applied physics on spectroscopy
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8/16/2019 Applied Physics on Spectroscopy
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Applied PhysicsOn
UV-vis Spectrophometry
Kuwat Triyanahttp://triyana.staff.ugm.ac.id
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EM Spectrum
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Ultra Violet (UV)
• Three regions of UV
– UV-A:• long-wave UV, near-ultraviolet, black light, or Wood’s light
• between 320 and 400 nm
– UV-B:• medium-wave UV
• between 280 and 320 nm
– UV-C :• short-wave, far ultraviolet, or germicidal UV
• between 180 and 280 nm
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Absorption Spectroscopy Introduction
A.) Absorption: electromagnetic (light) energy is transferred to atoms, ions, or molecules
in the sample. Results in a transition to a higher energy state.
- Transition can be change in electronic levels, vibrations, rotations, translation, etc.
- Concentrate on Molecular Spectrum in UV/Vis (electronic transition)
- Power (P): energy of a beam that reaches a given area per second
- Intensity (I): power per unit solid angle
- P and I related to amplitude2
Eo
E1
h Energy required of photonto give this transition:
h = DE = E1 - Eo
(excited state)
(ground state)
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Absorption and Emission
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Molecular Absorption and Emission
Atoms:
Electronic states only
Molecules:
Vibrational states within electronic states
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B.) Terms:
1.) Beer’s Law: A = bc
The amount of light absorbed (A) by a sample is dependent on the path
length (b), concentration of the sample (c) and a proportionality constant (e
– molar absorptivity)
Amount of light absorbed is dependent on frequency ( )
c
Absorbance is directly proportional to concentration Fe+2
Increasing Fe+2 concentration
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B.) Terms:
1.) Beer’s Law: A = bc
Transmittance (T) = I/Io %Transmittance = %T = 100T
Absorbance (A) = log10
Io/I
No light absorbed-
% transmittance is 100% absorbance is 0
All light absorbed-
% transmittance is 0% absorbance is infinite
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Relationship Described in Terms of Beer’s Law
A = Absorbance = ebc = -log(%T/100)
e = molar absorptivity: constant for a compound at a given frequency ()
or wavelength (l)
units of L mol-1 cm-1
b = path length: cell distance in cm
c = concentration: sample concentration in moles per liter.
Therefore, by measuring absorbance or percent transmittance at a given
frequency can get information related to the amount of sample (c) present
with an identified e and l.
Note: law does not hold at high
concentrations, when A > 1
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C.) Components of an Instrument for UV/Vis Absorbance Measurements:
1.) Basic Design:
Hitachi Instruments U-3010
Light Source, l selector, Sample cell holder, Detector (amplifier, recorder)
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a) Desired Properties of Components of UV/Vis:
Light Source Selector
Creates Proper l Narrow Bandpass:
Stable: Selects Desired l
Constant P Large Light Throughput:
Good Precision Increase P
Intense: Increase P
Easier to See Absorbance
Samp le Cell Holder Detector Fixed Geometry: Stable
Constant b Sensitive to l of Interest
Transmits l of Interest:
Increase P
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Molecular Spectra
400 500 600 700
l (nm)
Absorption Emission
I n t e n s i t y
Bands vs. Lines (atoms)
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Absorbed vs. Observed
Color absorbed → Complementary color observed
R
BG
B
YG BV
VY
RVO
G
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Spectroscopy Terms Describing
Absorption (Beer’s Law)
• Consider a beam of lightwith an (initial) radiantintensity Io
• The light passes through asolution of concentration (c)
• The thickness of thesolution is “b” cm.
• The intensity of the lightafter passage through thesolution (where absorptionoccurs) is I
I0 hv I
b
C o n c e n t r a t
i o n
( c )
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We Define
• Transm ittance (T) = I/I0 (units = %)
• Absorbance (A ) (units = none)
– A = log (I0/I)
– A = -log (T) = log (1/T)
– A = abc (or εbc) <--- Beer’s Law • a = absorptivity (L/g cm)
• b = path length (cm)
• c = concentration (g/L)
• ε = molar absorptivity (L/mol cm) – Used when concentration is in molar units
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Other expression of Lambert’s Law of Absorption
xe
I
I
I x
I
=
=
0
d
d
• The intensity I 0 if a beam of light decreases exponentially as it passes though a uniform
absorbing medium with the linear decay constant α .
