Spectrophotometry
MBLG1001 second session
Abs = 0.51
Introduction
Many compounds absorb ultraviolet (UV) or visible (Vis.) light. The diagram below shows a
beam of monochromatic radiation of radiant power P0, directed at a sample solution. Absorption
takes place and the beam of radiation leaving the sample has radiant power P.
The amount of radiation absorbed may be measured in a number of ways:
Transmittance, T = P / P
0
% Transmittance, %T = 100 T
Absorbance,
A = log10 P0 / P
A = log10 1 / T
A = log10 100 / %T
A = 2 - log10 %T Abs = log(I0/IT)
10Abs = (I0/IT), inverting
10-Abs = (IT/I0),
100* 10-Abs = 100*(IT/I0),
102-Abs = %T
ln Io/I = ax or log Io/I = a/2.303 (x)…… 4
Log Io/I = a’ (x) is called extinction coefficient
Look at slide 2 log Io/I is the absorbance.
This is the Lambert law
The law is extended by Beer, dI/dx is proportional to concentration
c and dI/dx is proportional to I and c, dI/dx = bcx b is the molar
absorption coefficient, b is a constan.
If c is in mol/L b will be in molar absorption coefficient .
log Io/I = b/2/303 cx
log Io/I =bcx and bcx is the absorbance (A)
log Io/I = A
Transmittance to Absorbance
The amount of radiation absorbed can be measured by the
transmittance. The relation between transmittance (T) and
absorbance (A) is given by the Beer-Lambert law (Beer’s law).
According to Beer’s Law, the absorbance is formulated as:
• The last equation, A = 2 - log10 %T , is worth
remembering because it allows you to easily calculate
absorbance from percentage transmittance data.
• The relationship between absorbance and transmittance
is illustrated in the following diagram:
•
• So, if all the light passes through a solution without any
absorption, then absorbance is zero, and percent
transmittance is 100%. If all the light is absorbed, then
percent transmittance is zero, and absorption is infinite.
Limitations of the Beer-Lambert law
The linearity of the Beer-Lambert law is limited by chemical and instrumental factors. Causes of nonlinearity
include:
•deviations in absorptivity coefficients at high concentrations (>0.01M) due to
electrostatic interactions between molecules in close proximity
•scattering of light due to particulates in the sample
•fluoresecence or phosphorescence of the sample
•changes in refractive index at high analyte concentration
•shifts in chemical equilibria as a function of concentration
•non-monochromatic radiation, deviations can be minimized by using a
relatively flat part of the absorption spectrum such as the maximum of an
absorption band
•stray light
Construction of Spectrophotometry
The Spectrophotometer
Photomultiplier Tube (PMT)
Introduction
Photomultiplier Tubes (PMTS) are light detectors that are useful in low intensity applications such as
fluorescence spectroscopy. Due to high internal gain, PMTs are very sensitive detectors.
Design
PMTs are similar to phototubes. They consist of a photocathode and a series of dynodes in an evacuated
glass enclosure. Photons that strikes the photoemissive cathode emits electrons due to the photoelectric
effect. Instead of collecting these few electrons (there should not be a lot, since the primarily use for PMT
is for verly low signal) at an anode like in the phototubes, the electrons are accelerated towards a series
of additional electrodes called dynodes. These electrodes are each maintained at a more positive
potential. Additional electrons are generated at each dynode. This cascading effect creates 105 to 107
electrons for each photon hitting the first cathode depending on the number of
dynodes and the accelerating voltage. This amplified signal is finally collected at
the anode where it can be measured.
Schematic of a PMT
•
Cross section of a photomultiplier tube
Schematic of a wavelength-selectable, single-beam UV-Vis spectrophotometer
Before the prac class begins
experimental work….
• The spectrophotometer actually measures
transmittance: %T = (IT/I0)*100.
• This is the intensity of light transmitted
through the solution (IT) divided by the
intensity of light entering the solution, the
incident light (I0). The problem with this
measurement is that as the solution becomes more
concentrated the %T decreases. The relationship is also
not linear.
