components to be separated by hydrogen bonding and polar-polar interactions.
Compounds that act in this way are firmly attracted to the stationary phase, and thus
move very slowly along the stationary phase as the mobile phase passes.
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Figure 2 Structure of cellulose
Activity 2: Chromatography with different stationary phases
The mobile phase and the stationary phase are the two most important things
that control the effectiveness of a separation. In this activity you will perform
chromatography of pen inks by using different types of stationary phases, and see
how the results are affected.
Procedure
1. Obtain a thin-layer chromatographic
plate from your teacher. The plate
should be cut to the size of 2.5 cm
x 5 cm. Spot the plate with different
coloured pen inks as what you did in
Step 4 of the previous activity.
2. Put the plate into a beaker that
contains a shallow layer of ethyl
acetate. Cover the beaker and wait
until the solvent rises to the top of
the plate.
3. Take the plate out of the beaker, and
compare the result with that obtained
with the piece of filter paper.
Figure 3 TLC in a beaker that contains
ethyl acetate.
Silica gel
Figure 4 Two sides of a TLC plate
Aluminium backing plate
Chemistry
The plate that you have just used is called a TLC plate (TLC = thin layer
chromatography). The plate has two sides; one is white and one is silvery. The white
side is actually a thin layer of adsorbent that serves as the stationary phase. This
absorbent is silica gel, the same substance that you will find inside the small packs
of desiccant in food packages. The silvery side is an aluminium sheet, providing the
surface for the absorbent to be coated on, and rigidity for the plate.
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This technique is known as thin layer chromatography because the stationary phase
is in the form of a thin layer. As it is very easy to carry out and effective for separating
many different kinds of compounds, this technique is very often used in chemical
laboratories for the quick analysis of compositions of mixtures.
? Question and Discussion
Compare the results that obtained with TLC plate and paper. Try to draw a conclusion
and explain which stationary phase material is more polar and thus interacts more
strongly with polar compounds.
Activity 3: Paper chromatography / TLC with different solvents
In this activity, you will repeat the paper chromatography and TLC of pen inks using
different solvents to see how the results will be affected.
Procedure
1. Spot several pieces of filter paper and TLC plates with different coloured pen inks
as you did in Step 4 of Activity 1.
2. Run the paper chromatography and TLC using the following solvents, one by one.
• n-Hexane
• Ethanol
3. Compare the results you obtained with the different solvents (mobile phases).
Which of the solvents give better (or worse) separation than ethyl acetate?
From the results obtained, you will be able to see that some solvents can move
the sample spots to higher positions than others. Some solvents can give good
separation for the dye spots but some cannot. One can adjust the performance of a
chromatographic separation by choosing a suitable mobile phase (solvent). In paper
chromatography and TLC, the polarity of the solvent used and the solubility of the
sample in the solvent are both important. In general, polar solvents interact better
with the polar components present in the sample and can bring the components to
higher positions. Non-polar solvents interact weakly with the polar components, and
cannot effectively compete with the strong interactions between the components
and the stationary phase, which makes the polar components move much more
slowly.
Chemistry
For effective separation, the sample mixture should be soluble in the mobile phase.
Otherwise, the mobile phase cannot bring along the components as it moves up the
stationary phase and the sample will remain at the starting point.
1. Compare the results you obtained in this activity with those you obtained in
Activity 1. Try to arrange the solvents in an increasing order of polarity.
2. The following solvents are listed according to an increasing order of polarity:
n-Hexane, ethyl acetate and ethanol.
Is this order the same as you deduced in Question 1? Try to propose an
explanation for any discrepancy that you have observed.
As a simple rule of thumb to help you decide on the best solvent to use as the mobile
phase for a particular chromatographic separation, solvent selection depends on the
polarity of the samples to be separated. Polar solvents (e.g. ethanol) are required for
separating polar substances, while non-polar solvents (e.g. n-hexane) are suitable
for separating non-polar mixtures.