Restatement: In a uniform absorbing medium, the intensity of a beam of light decreases
by the same proportion for equal path lengths traveled.
• The linear decay constant α is a characteristic of the medium. It has units of reciprocal
length. α is the path length over which the intensity is attenuated to 1/e.
α
I 0
I ( x )
x
Lambert described how intensity changes with
distance in an absorbing medium.
xe I x I = 0)(
l e I I = 0
I
l
xe I I
x I I
=
=
0
dd
The distance traveled through the medium is
called the path length.
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Lambert’s Law of Absorption (base 10)
xk x
e I
I ==100
Typically base 10 is used in photometry.
xk x I e I I == 1000
10ln =k
k is the path length over which the intensity is attenuated to 1/10.
xk
I
I =100
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Lambert’s Law Example If one slab of absorbing material of thickness l reduces the intensity of a beam of light
to half.
α I 0
2
110
0
== l k
I
I I
l
And three slabs will reduce the intensity of a beam of light to one eight .
Then two slabs of the same absorbing material will then reduce the
intensity of a beam of light to one quarter .
4
1
2
110
2
2
0
=
== l k
I
I α I 0 I
l
α
l
8
1
2
110
3
3
0
=
== l k
I
I α I
l
α
l
α I 0
l
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Beer’s Law Beer found that Lambert’s linear decay constant k for a solutionof an absorbing substance is linearly related to its
concentration c by a constant, the absorptivity ε , acharacteristic of the absorbing substance.
Restatement: The linear decay constant k is linear in concentrationc with a constant of proportionality ε .
(August Beer, 1825-1863)
A colored absorber has an absorptivity that is dependent on wavelength of thelight ε ( λ).
The absorptivity is the fundamental property of a substance. This is the property
that contains the observable spectroscopic information that can be linked to
quantum mechanics (also see absorption cross section.)
ck e =
Typical units are: k cm−1; c M (moles/liter); ε M−1cm−1
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Photometric Quantities
Transmittance (T)
Absorbance (A) (AKA optical density, O.D.)
0 I
I T = usually given in percent
T I
I A loglog
0
=
= by convention, base 10 logs are used
In photometry we measure the intensity of light and characterize its
change by and object or substance. This change is typicallyexpresses as percent transmittance or absorbance.
Frequently when your primaryinterest is the light beam
Used almost exclusively when your interest
concerns the properties of the material
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Beer-Lambert LawLambert’s and Beer’s Laws are combined to describe the
attenuation of light by a solution. It is easy to see how the twostandard photometric quantities can be written in terms of this law.
xc I I e = 100
xc A
T
I
I A
e =
=
= loglog
0
xcT
I
I T
e
=
=
10
0
Transmittance Absorbance
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Cross-Sections and Absorptivitythe connection to single particles and molecules
The absorption of light by particles (and single molecules) ischaracterized by an absorption cross section C . In this model theparticle is replaced by a perfectly absorbing sphere with a crosssectional area C . This cross section is a property of the particleand is not related to its geometric cross sectional area. Theconcentration of particles per unit volume is N .
=
=
=
3
3
cm
liter 1010ln
10ln
C N
NC k
NC
Ae
typical units are: C cm2; N cm−3
The cross section can be directly related to the
molar absorptivity. N A is Avagadro’s number.
units are: C cm2; N cm−3; N A mole−1; ε
M−1cm−1
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Efficiency
The absorption efficiency Q of a particle is the ratio of its absorption
cross section C to its geometric cross section C geo.
Absorption efficiency is dimensionless.
geoC
C Q =
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Extension to Scattering and
Extinction
Attenuation of light by absorption and scattering both obey
Lambert’s Law. Thus we can extend our treatment of absorption to
scattering and extinction. (Recall that extinction is the effect of
absorption + scattering.)
cxcx A
QQQ
C C C
scaabsext
scaabsext
scaabsext
scaabsext
e e e
e e e
==
=
=
= The scattering efficiency can be much larger than unity.