• Absorbance is actually log10I0/IT
Now, suppose we have a solution of copper sulphate (which appears blue
because it has an absorption maximum at 600 nm). We look at the way in
which the intensity of the light (radiant power) changes as it passes through
the solution in a 1 cm cuvette. We will look at the reduction every 0.2 cm as
shown in the diagram below. The Law says that the fraction of the light
absorbed by each layer of solution is the same. For our illustration, we will
suppose that this fraction is 0.5 for each 0.2 cm "layer" and calculate the
following data:
Question : What is the significance of the molar absorbtivity, e ?
Answer : To begin we will rearrange the equation A = εbc :
ε = A / bc
In words, this relationship can be stated as "ε is a measure of the amount of light absorbed per
unit concentration".
Molar absorbtivity is a constant for a particular substance, so if the concentration of the solution
is halved so is the absorbance, which is exactly what you would expect.
Let us take a compound with a very high value of molar absorbtivity, say 100,000 L mol-1 cm-1,
which is in a solution in a 1 cm pathlength cuvette and gives an absorbance of 1.
Therefore, c = 1 / 100,000 = 1 ´ 10-5 mol L-1
Now let us take a compound with a very low value of e, say 20 L mol-1 cm-1 which is in solution
in a 1 cm pathlength cuvette and gives an absorbance of 1.
e = 1 / 1 ´ c Therefore, c = 1 / 20 = 0.05 mol L-1
Question : What is the molar absorbtivity of Cu2+ ions in an aqueous solution of
CuSO4 ? It is either 20 or 100,000 L mol-1 cm-1
Answer : I am guessing that you think the higher value is correct, because copper
sulphate solutions you have seen are usually a beautiful bright blue colour.
However, the actual molar absorbtivity value is 20 L mol-1 cm-1 ! The bright blue
colour is seen because the concentration of the solution is very high.
b-carotene is an organic compound found in vegatables and is responsible for the
colour of carrots. It is found at exceedingly low concentrations. You may not be
surprised to learn that the molar absorbtivity of b-carotene is 100,000 L mol-1 cm-1 !
The principle of the UV absorbance method is that nucleic acids (DNA
or RNA) contain conjugated double bonds in their purine and
pyrimidine rings that have a specific absorption peak at 260 nm. The
maximum absorbance of nucleic acids occurs at a wavelength of 260
nm
Proteins in solution absorb ultraviolet light with absorbance
maxima at 280 and 200 nm. Amino acids with aromatic rings
are the primary reason for the absorbance peak at 280 nm.
Due to the presence of tyrosine and tryptophan, proteins and
peptides containing these aromatic amino acids absorb UV light at a
wavelength of 280 nm. Each of these residues has distinct
absorption and emission wavelengths and varies in quantum yields.
Phenylalanine and disulfide bonds also contribute in the absorption
at this wavelength but since it is relatively insignificant, it can only be
observed in the absence of both tryptophan and tyrosine.
At a wavelength of 260 nm, the average extinction coefficient for doublestranded DNA is 0.020 (μg/ml)−1 cm−1, for single-stranded DNA it is 0.027
(μg/ml)−1 cm−1, for single-stranded RNA it is 0.025 (μg/ml)−1 cm−1 and for
short single-stranded oligonucleotides it is dependent on the length and
base composition.
A=εbc
1=0.02x1xc
C=1/0.02x1 = 50 µg/ml
A 260/280 ratio of ~1.8 is generally accepted as “pure” for DNA; a
ratio of ~2.0 is generally accepted as “pure” for RNA. Abnormal
260/280 ratios usually indicate that a sample is contaminated by
residual phenol, guanidine, or other reagent used in the extraction
protocol, in which case the ratio is normally low.
Relationship between %Transmittance and
light path length and concentration
%T
100
100
80
80
60
%T
60
40
40
20
20
0
length
0
concentration
Transmittance = IT/I0 = e-acl
where a is an extinction constant, c is
concentration and l is light path length
Absorbance increases linearly
with concentration
From IT/I0 = e-acl
I have used a here to describe a constant that is proportional to the extinction
coefficient. As you can see from the maths it is not the mMolar extinction
coefficient
taking logs of both sides and inverting
Ln (I0/IT) = acl
Converting to Log10
Log10(I0/IT) =
ecl
Hence the Beer-Lambert Law
A = ecl
Before the prac class begins
experimental work….
