Introduction to beer lambert law

Introductory description and the limitation of Beer-Lambert law.

1. Page 1 of 6
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ANALYTICAL CHEMISTRY PART I
Lambart Law: This law states that when monochromatic light passes through a transparent
medium, the rate of decrease in intensity with the thickness of the medium is proportional to the
intensity of the light.
This is equivalent to stating that the intensity of the emitted light decreases exponentially as the
thickness of the absorbing medium increases arithmetically, or that any layer of given thickness
of the medium absorbs the same fraction of the light incident upon it.
The law can be expressed as follows-
−
�??????
��
∝??????
−
�??????
��
=????????????………………………………(??????)
Where, I= Intensity of the incident light of wavelength λ
l= Thickness of the medium
a= Proportionality factor
Now, Integrating above equation,
∫−
�??????
??????
=??????∫��
??????
??????
0

��,��??????
????????????−��??????
????????????
0=−??????�
��,��??????
??????
??????
??????
0
=−??????�
��,
I
??????
0
=�
−��

��,??????=??????
0�
−��
………………………………………… ( ii)
Beer studied the effect of concentration of the colored constituent in solution upon the light
transmission or absorption. He found the same relationship between transmission and
concentration as Lambert had discovered between transmission and thickness of the layer, i.e.
the intensity of a beam of monochromatic light decreases exponentially as the concentration of
the absorbing substance increases arithmetically.
??????=??????
0�
−��
…………………………………………(??????????????????)
From combination of equation (ii) and (iii), we get,
??????=??????
0�
−���

��,
??????
??????
0
=�
−���

or,

2. Page 2 of 6
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��,
??????
??????
0
=10
−0.4343���

��,
??????
??????
0
=10
−??????��

��,��??????
??????
0
??????
=??????��……………………………………(????????????)
??????=??????????????????; ∵ ??????=��??????
??????
0
??????

Here, ε= Molar absorption coefficient
Here, equation (iv) is the fundamental of colorimetry and spectrophotometry, and is often
spoken as the Beer-Lambert Law. The value of ε will depend upon the method of expression of
the concentration.
Again, Since the fraction of radiant energy transmitted decays exponentially with path length,
we can write it in exponential form-
�=
??????
??????
0

Where, T is called the transmittance, the fraction of radiant energy transmitted.
The percent transmittance is given by,
%�=
??????
??????
0
×100=100�…………………………...(??????)
??????=��??????
??????
0
??????

��,??????=log (
1
�
)
��,??????=log (
100
%??????
) (from equation v)
��,??????=log(
10
2
%�
)
��,??????=2log10−log%T
∴??????=??????−??????????????????%??????
And, ??????????????????%??????=??????−??????

Question: Why do we prefer to express the Beer-Lambert law using absorbance as a measure
of the absorption rather than %T?
Answer: From Beer-Lambert equation, it is found that-
??????=??????−??????????????????%??????

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Now, suppose, a solution of copper sulfate (which appears blue because it has an absorption
maximum at 600 nm). 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. The reduction every 0.2 cm is shown
in the diagram below.
The Law says that the fraction of the light absorbed by each layer of solution is the same. For
this illustration, it will be supposed that this fraction is 0.5 for each 0.2 cm "layer" and will be
calculated the following data:
Path length / cm 0 0.2 0.4 0.6 0.8 1.0
%T 100 50 25 12.5 6.25 3.125
Absorbance 0 0.3 0.6 0.9 1.2 1.5










A = εcl, shows that absorbance depends on the total quantity of the absorbing compound in the
light path through the cuvette. If absorbance is plotted against concentration, a straight line is
obtained passing through the origin (0,0).
The linear relationship between concentration and absorbance is both simple and
straightforward, which is why it is preferred to express the Beer-Lambert law using absorbance
as a measure of the absorption rather than %T.

Limitations to Beer-Lambert's law
The linearity of the Beer-Lambert law is limited by chemical and instrumental factors. Causes of
nonlinearity include:
a) Deviations in absorptivity coefficients at high concentrations (>0.01M) due to
electrostatic interactions between molecules in close proximity

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b) Scattering of light due to particulates in the sample

c) Fluorescence or phosphorescence of the sample

d) Changes in refractive index at high analyte concentration

e) Shifts in chemical equilibria as a function of concentration
f) 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

g) Stray light


Example: A sample in a 1.0-cm cell transmits 80% light at a certain wavelength. If the
absorptivity of this substance at this wavelenth is 2.0, what is its concentration?
Solution: The percent transmittance is 80%, and so T =0.80.
Also, l= 1 cm and ε= 2 cm
-1
gm
-1
L

So,
??????=log(
1
�
)=??????��
��.log(
1
0.80
)=2×1×�
��,0.0969=2�
��,�=
0.0969
2
=0.04846≈0.05 ??????/??????
Answer: 0.05 g/L.

Example: A solution containing 1.00mg iron (as the thiocyanate complex) in 100mL was
observed to transmit 70.0% of the incident light compared to an appropriate blank.
(a) What is the absorbance of the solution at this wavelength?
(b) What fraction of light would be transmitted by a solution of iron four times as
concentrated?
Solution:
(a) Here, Percent transmittance is 70%, and so T =0.70
??????=log(
1
�
)=��??????(
1
0.70
)=log1.43
∴??????=0.155

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(b) According to Beer’s law, absorbance is linearly related to concentration. If the original
solution has an absorbance of 0.155, a four times more concentrated solution will have
an absorbance four times as much or 4×0.155=0.620.

The transmittance T will be: T =10−0.620 =0.240

Answer: (a) 0.155
(b) 0.240

Example: Amines, RNH2, react with picric acid to form amine picrates, which absorb strongly
at 359nm (ε=1.25×10
4
). An unknown amine (0.1155g) is dissolved in water and diluted to
100mL. A 1-mL aliquot of this is diluted to 250mL for measurement. If this final solution
exhibits an absorbance of 0.454 at 359nm using a 1.00-cm cell, what is the formula weight of the
amine?
Solution: Here,
ε=1.25×10
4
cm
-1
gm
-1
L
l = 1 cm
A= 0.454
We know that, A= εcl
��,0.454=1.25×10
4
×1×c
�=3.63×10
−5
���/??????
(3.63×10
−5
���/??????)(0.250 ??????)
1��
×100 �??????=9.08×10
−4
���
∴
0.1155g
9.08×10
−4
���
=127.2 ??????/���

Answer: 127.2 g/mol

Example: Chloroaniline in a sample is determined as the amine picrate as described in last
Example. A 0.0265-g sample is reacted with picric acid and diluted to 1L. The solution exhibits
an absorbance of 0.368 in a 1-cm cell. What is the percentage chloroaniline in the sample?
Solution: Here,
ε=1.25×10
4
cm
-1
gm
-1
L
l = 1 cm

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A= 0.386
We know that, A= εcl
��,0.386=1.25×10
4
×1×c
�=2.94×10
−5
���/??????
(2.94×10
−5
���/??????)×(127.6 ??????/���) =3.75×10
−3
?????? �ℎ����??????�??????�??????��
3.75×10
−3
?????? �ℎ����??????�??????�??????��
2.65×10
−2
?????? �??????����
×100%=14.2%


Answer: 14.2%