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Activity 4: Analysis of some pain-killing drugs using thin layer chromatography
In this activity, you will use thin
layer chromatography to find out the
ingredients present in some common
pain-killing drugs. The following
figure shows the chemical structures
and names of the common active
ingredients present in the drugs.
Figure 5 Pain-killing drugs
Figure 6(a) Aspirin
Figure 6(b) Acetaminophen
Figure 6(c) Caffeine
Preparation of standard solutions for reference
(1) Aspirin:
Dissolve about 0.1 g of aspirin in 3 mL of methanol in a 5-mL
vial (or a small test tube)
(2) Caffeine:
Dissolve about 0.02 g of caffeine in 3 mL of methanol in a 5-mL
vial (or a small test tube)
(3) Acetaminophen: Dissolve about 0.02 g of acetaminophen in 3 mL of methanol in
a 5-mL vial (or a small test tube)
Thin layer chromatographic analysis
Two solvents are provided for running TLC for the standard solutions:
n-hexane and ethanol.
Did you encounter any difficulty in carrying out this activity? Could you tell the
positions of the sample spots on the TLC plates?
How can colourless mixtures be analysed?
In the chromatographic experiments with pen inks, the results can be clearly seen
on the paper or TLC plate because the sample spots are coloured. However, only
a small fraction of the organic compounds that exist can absorb visible light and be
seen directly. In other words, most organic compounds are colourless. How could
you find out the positions of colourless sample spots against a white background?
Chemistry
1. Obtain a 3 cm x 8 cm TLC plate, and apply the three standard solutions to the
plate using capillary tubes.
2. Run the TLC using n-hexane as the mobile phase.
3. Repeat the TLC analysis with the other solvent, and find out which solvent gives
the best separation for the three standard compounds.
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Things to do
Now place the TLC plates that have
the drug samples spotted on them
under an ultraviolet lamp. You should
use a lamp that gives UV radiation at
254 nm. What do you see?
Note: Be very careful when you are
using the UV lamp, and consult your
teacher before turning it on. UV light is
high energy radiation and can cause
serious burns to your skin and eyes.
Never look at a UV lamp directly.
Figure 7 Ultraviolet lamp
Visualising colourless compounds on TLC plates with UV light
Now you should see the TLC plates glow brightly under the UV light, and some small
dark spots on the plates. Why do the plates grow brightly? What are those spots?
Analyse the three standard solutions of common active ingredients present in drugs
with TLC plate and n-hexane:
Aspirin
Aspirin
Acetaminophen
Acetaminophen
Caffeine
Caffeine
Chemistry
Analyse the three standard solutions of common active ingredients present in drugs
with TLC plate and ethanol:
Acetaminophen
Acetaminophen
Caffeine
Aspirin
Aspirin Caffeine
Figure 8 TLC Plate under ultraviolet lamp
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What you have done is a typical visualisation technique for colourless samples on
TLC. Although most organic compounds are colourless, some of them do absorb
UV. The plates glow under the UV lamp because they are coated with a fluorescent
dye. The dye absorbs UV, transforms it into visible light and then emits that light.
This explains why the plates grow brightly when irradiated with UV. If organic
compounds that can absorb UV are present on a plate, the UV will be absorbed by
those compounds and will not reach the fluorescent dye, causing some dark spots
to form on the plate. Put simply, the dark spots mark the locations of the samples.
Mark out the locations of the sample spots on the plates with a pencil. Decide which
solvent gives the best separation for the three standard compounds. Use the chosen
solvent system for analysing the sample solutions.
Preparation of sample solutions for comparison
1. You will be given two unknown drug sample tablets for analysis. Your teacher
may have crushed the two tablets into powder before handing them to you, so
you cannot tell the brands of the tablets from the marks impressed on them, to
make the work more challenging.
2. For each sample, grind the tablet into powder and dissolve about 0.1 g of the
sample powder into 3 mL of methanol in a 5 mL vial (or in a small test tube).