• Absorbance increases linearly with
concentration as predicted by the BeerLambert Law
A = ecl
• Explain why the working range of a
spectrophotometer is 0.1 – 1.0. Remember
Abs is a log scale. An absorbance of 1.6 is 2%
light transmitted while an absorbance of 2 is 1%
light transmitted. The class specs can not
accurately distinguish 1% from 2%. An
absorbance of 1.0 is 10% transmitted light.
Experiment 1: Identifying a
compound by spectrophotometry
• If a compound absorbs light its absorption
spectrum is a unique property of that
compound.
• The molecular structure is responsible for
the absorption properties
• The most common feature of absorbing
compounds are conjugated double bonds,
often as an aromatic ring
Experiment 1: Identifying a
compound by spectrophotometry
• This amount of absorbed energy (DE) will
determine the l of light absorbed.
• The DE is inversely proportional to the
wavelength of light absorbed ie. DE = hc/l,
where h is Planck’s constant and c is the
velocity of light. (Remember this from
physics!??)
Common Absorbing Biochemicals
O
• The bases of nucleic
acids
N
NH2
O
N
H3C
N
H
N
H
Thymine
O
N
H
N
NH2
Guanine
N
NH
NH
NH2
N
N
Adenine
N
H
Cytosine
O
Nucleic Acid Absorption Properties
e (mM-1cm-1)
Guanine
lmax
(nm)
275
Adenine
260
12.9
Cytosine
265
5.8
Thymine
258
8.0
Base
8.0
Absorption Spectrum: Guanine
Absorption Spectrum: Adenine
0.5
1.0
0.4
0.8
Absorbance
Absorbance
Nucleic Acid Absorption Properties
0.3
0.2
0.1
0
220
240
260
280
300
0.6
0.4
0.2
0.0
220
320
Wavelength (nm )
0.4
0.5
Absorbance
Absorbance
0.6
0.3
0.2
0.1
260
280
Wavelength (nm )
280
300
320
Absorption Spectrum: Thymine
0.5
240
260
Wavelength (nm )
Absorption Spectrum: Cytosine
0.0
220
240
300
320
0.4
0.3
0.2
0.1
0.0
220
240
260
280
300
Wavelength (nm )
320
Amino Acid absorption Properties
H2N
CH
C
O
O
O
H2N
OH
CH
C
H2N
OH
C
CH2
CH2
CH2
CH
N
NH
HN
Histidine
OH
Tryptophan
Tyrosine
O
H2N
CH
C
CH2
OH
O
H2N
CH
C
CH2
SH
Cysteine
Phenylanine
OH
OH
The Dyes
• The dyes chosen for this experiment are
different colours (A – F)
• Each pair of students will be assigned a
dye by the demonstrator. They use this
dye for the whole practical.
• Students should take note of the colour of
their dye and record the concentration
(mM) on the bottle
Obtaining a Spectrum for the dye.
• Using the Shimadzu spectrophotometers
in spectrum mode (mode 2 on main menu)
place a 1 mL plastic cuvette full of H2O in
the holder and obtain a baseline correction
(F1). This will take some time.
• Then, using the same cuvette, fill it with
the dye solution and obtain a spectrum.
Find the peaks. If you are unclear how to
do this practise beforehand.
Obtaining a Spectrum for the dye.
• The reason for doing this is to find the
absorbing region of the dye. It takes a long
time to obtain a spectrum from 600 nm to
350 nm. A quick narrowing of the range is
to be encouraged.
• Get students to consider the colour of the
solution and how this might give clues to
the absorption minima and maxima
The relationship between colour
and absorption
• A compound will be yellow if it reflects light
in the yellow wavelengths and absorbs
light of other wavelengths.
• Yellow compounds (often red crystals)
usually absorb in the blue range ~450 –
350 nm and have an absorption minimum
>550 nm
Dye C: Riboflavin
Absorption Spectrum: Dye C
1.2
Dye C is yellow red and has an
absorption
minimum in the
yellow/red region
Absorbance
1.0
0.8
0.6
0.4
0.2
0.0
350
400
450
500
w avelength (nm )
550
600
Obtaining the spectrum
• Once the baseline is corrected and the
absorbing range determined, find the
absorbance of the dye every 10 nm within
the absorbing range. Every 50 nm will do
outside the absorbing range.