Extinction paradox: Qext = 2 (Qabs = 1; Qsca = 1)
for an perfectly absorbing particle very large compared
to the wavelength of light.
Note:
• All of these quantities are in general wavelength dependent.
•Our discussion has not included the mechanism (cause) of absorption and scattering.
•There are many different mechanisms that cause of absorption and scattering.
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Instrumentation
• Spectrometer: measures I vs λ.Simply measures the spectrum of the light (e.g. emission spectroscopy).
• Spectrophotometer: measures I /I 0 vs λ.Measures how the sample changes the spectrum of the light (e.g.
transmission, reflection, scattering, fluorescence).All spectrophotometers contain a spectrometer.
• -meter: the detector is electronic
• -graph: light intensity recorded on film
• photometer: measures I /I 0 without λ selection.
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OU NanoLab/NSFNUE/Bumm & Johnson
The Spectrophotometer
Measures absorbance as a function of wavelengthComponents: light source, monochromator, sample cell, detector,
optical system.
monochromator sample cell
detector
light source
s l i t
d
i f f r a c t i o n
g r a t i n g
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balance the
forces:
Computer controlled
acquisitionof absorption spectra
Cary 50 UV-Vis Spectrophotometer
sample
detector
lightsource
monochromator
Can you find thediffraction grating and theslit?
www.varianinc.co
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Making a Measurement with the Cary 50
• First, measure the baseline using a blank sample. This is raw I 0.The blank sample is the cuvette with deionized water(everything but your nanoparticles). This corrects for anyabsorption due to the cuvette, water, and variations of the lightintensity of the light source, monochromator, etc.
• Second, measure the zero by inserting the beam block . Thiscorrects the instrument for the detector background.
• Third, measure your sample. This is the raw I . The Cary 50automatically calculates the corrected intensities (I and I 0) bysubtracting the zero from each of the raw intensities.
• Subsequent measurements do not require re-measuring the
blank and zero, simply repeat step 3.
T A
zero I raw
zero I raw
I
I T
log
00
=
==
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Transmittance
T => transmittance
IT = -----
Io
Io I
b
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I0 = 10,000 I = 5,000
5.010000
5000
0
=== P
P T
-b-
Example
A = -log T = -log (0.5) = 0.3010
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Beer’s Law
A = abc = ebc
where T => transmittance
%T => percentage transmittance
I => transmitted Iower of radiation
Io => incident power of radiation
A => absorbance
a => absorptivity
b => path length
c => concentration
e => molar absorptivity, extinction coefficient
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Beer’s Law
A = abc = ebc
A
c
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Beer’s Law A = ebc
Path Length Dependence, b
Readout
Absorbance
0.82
Source
Detector
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Beer’s Law A = ebc
Path Length Dependence, b
Readout
Absorbance
0.62
Source
Detector
b
Sample
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Beer’s Law A = ebc
Path Length Dependence, b
Readout
Absorbance
0.42
Source
Detector Samples
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Beer’s Law A = ebc
Path Length Dependence, b
Readout
Absorbance
0.22
Source
Detector Samples
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Beer’s Law A = ebc
Wavelength Dependence, a
Readout
Absorbance
0.80
Source
Detector
b
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Beer’s Law A = ebc
Wavelength Dependence, a
Readout
Absorbance
0.82
Source
Detector
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Beer’s Law A = ebc
Wavelength Dependence, a
Readout
Absorbance
0.30
Source
Detector
b
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Beer’s Law A = ebc
Wavelength Dependence, a
Readout
Absorbance
0.80
Source
Detector
b
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Non-Absorption Losses
"Reflection andscattering losses."
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Limitations to Beer’s Law
• Real – At high concentrations charge distribution effects
occur causing electrostatic interactions betweenabsorbing species
• Chemical – Analyte dissociates/associates or reacts with solvent
• Instrumental
– ε = f(λ); most light sources are polychromatic notmonochromatic (small effect)
– Stray light – comes from reflected radiation in themonochromator reaching the exit slit.