3. Cap the vials and shake them gently to make the powder dissolve into the solvent.
4. Place the vial on the bench and wait for the residual solid to settle down. Spot
the clear solutions onto a TLC plate and run the TLC. Compare the TLC results
with those obtained with standard solutions to identify the compounds present in
the samples.
5. Try to identify the brand names of the two given samples by comparing the
experimental results with the ingredients listed on the drug packages.
You have just finished a simple experiment in qualitative analysis. You first performed
TLC analysis with standard compounds to find out the best mobile phase for carrying
out the analysis, and then identified the positions of the sample spots on the TLC
plate. You would then use this set of data as a reference for identifying the active
ingredients in the unknown drug samples. It is important to note that to ensure
the reliability of your results, all TLC runs should be performed using the same
experimental conditions (i.e. the same types of stationary and mobile phases, the
same size of TLC plate, the TLC run at the same temperature, etc.)
Chemistry
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B. Column Chromatography
Thin layer chromatography is a very good and convenient way of analysing the
composition of a mixture. One advantage is that only a very tiny amount of sample
is required to carry out an analysis. This is particularly important when you have
only a very limited amount of sample, or the concentrations of the components in the
sample solution are quite low. However, the method is not as practical if you want
to separate a mixture and collect the separated components. For example, if you
have one gram of sample and you want to separate it, you cannot do that with TLC.
TLC is basically a technique that allows you to carry out analytical work. You need
a preparative technique to perform tasks at a preparative scale.
Chemistry
Shown in the image on the right is a
cylindrical glass column with its inside
packed with silica gel, acting as the
stationary phase. To separate a mixture,
the sample solution is first applied to the top
of the stationary phase. A suitable solvent
(the mobile phase) is then added to the
column and allowed to pass through the
stationary phase slowly and continuously,
eluting the sample mixture down the
column. As each component is retained to
a different degree by the stationary phase,
they would pass through the column at
different speeds and come out from the
bottom of the column at different times.
The separated, components are then
collected in fractions separately.
Thin layer chromatography and paper
chromatography are very convenient
techniques for analytical work because
they can work with only micrograms (mg)
of mixtures. A very small drop of a diluted
sample solution is sufficient for carrying
out an analysis. Column chromatography,
in contrast, is very suitable for preparative
work. Up to 10 g of sample mixture can be
separated easily, depending on the size of
the column and the amount of absorbent
(stationary phase) used.
Figure 9 Column with silica gel
Activity 5: Isolation of plant pigments from spinach by column chromatography
Many plants exhibit bright and attractive colours because they contain certain plant
pigments. Plant leaves, for example, usually contain two major types of pigments
known as chlorophylls and carotenoids. Chlorophylls are green and are responsible
for carrying out photosynthesis. They absorb light and transform the light energy
into chemical energy. Carotenoids are plant pigments that are usually yellowish,
orange or red. Two of the most commonly known carotenoids are carotenes (which
make carrots look orange) and lycopene (which make tomatoes red).
Spinach leaves contain chlorophylls and b-carotene (one type of carotene, which
is the precursor to vitamin A), and small amounts of other pigments. The two
major pigments can be extracted and isolated from spinach leaves by column
chromatography.
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The isolation of chlorophylls and b-carotene from spinach leaves is a very common
experiment for senior secondary or elementary undergraduate classes. A large
number of well-written experimental procedures are downloadable from the web.
Here is an example:
http://www.uwlax.edu/faculty/koster/Spinach.htm.
In this experimental procedure, the stationary phase used is silica gel, a polar
substance that retains polar compounds better than non-polar compounds. The mobile
phases used are mixtures of n-hexane with varying amounts of propanone (acetone)
added. The purpose of mixing n-hexane with different amounts of propanone is to
adjust the mobile phase so that it has the desired polarity. Propanone is a more
polar solvent than n-hexane. Increasing the amount of propanone in the solvent
mixture can increase the solvent’s polarity.
1. Shown below are the chemical structures of chlorophyll a (a common chlorophyll)
and b-carotene. Can you identify which compound is more polar by looking at
their chemical structures?