• If the baseline correction is done you don’t
have to zero every time you change
wavelengths. This saves heaps of time.
Obtaining the spectrum
• Students must, by the next lab, plot the
spectrum. It would be a good idea to get
them to this now if there are free
computers. Otherwise get them to identify
the dye by the peaks, comparing to the
sample spectra at the back of this section
of the lab manual.
• From the concentration on the bottle
estimate the extinction coefficient.
Obtaining the spectrum
• Make sure you are very familiar with the
Excel chart drawing process as you will
need to help the students here. Practise
with the spectro.xls spreadsheet provided.
It has the raw data obtained for riboflavin
and the worked solution.
• Check out what is expected graph-wise in
the worked solution. Students should have
practised much of this with the Excel task
in the last practical.
Calculating the Extinction
Coefficient
• This comes directly from the relationship
A = ecl,
Where e is the extinction coefficient
expressed in the units of c, the
concentration. In this experiment the conc
units will be mM so e will have the units
mM-1cm-1. Round the value off to 1 dec pl.
Discussion
• Predict which of the following parameters would
change with dilution? How would they change?
• The number of peaks
• The l max
• Al1/Al2
• Absorbance at l max
• Extinction Coefficient
• Transmittance at lmax
• Can you predict what would happen to the
absorption spectrum if you diluted your dye with
another dye?
Experiment 2: The Standard Curve
• Although identifying a compound by
spectro is a useful property,
spectrophometry is used more often to
measure the concentration of a
compound.
• Sometimes the extinction coefficient can
be used directly. This occurs when the
compound of interest has an intrinsic
absorbance.
Experiment 2: The Standard Curve
• If the compound of interest does not have its
own intrinsic absorbance then a coloured
derivative must be made by reacting it with
reagents. Then a standard curve must be
produced.
• In today’s practical students will gain experience
at producing and using a standard curve, even
though in this situation you would normally use
the extinction coefficient.
Experiment 2: The Standard Curve
• Using the same solution as the one used
to obtain the spectrum, get the students to
dilute it so that there are at least 5 points
for the line. My suggestion is 200, 400,
600, 800, 1000 uL of dye, then make each
up to 1 mL with water.
• Mix well and obtain the absorbances at the
lmax.
The standard curve
Absorbance @ 445 nm
Standard Curve: Riboflavin
1.2
y = 0.0125x
1.0
R = 0.9999
2
0.8
0.6
0.4
0.2
0.0
0
20
40
60
[Riboflavin] (nmol/mL)
80
100
The standard curve
Absorbance @ 445 nm
Standard Curve: Riboflavin
1.2
y = 0.0125x
1.0
R = 0.9999
2
0.8
Plot the
concentration in
nmol/tube or
nmol/mL
0.6
0.4
0.2
0.0
0
20
40
60
[Riboflavin] (nmol/mL)
80
100
The standard curve
• The main point of confusion in this task is how to
plot the concentration. The purists would plot it
in mM or uM and this would be correct BUT
confusing for the students when they come to
back calculate with it.
• Instead plot it in nmoles per tube which in this
case is nmol/mL. Note that the riboflavin
concentration range is ~0 – 80 nmol/mL. It is not
“neat” due to the concentration of the starting dye
solution 0f 0.0836 mM.
How to use the standard curve
• The standard curve is used to find the
concentration of an unknown solution of
riboflavin.
• This practical session has 2 different
unknowns the students must determine
the concentration of; one which is in the
working range of the spectrophotometer or
standard curve i.e. 0.1 – 1.0, the other is
outside the range.
Unknown 3a
• This unknown can be directly determined
by measuring its absorbance without
dilution.
• However it is always good practise to do at
least one dilution when estimating a
concentration.
• The obvious dilution is a 1 in 2 dilution.
This is your chance to introduce dilutions
to the students.
Dilutions
• A 1 in 2 dilution is 1 part riboflavin
unknown C1 and 1 part H2O.