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Chemical Limitations
A reaction is occurring as you record
Absorbance measurements
Cr 2O72- + H2O 2H+ + CrO4
2-
A550 A446
concentration concentrationwavelength
400 500300
CrO42-
Cr 2O72-
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Instrumental Limitations - ε = f(λ)
• How/Why does εvary with λ?
• Consider awavelength scan fora molecular
compound at twodifferent wavelengthbands
• In reality, a
monochromator cannot isolate a singlewavelength, butrather a smallwavelength band
Larger the Bandwidth – larger deviation
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Instrumental Limitations – Stray Light
• How does stray light effect Absorbanceand Beer’s Law?
• A = -log I/Io = log Io/I
• When stray light (Ps) is present, theabsorbance observed (Aapparent) is the sum
of the real (Areal) and stray absorbance
(Astray)
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Instrumental Limitations – Stray Light
• Aapp = Areal + Astray =
• As the analyte concentration increases([analyte]↑), the intensity of light exiting the
absorbance cell decreases (I↓)
• Eventually, I < Is
o s
s
I Ilog
I I
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Instrumental Limitations – Stray Light
• Result – non-linearabsorption (Analyte
vs. Conc.) as a
function of analyteconcentration
– Similar to
polychromatic light
limitations
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Solution too Concentrated
• Refractive index changes with larger
concentrations
A e e n n
n2 + 2
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Allura Red
Formula: C18H14N2Na2O8S2
Molar mass: 496.42 g/mol
2Na+
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During Absorption…
Starting e- arrangement: After photon absorption:
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Absorption is random
•Photons collide with molecules•Certain probability of absorption
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Transmittance, T
I0 I1 TII
0
1 =
T: ratio of light “in” vs. “out”
= fraction of light passing through
Depends on:
# of molecules b & c
molecules’ identity e
Pathlength, b
Concentration,c
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Transmittance and Pathlength
1
2
0
1
II
II =
For cell with same pathlength: constant T
1
2
II
I0 I1
0
1
II
I2
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Transmittance and Pathlength
1
2
I
I
0
1
I
I
2
0
1
1
2
0
1
0
2
I
I
I
I
I
I
I
I
==I0
For cells with same pathlength: constant T
Double pathlength: square T
I2
#1 #2
2
121total TTTT ==
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T and Pathlength
1- b
b
I
I
0
1
I
I
1- b
b
0
1
b
0
1
0
b
I
I
I
I
I
I
I
I
=
=
I0
For b cells (b pathlengths = 1cm)
Tb
Ib
#1 #b
b1
b
1total TTTT ==
…
b in power
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Concentration
• # photons absorbed depends on #molecules in path
I0 I1
Transmittance and
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Transmittance and
Concentration
1
2
I
I
b
0
cm1
I
IT
=
0
1
I
I
I0 I1 I2
1.0 M reference solution gives
2.0 M: Double concentration,c→“Double pathlength, b”…
bc
0
1Mcm,1 bc
1I
ITT
==
…
bc in powerpathlength and
concentration
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Molar Extinction Coefficient, e
e
Probability of absorption
– Specific to molecule
– Function of l
lmax Most efficient absorption
– Peak of absorption curve
–“Best l” for experiment
T: concentration c & pathlength b What about: molecular identity?
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b, c, and e
εcb- bcε-
bc
0
1Mcm,11010
I
IT ==
=
I
Ilogε
0
1Mcm,1
=
e in powermolar extinction coeff.