Figure 10 (a) Chlorophyll a
Chemistry
Figure 10 (b) b -carotene
b-Carotene is less polar than chlorophyll, so it will be retained less strongly by
the stationary phase in the column, and will move faster along the column. A less
polar solvent is needed to elute b-carotene out. During the early stage of the
separation process, the column is eluted with hexane and then 90/10 mixture of
hexane and propanone to bring the less polar b-carotene out. The 90/10 mixture
of hexane and propanone is polar enough to elute b-carotene, but not the more
polar chlorophylls.
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When the yellow fraction of b-carotene is eluted out and collected, the mobile
phase is changed to 70/30 mixture of hexane and propanone, and then to pure
propanone, to carry the polar chlorophylls down the column. This technique is
called stepwise elution or gradient elution. The polarity of the mobile phase is
increased gradually to decrease the retention of the slow-moving components.
Column chromatography for colourless mixtures
The isolation of chlorophylls and b-carotene from spinach leaves is easy to carry
out because the desired fractions are pigments and are thus coloured substances.
When the fractions are moving down along the column, their positions can be easily
seen. It would be a more challenging exercise if the fractions were colourless, which
would mean that you could not tell their positions by simply looking at the column
and would not know when to start and stop collecting the desired fractions. That kind
of separation is not uncommon, because most organic compounds are colourless.
The problem can be solved by collecting the colourless solution that eluted out
from the column in several fractions of a fixed volume (e.g., 5 mL for each fraction),
and then analysing the composition of each fraction by thin layer chromatography.
By knowing what compounds are present in each fraction, you would know which
fractions should be collected to obtain the desired compounds.
Application of chromatography in modern instrumental analysis
Although thin layer chromatography is a very good and convenient method for
analysing the composition of a mixture, it is not applicable if the concentration of
the analyte is very low, and it is not a quantitative analytical method. TLC can tell
what compounds are present in a mixture, but cannot accurately tell the amount
of compounds in the mixture. Based on the principles of chromatography, several
kinds of modern analytical instruments have been developed to carry out delicate
qualitative and quantitative chemical analysis.
C. Gas Chromatography (GC)
Chemistry
Shown below is a photo of a typical gas chromatography system (a Gas
Chromatograph), and a schematic diagram showing its components.
Figure 11 Gas Chromatograph
Figure 12 Schematic diagram of Gas Chromatograph
This technique is known as gas chromatography because the mobile phase used
is a gas, usually an unreactive gas such as helium or nitrogen. The gas carries
the sample mixture through a long and narrow column in which the separation is
achieved. Inside the column, a special viscous liquid, serving as the stationary
phase, is coated on the inner surface. The stronger the interaction between the
sample compound and the liquid stationary phase, the stronger the retention, and a
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longer time is needed for the carrier gas to carry the compound completely through
the column. A detector is installed at the end of the column to detect the compounds
that emerge from the end. The time taken for a compound to reach the column’s end
is called the retention time. Compounds are identified by their retention times in a
GC run.
The column is housed in a programmable oven that provides precise temperature
control. The retention time of a compound is affected (and can be controlled) by
the column temperature; increasing the temperature can weaken the interaction
between the sample compound and the stationary phase, allowing the sample to
move faster along the column and thus creating a shorter retention time.
The data obtained from a gas chromatograph is often presented in the form of
a graph, as shown below. The graph, known as a chromatogram, has its x-axis
representing the time that passed after the sample is injected into the column, and
y-axis representing the signal intensity recorded by the detector. The components
present in the sample appear as peaks on the chromatogram. The time marked
by the apex of a peak that corresponds to a particular compound is the retention
time of this compound. The area of the peak shows the amount of that compound
present in the mixture; a concentrated sample gives a larger peak area. This allows
quantitative analysis to be carried out.
Signal
Benzophenone
Impurity peaks
from solvent
Caffeine
Retention time (min)
Gas chromatography is a very useful technique for analysing compounds that are
relatively volatile and thermally stable. If the compounds are non-volatile, they will
stay permanently inside the column. If the compounds are not thermally stable (e.g.,
sucrose), they will decompose inside the column. In both cases, the delicate and
expensive column would be permanently damaged.