• If you wanted to make up a 1 in 2 dilution
of unknown C1 which could be easily read
off the standard curve you would make it
up to 1 mL.
• This would mean 500 uL of unknown C1
and 500 uL H2O.
Experiment 3a:Unknown C1
Slope
0.01245 Intercept
0.000571
Unknown C1
Dilution factor
1
2
A445
0.829
0.417
[Riboflavin]
(nmol/mL)
66.5
33.4
Original Conc.
(nmol/mL)
66.5
66.9
Average
(nmol/mL)
66.7
From standard curve
or using
INTERCEPT
function in Excel
From standard curve
or using SLOPE
function in Excel
Slope
0.01245 Intercept
0.000571
Unknown C1
Dilution factor
1
2
A445
0.829
0.417
[Riboflavin]
(nmol/tube=mL)
66.5
33.4
Original Conc.
(nmol/mL)
66.5
66.9
Average
(nmol/mL)
66.7
Average of 2
values
[Riboflavin]
(nmol/mL) =
(A445intercept)/slope
[original] =
[riboflavin]*
dilution
factor
Quick tips
• You can use the extinction coefficient
obtained in the first experiment or the
standard curve. Get the students to try
both methods.
• To get from the Absorbance to the
concentration you solve the equation of
the standard curve for x; you know the y
value (Abs) and you want to find out the x
value (conc.)
Why we express the
concentration in nmol/tube
• Provided you make the unknown dilutions
to the same volume as the standards you
can directly work out how much there is in
the tube straight from the graph.
• In the next exercise, unknown C2 or C3, it
will be a real advantage
Discussion Q from this section
• Where does the extinction coefficient fit in to the
std curve?
– It is the gradient, but the units of the ext. coefficient
are in the conc. Units on the x-axis
• What would happen to the absorbance
response and the equation of the line if:
– you measured the absorbance at a wavelength other
than the lmax?
– It would be linear but the gradient i.e. the ext.
coefficient would be lower See the varyQ worksheet
in the spectro.xls
What would happen to the absorbance
response and the equation of the line if:
• you expressed the concentration in different units (try M
and % (w/v), obtaining the molecular weight of the dye
from your demonstrator)?
– The equation changes, in particular the gradient. The
easiest way is to try this. Use the data given in the
spectro spreadsheet.
– you made the dye solutions up in 3 mL instead of 1
mL?
– No difference, the absorbance is dependent on the
concentration, which is volume independent. The only
problem is if the cuvette is not full enough to cover
light source
– you used cuvettes with a 2 cm light path instead of 1
cm? The slope would be double as the absorbance
would double for each tube
Experiment 3b: Unknown outside
the working range of the
spectrophotometer
• What concentration of riboflavin gives an
absorbance of 0.5? (always aim for the middle of
the range)
Using A = ecl……0.5 = 12.5*c*1
Conc = 0.04 mM à 40 uM à 40 nmol/mL
• So we need 40 nmoles of riboflavin in 1
mL to get an absorbance of 0.5
Experiment 3b:
• Now how much of our unknown do we
need to add to give an absorbance of 0.5?
• Our unknown, C2, lies between 0.5 and 1
mM. Undiluted this unknown would give an
absorbance between 6 and 12…way too
high!
• So we need to dilute our unknown…but by
how much
Experiment 3b:
• Let’s consider the upper end of the range,
1 mM.
• If our unknown is 1 mM, which is
1 umol/mL or 1 nmol/uL then…..
• as we need 40 nmol/mL to give an
absorbance of 0.5 so we would need to
add 40 uL (40/1).
Experiment 3b:
• Then let’s consider the lower end of the
range, 0.5 mM.
• If our unknown is 0.5 mM, which is
0.5 umol/mL or 0.5 nmol/uL then…..
• as we need 40 nmol/mL to give an
absorbance of 0.5 so we would need to
add 80 uL (40/0.5).
Experiment 3b:
• So we need to add between 40 and 80 uL
of our unknown and make these up to
1 mL à mix well …….
• Then measure the absorbances of our
samples.