ε
0
1Mcm,110
I
I =
Therefore:
bc
0
1Mcm,1 bc
1I
ITT
==
bc in powerpathlength and concentration
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Absorbance
Beer’s Law : Defines absorbance, A
εbc A
A
T A
εbc
=
=
=10log
log
A : Directly proportional to concentration
high A ≡ low T
Part 1 Spectral Profile of Allura
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Part 1 Spectral Profile of Allura
Red, lmax
•Use stock solution (record conc.) cstock
•Record %T, 400 – 700 nm %T
– 10 nm intervals near lmax
– 20 nm intervals elsewhere
•Record cell width, b = pathlength b
•Calculate A A
•Plot A vs. l
•Determine lmax l
max
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Part 1: lmax
400 500 600 700l (nm)
Absorption Transmittance
I n t e n s i t y
Find l where
%T is lowest
A is highest
This is not
necessarily
Allura Red
(but would
appear red)
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Spectral Profilel (nm) Absorbance
400 0.078
420 0.130
440 0.268 460 0.555
480 0.966
490 1.189
500 1.383
510 1.485
520 1.515
530 1.459
540 1.217
560 0.395
580 0.063
600 0.019
620 0.015
640 0.010
660 0.009
680 0.006
700 0.000
Spectral profile - Allura Red
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
400 450 500 550 600 650 700
wavelength (nm)
A b s o r b a n c
e
Part 2 Absorbance for Various
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Part 2 Absorbance for Various
Concentrations
•Stock solution, 4 dilutions, and blank• n1 = n2
• M1 . V1 = M2
. V2
•
•Determine %T at lmax for each
%T
•Calculate A
A
•Plot A vs. conc (Beer’s Law plot)
•Slope = eb
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Part 2 Concentrations•1 blank: pure DI water M0 = 0.00 M allura red
•1 stock: 0.002% allura red M5 = 4×10-5 M
•4 dilutions, each by ½:
– (4×10-5 M)(25.00 mL) = (M1 )(50.00 mL) M1 = 2×10-5 M – (2×10-5 M)(25.00 mL) = (M2 )(50.00 mL) M2 = 1×10-5 M
– (1×10-5 M)(25.00 mL) = (M3 )(50.00 mL) M3 = 5×10-6 M
– (5×10-6 M)(25.00 mL) = (M4 )(50.00 mL) M4 = 2×10-6 M
M solution L
solution g
red allura g
red alluramol
solution g
red allura g 5
001.0
1
42.496
1
100
002.0104%002.0 ==
diluteed concentrat dilutediluteed concentrat ed concentrat nnV M V M ==
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Beer’s Law
e,b constant
xm ==
=
y
cεbA
TlogA
A === 1010
%100
%TT εbc
c varied
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Beer’s Law Plot
cεbA =
x y =m
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
Molarity
A b s o r b a n c e
y = 1.9 x
yD
εb=D
D=
x
y slope
xD
B ’ L Pl l
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Beer’s Law Plot example
Dilution Factor M (mol/L) T A = -logT
0 0 1.00 0.000
1/16 2.E-06 0.87 0.060
1/8 5.E-06 0.73 0.137
1/4 1.E-05 0.58 0.240
1/2 2.E-05 0.32 0.495
1 4.E-05 0.10 1.000
Stock M 4E-05 mol/L
A = e×b×c
Slope = D
A/D
c Here:
e×b = 24943
A 1-cm cell: e
24943 cm-1M-1
Absorbance vs. Concentration - Allura Red
y = 24943x
R 2 = 0.9995
0.000
0.200
0.400
0.600
0.800
1.000
1.200
0.E+00 1.E-05 2.E-05 3.E-05 4.E-05
Molarity (mol/L)
A b s o r b a n c e
Part 3 Allura Red Concentration in
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Part 3 Allura Red Concentration in
Mouthwash
•Use 1:25 dilution
•
•Determine %T at lmax %T
•Calculate AA
•Measure b b
•Find concentration, c c
soεbc A =b
Ac
e
=
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Part 3 Example
Mouthwash trials (1:25 dilution): T A M
dilute Mconcentrated
0.68 0.167 7E-06 1.7E-04
0.70 0.155 6E-06 1.6E-04 0.65 0.187 8E-06 1.9E-04
0.69 0.161 6E-06 1.6E-04
Mdilute= A/(b×e ) if cell length b is constant
Average Mconcentrated = 1.7×10-4 M = 2×10-4 M
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Report• Abstract
• Data/Results
• Sample calculations including:
– Absorbance from transmittance
– Dilution
– Slope and extinction coefficient
– [Allura red]cuvette and [Allura red]mouthwash
– # allura red molecules in 1 mL mouthwash
• Discussion/review questions
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