Shown below are chromatograms of:
(a) pure trans-stilbene;
(b) pure cis-stilbene;
(c) pure chalcone; and
(d) a mixture containing trans-stilbene, cis-stilbene, chalcone and an unknown compound.
Chemistry
Figure 13 A gas chromatogram obtained for analysing the caffeine content in a soft drink.
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Peak Area
7.848
5285989
Figure 14(a)
Chromatograms of pure trans-stilbene
6.971
Chemistry
Peak Area
9.021
2846255
Figure 14(c)
Chromatograms of pure chalcone
826178
Figure 14(b)
Chromatograms of pure cis-stilbene
Peak R
Peak Area
Retention
(min)
etention time (min)
Peak
Area
Peak Area
A
6.967
948313
B
7.830
1257767
C
8.241
821377
D
9.008
903144
Figure 14(d)
Chromatograms of a mixture contains
trans-stilbene, cis-stilbene and chalcone,
and also an unknown compound
* trans-stilbene, cis-stilbene and chalcone are simple organic compounds, and their
chemical structures are shown below.
Figure 15(a) trans-stilbene
Figure 15(b) cis-stilbene
Figure 15(c) chalcone
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1. From the chromatograms (a), (b) and (c), find out the retention time of transstilbene, cis-stilbene and chalcone, respectively.
2. Which compound interacts most strongly with the stationary phase of the column?
Which compound interacts least strongly?
3. Identify the peaks that correspond to trans-stilbene, cis-stilbene and chalcone in
the chromatogram (d).
4. Compare the areas of the peaks in chromatogram (d). Which compound apparently
has the highest concentration?
D. High Performance Liquid Chromatography (HPLC)
As mentioned previously, GC is not suitable
for analysing samples that are non-volatile
or easily decomposed by heat. These kinds
of samples can be analysed by another
technique known as High Performance
Liquid Chromatography (HPLC). Shown
below is a photo of a typical HPLC system
and a schematic diagram showing its major
parts. The main difference between GC
and HPLC is the mobile and stationary
phases used. In principle, HPLC is similar
to column chromatography. Solvents, used
as the mobile phase, are pressurised with a
pump and forced to pass through a column
densely packed with the solid stationary
phase. As in GC, a detector is connected
to the end of the column to electronically
detect the emergence of compounds. The
data obtained from a HPLC system are also
presented in the form of chromatograms,
showing the signal intensity recorded by the
detector against the time that passed after
sample injection.
Figure 16 HPLC system
Chemistry
Figure 17 Schematic diagram of a HPLC system
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mAU
Retention time = 12.941 min
Peak area = 1992
175
150
125
100
75
50
25
0
0
2
4
6
8
10
12
14
min
Figure 18 Sample HPLC chromatogram
HPLC gives much better separation performance than column chromatography
(which is why it is called high performance liquid chromatography) because the
particle sizes of the absorbents used for HPLC are much smaller than those used
for column chromatography. As the particle sizes are reduced, the surface area of
the stationary phase for interacting with samples is greatly increased. Hence, HPLC
can achieve better separation with a shorter column than column chromatography.
Analysing a chromatogram – quantitative analysis
Shown below are chromatograms obtained with a HPLC system. The chromatogram
on the top was obtained from a 10 ppm* solution of cinnamaldehyde, and the
chromatogram at the bottom was obtained from a 10 ppm solution of cinnamic acid.
Cinnamic acid (t R** = 11.3 min) has a shorter retention time than cinnamaldehyde (t R
= 12.9 min), meaning that cinnamaldehyde is retained more strongly than cinnamic
acid by the HPLC column.