0.396
0.490
0.594
0.681
0.774
Results
Unknown C2 Range (0.5 - 1.0 mM)
Unknown
(uL)
A445
[Riboflavin]
(nmol/mL)
[Riboflavin]
(nmol/uL)
[Riboflavin]
(mM)
40
0.396
50
0.49
60
0.594
70
0.681
80
0.774
31.76
39.31
47.66
54.65
62.12
0.79
0.79
0.79
0.78
0.78
0.79
The back calculations
• The volume of the unknown (uL) and the
A445 are entered in directly as data.
• To calculate the [Riboflavin] (nmol/mL) you
solve the standard curve equation for x. This
gives the #nmoles of riboflavin in each cuvette.
31.76
39.31
47.66
54.65
62.12
The back calculations
From the [Riboflavin] (nmol/mL) we need to know
the concentration in the original unknown C2
31.76
39.31
47.66
54.65
?
62.12
The back calculations
Each cuvette has a known volume of the
original unknown added and we know
#nmoles of riboflavin in each cuvette
40 uL
50 uL
60 uL
70 uL
80 uL
31.76
nmoles
39.31
nmoles
47.66
nmoles
54.65
nmoles
62.12
nmoles
If we simply divide the #nmoles by the volume
added in uL we have the original concentration
of the unknown riboflavin in nmol/uL which is
mM. After all we only added two solutions to each cuvette; water
and riboflavin. We hope the nmoles came from the dye not the
water.
40 uL
0.79
mM
50 uL
60 uL
70 uL
80 uL
0.79
mM
0.79
mM
0.78
mM
0.78
mM
Average = 0.79 mM
For unknown C3: range 1 – 2 mM
• The upper range:
2 mM à 2 umol/mL à 2 nmol/uL
• So we need to add 40/2 = 20 uL
• The lower range:
• 1 mM à 1 nmol/uL
• So we need 40/1 = 40 uL
Results
Unknown C3 (1 - 2 mM)
Unknown
(uL)
A445
[Riboflavin]
(nmol/mL)
[Riboflavin]
(nmol/uL)
[Riboflavin]
(mM)
20
25
30
35
40
0.443
0.539
0.657
0.755
0.869
35.54
43.25
52.72
60.59
69.75
1.78
1.73
1.76
1.73
1.74
1.75
Integrating this equation from z = 0 to z = b gives:
or
Since N (molecules/cm3) * (1 mole / 6.023x1023 molecules) * 1000
cm3 / liter = c (moles/liter) and 2.303 * log(x) = ln(x), then
or
where
=
* (6.023x1020 / 2.303) =
* 2.61x1020
Typical cross-sections and molar absorptivities are:
(cm2
(M-1 cm-
)
1)
absorption - atoms
10-12
3x108
absorption - molecules
10-16
3x104
absorption - infrared
10-19
3x10
Raman scattering
10-29
3x10-9
Double Beam UV.VIS
Spectrophotometer
diagram of a double-beam UV-Vis. spectrophotometer
Question : What is the significance of the molar absorbtivity, e ?
Answer : To begin we will rearrange the equation A = e bc :
e = A / bc
In words, this relationship can be stated as " e is a measure of the amount of light absorbed per unit
concentration".
Molar absorbtivity is a constant for a particular substance, so if the concentration of the solution is
halved so is the absorbance, which is exactly what you would expect.
Let us take a compound with a very high value of molar absorbtivity, say 100,000 L mol-1 cm-1, which is
in a solution in a 1 cm pathlength cuvette and gives an absorbance of 1.
e=1/1xC
Therefore, c = 1 / 100,000 = 1 x 10-5 mol L-1
Now let us take a compound with a very low value of e, say 20 L mol-1 cm-1 which is in solution in a 1 cm
pathlength cuvette and gives an absorbance of 1.
e=1/1xc
Therefore, c = 1 / 20 = 0.05 mol L-1
The answer is now obvious - a compound with a high molar absorbtivity is very effective at absorbing
light (of the appropriate wavelength), and hence low concentrations of a compound with a high molar
absorbtivity can be easily
which means high Sensitivity
Crystal Violet Absorption Spectrum
1 .4
Absorbance
1 .2
1
0 .8
0 .6
lmax
0 .4
0 .2
0
200
250
30 0
35 0
400
450
500
wavelength, nm
550
600
650
700
750