chapter 2 - aneeminiechem08028

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Transcript chapter 2 - aneeminiechem08028

Quantitative
Model of light interaction with matrix
and how spectrophotometer works
Transmittance (T) – the fraction of incident
radiation transmitted by the medium:
T = I/I0
Absorbance (A) – the amount of incident
radiation absorb by the medium and
expressed by:
 A = log(1/T) = - logT = log P0/P
BEER LAMBERT LAW
Light
I0
I
Glass cell filled with
concent rat ion of solut ion (C)
As the cell thickness increases, the intensity of
I (transmitted intensity of light ) decreases.
T=
T- Transmittance
I
I0
I0 - Original light intensity
I- Transmitted light intensity
I
% Transmittance = 100 x
I0
T=
%T
100
Absorbance (A) = Log
100
%T
A = 2 – Log %T
%T= antilog
(2 – A )


Beer’s Law (Beer – Bourguer – Lambert
Law)
Radiation energy decays exponentially
T = P/P0 = 10-kb
Where:
 k – constant
 T – the transmittance – the fraction of
 transmitted radiant energy
 b – the pathlength of the medium
A  CL = KCL by definition and it is called
the Beer Lambert Law.
A = KCL
K = Specific Extinction Coefficient ---- 1 g
of solute per liter of solution
A = ECL
E = Molar Extinction Coefficient ---Extinction Coefficient of a solution
containing 1g molecule of solute per
1 liter of solution
E =
Absorbance x Liter
Moles x cm
E differs from K (Specific extinction
Coefficient) by a factor of molecular weight.
UNITS
A = ECL
A = No unit (numerical number only)
Lit er
E =
Cm x Mole
L = Cm
C = Moles/Liter
A = ECL = (
A = KCL
A = No unit
Liter
Cm x Mole
)x
Mole
x Cm
Liter
C = Gram/Liter L = Cm
Liter
K=
Cm  Gram
Lit er
A = KLC = (
Cm x Gram
Gram
)x
Lit er
x Cm
STEPS IN DEVELOPING A SPECTROPHOTOMETRIC
ANALYTICAL METHOD
1. Run the sample for
spectrum
2. Obtain a monochromatic
wavelength for the
maximum absorption
wavelength.
3. Calculate the concentration
of your sample using Beer
Lambert Equation: A = KCL
Absorbance
2.0
0.0
200
250
300
350
400
W avelengt h (nm )
450
SPECTROPHOTOMETR READINGS
ULTRAVIOLET SPECTRUM
Slope of Standard Curve =
A
C
Absorbance at 280 nm
1.0
0.5
1
4
2
3
Concentration (mg/ml)
5
There is some A vs. C where graph is linear.
NEVER extrapolate beyond point known where
becomes non-linear.
SPECTROMETRIC ANALYSIS USING STANDARD CURVE
Absorbance at 540 nm
1.2
0.8
0.4
3
1
2
Concentration (g/l) glucose
4
Avoid very high or low absorbencies when drawing a standard curve.
The best results are obtained with 0.1 < A < 1. Plot the Absorbance vs.
Concentration to get a straight line
CELLS
UV Spectrophotometer
Quartz (crystalline silica)
Visible Spectrophotometer
Glass
IR Spectrophotometer
NaCl
LIGHT SOURCES
UV Spectrophotometer
1.
Hydrogen Gas Lamp
2.
Mercury Lamp
Visible Spectrophotometer
1.
Tungsten Lamp
IR Spectrophotometer
1.
Carborundum (SIC)
UV SPECTROMETER APPLICATION
Protein
Amino Acids (aromatic)
Pantothenic Acid
Glucose Determination
Enzyme Activity (Hexokinase)
VISIBLE SPECTROPHOTOMETER APPLICATION
Niacin
Pyridoxine
Vitamin B12
Metal Determination (Fe)
Fat-quality Determination (TBA)
Enzyme Activity (glucose oxidase)
EXAMPLES
1.
Calculate the Molar Extinction Coefficient ε at 351 nm for
aquocobalamin in 0.1 M phosphate buffer. pH = 7.0 from the
following data which were obtained in 1 Cm cell.
SolutionC x 105 M
Io
I
A
2.23
100
27
B
1.90
100
32
2.
The molar extinction coefficient (ε) of compound
riboflavin is 3 x 103 Liter/Cm x Mole. If the absorbance reading
(A) at 350 nm is 0.9 using a cell of 1 Cm, what is the
concentration of compound riboflavin in sample?
3.
The concentration of compound Y was 2 x 10-4
moles/liter and the absorption of the solution at
nm using 1 Cm quartz cell
was 0.4. What is
molar extinction coefficient of compound Y?
300
the
3. Complete the following table.
[X](M)
Absorbance Transmittance(%)
30
0.5
2.5 x 10-3
0.2
4.0 x 10-5
50
2.0 x 10-4
ε(L/mole-cm)
2000
2500
L(cm)
1.00
1.00
1.00
5000
150
[X](M) = Concentration in Mole/L
4. The molar absorptivity of a pigment (molecular weight 300) is 30,000 at 550 nm.
What is the absorptivity in L/g-cm.!
5. The iron complex of o-phenanthroline (Molecular weight 236) has molar absorptivity
of 10,000 at 525 nm. If the absorbance of 0.01 is the lowest detectable signal, what
concentration in part per million can be detected in a 1-cm cell?
A = 0.01
ε= 10000L / mole x cm
L = 1cm
A = ECL
0.01= 10000L/mole X Cm X C (Concentration) x 1Cm
C = mole / Liter
C = X mole / Liter = X mole (236 g/mole) / Liter (1000 Cm3) x PPM
(10-6 g/Cm3)
= X mole (236 g / mole) / Liter x 1 Liter / 1000 Cm3 x ( PPM) 10-6
g / Cm3)
=x PPM
PPM = 1ug / Cm3
1ug = 10-6 g

Multi-component mixture:

AT = A1 + A2 +A3 + An

AT = ε1bc1 + ε2bc2 + ε3bc3 + εnbcn
Where: 1,2,3,….,n – refer to absorbing
components
Potassium dichromate and potassium permanganate in
1 M H2SO4 has absorbance spectra overlap (overlab).
K2Cr2O7 have maximum absorbance at 440 nm and
KMnO4 545 nm. Mixture of both substances have
been analyzed by measuring the absorbance of
solution at the second wavelength with the following
results: A 440 = 0.405, A545 = 0.712. 1.0 cm cell
used. Absorbance K2Cr2O7 pure solution (1 x 10-3
M) and KMnO4 (2 x 10-4 M) in 1 M H2SO4 by using
the same cell are as follows: A Cr, = 0.374 440, A Cr,
545th = 0.009, A Mn, 440 = 0.019, A Mn, 545 =
0.475
Calculate the concentration of dichromate and
permanganate in the sample solution?
Deviations from the direct
 proportionality (b=const)
 Instrumental Deviations
 Chemical Deviations
Deviations from the direct proportionality (b =
const)
 Valid at concentration usually below
0.01M


Molecules interference – distance between
molecules affects charge distribution of
molecules (ions)
Effect of refractive index –concentration affects
refractive index ε is affected (low, less
important)
Chemical Deviations
Association, dissociation and reaction with solvent and
other molecules
Instrumental Deviations
 monochromatic radiation – quality of
monochromator and control of bandwidth
and slit

Instrumental noise – accuracy of
measurement of transmittance – quality of
detector