Chemistry
Figure 19(a) Cinnamaldehyde
Figure 19(b) Cinnamic acid
mAU
Retention time = 12.941
Peak area = 968.5
175
150
125
100
75
50
25
0
0
2
4
6
8
Figure 20 Chromatogram of 10 ppm of cinnamaldehyde
10
12
14
min
87
mAU
Retention time = 11.261
Peak area =730.3
140
120
100
80
60
40
20
0
0
2
4
6
8
10
12
14
min
Figure 21 Chromatogram of 10 ppm of cinnamic acid
Note that although the concentrations of the two sample solutions are the same
(both are 10 ppm), the areas of the two sample peaks in the chromatograms are
actually quite different.
Cinnamaldehyde (peak area = 968.5) gave a stronger
peak than cinnamic acid (peak area = 730.3). The HPLC detector exhibits different
sensitivities to different compounds; it has a higher sensitivity to cinnamaldehyde
than to cinnamic acid, and thus signal intensity for cinnamaldehyde is higher. In
actual chemical analysis, the relationship between the concentration of the analyte
and the intensity of the detector response is expressed by a (linear) calibration
graph. For HPLC analysis, it involves measuring the peak areas with a series of
standard solutions of different concentrations, and then constructing a calibration
graph. The HPLC chromatogram of the sample is then obtained. The concentration
of the analyte in the sample is determined from the corresponding peak area in the
sample’s chromatogram and the calibration graph.
* ppm = parts per million – a unit that measures concentration. 1 ppm = 1 mg of sample in 1 L of solution.
** t R = retention time
Application of HPLC in chemical analysis – a case study
Cinnamon is a spice commonly used in cooking. It is also a common ingredient in
Chinese herbal medicine. A group of students knew that cinnamaldehyde, a major
ingredient present in the essential oil of cinnamon, has antimicrobial effects. They
anticipated that the medicine property of cinnamon possibly arises from the presence
of cinnamaldehyde. They also knew that cinnamaldehyde is a chemically reactive
compound that can be oxidised into cinnamic acid.
1. Shown below are the HPLC data obtained from standard cinnamaldehyde
solutions. Use the data to construct a calibration graph (plot the peak areas
against the concentrations of cinnamaldehyde).
Retention time of cinnamaldehyde = 12.9 min
Concentration of cinnamaldehyde
Peak Area
2 ppm
198.1
4 ppm
399.2
10 ppm
20 ppm
968.5
2140.6
50 ppm
5313.7
Chemistry
The students worked on a research project to find out the best way of extracting
cinnamaldehyde from cinnamon bark. They used HPLC to analyse the amounts of
cinnamaldehyde and cinnamic acid obtained from different extraction methods.
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2. Make a best straight line for the plot (you can do this with the help of spreadsheet
software such as Microsoft Excel).
3. Construct a calibration graph for the HPLC data obtained for cinnamic acid.
Make a best straight line for the plot.
Retention time of cinnamic acid = 11.3 min
Concentration of cinnamic acid
Peak Area
2 ppm
120.4
4 ppm
286.2
10 ppm
20 ppm
730.3
1492.8
50 ppm
3576.7
4. The group of students performed six extractions using different experimental
conditions. Shown below are the HPLC results that they obtained.
Method 1: Heating cinnamon in boiling n-hexane for 1 hour.
mAU
Peak B
20
15
10
Peak A
5
0
-5
-10
0
2
4
6
8
10
12
14
min
Peak Area
Compound
Peak A
11.113
12.3
Cinnamic acid
Peak B
12.804
242.2
Cinnamaldehyde
Method 2: Heating cinnamon in boiling n-hexane for 1 hour, and then removing
the solvent (n-hexane) from the extract solution by rotary evaporator.
Chemistry
mAU
Peak B
20
10
Peak A
0
-10
0
2
4
6
8
10
12
14
Peak Area
Compound
Peak A
11.066
25.1
Cinnamic acid
Peak B
12.819
329.1
Cinnamaldehyde
min
89
Method 3: Heating cinnamon in boiling ethanol for 1 hour.
mAU
Peak B
200
175
150
125
100
75
50
Peak A
25
0
0
2
4
6
8
10
12
14
Peak Area
Compound
Peak A
11.092
106.3
Cinnamic acid
Peak B
12.806
2117.1
Cinnamaldehyde
min
Method 4: Heating cinnamon in boiling ethanol for 1 hour, and then removing the
solvent (ethanol) from the extract solution by rotary evaporator.
mAU
Peak B
150
125
100
75
50
25
Peak A
0
0
2
4
6
8
10
12
Peak Area
14
min
Compound
Peak A
11.101
78.2
Cinnamic acid
Peak B
12.813
1753.9
Cinnamaldehyde
Method 5: Heating cinnamon in boiling water for 1 hour.
mAU
Peak A
2
Chemistry
4
Peak B
0
-2
-4
-6
-8
-10
0
2
4
6
8
10
12
14
Peak Area
Compound
Peak A
11.157
71.3
Cinnamic acid
Peak B
12.832
50.6
Cinnamaldehyde
min
90
Method 6: Soaking cinnamon in water at room temperature for 1 hour.
mAU
Peak A
2
0
Peak B
-2
-4
-6
-8
-10
2
4
6
8
10
12
14
Peak Area
Compound
Peak A
11.090
6.9
Cinnamic acid
Peak B
12.842
25.2
Cinnamaldehyde
min
Complete the following table using the HPLC data.
Method
Peak area:
Cinnamaldehyde
Peak area:
Cinnamic acid
Concentration:
Concentration:
Cinnamaldehyde (ppm) Cinnamic acid (ppm)
1
2
3
4
5
6
5. Discuss the following questions
(a) Which method could extract the most cinnamaldehyde?
(b) Which method could extract the most total cinnamaldehyde and cinnamic acid?
(c) Which method gave the smallest cinnamic acid/cinnamaldehyde ratio?
(d) Which is the best method for extracting cinnamaldehyde from cinnamon?
Chemistry
III. Activity Guidelines
Number of Sessions
Session 1
•
•
•
Carry out Activity 1 and Activity 2
Discuss the basic principles of chromatography
Discuss the effects of changing the stationary phase for a chromatographic
separation
Session 2
•
•
Carry out Activity 3
Discuss the effects of changing the mobile phase for a chromatographic separation
Session 3
•
•
91
Discuss the similarities and differences between thin-layer chromatography and
column chromatography
Session 4 (optional)
•
Session 5
•
•
Introduce the basic principles and applications of gas chromatography (GC) and
high performance liquid chromatography (HPLC)
Complete the exercise and discussion about analysing the gas chromatograms of
trans-stilbene, cis-stilbene and chalcone
Session 6
•
•
Discuss the application of GC and HPLC in quantitative analysis
Complete the exercise/discussion part of the Application of HPLC in chemical
analysis – a case study
IV. Assessment Guidelines
Your performance will be assessed according to three major components.
(1) Performance in lab activities (40%)
(2) Participation in class discussion (40%)
(3) Completing the activity: Applications of HPLC in chemical analysis – a case study (20%)
VI. References and Online Resources
References
Palleros, D. R. (2000). Experimental organic chemistry. New York: John Wiley & Sons.
Skoog, D. A., Holler, F. J., & Nieman, T. A. (1998). Principles of
instrumental analysis (5th ed.). Philadelphia: Saunders College Publishing.
Skoog, D. A., West, D. M., & Holler, F. J. (1996). Fundamentals of analytical
chemistry (7th ed.). Fort Worth: Saunders College Publishing.
Online Resources
Introduction to several common types of chromatographic techniques
http://www.chemguide.co.uk/analysis/chromatogrmenu.html
MIT 5.301 Chemistry laboratory techniques – column chromatography
http://www.youtube.com/watch?v=6fzBJ8nuuzk&feature=PlayList&p=57499F5778A
AB619&playnext_from=PL&index=5
Chemistry
Animation showing how TLC and GC work
http://www3.wooster.edu/chemistry/analytical/gc/default.html

Chromatography

Transcription

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