# Absorption Spectra and the Beer-Lambert Law

Pages: 6 (1007 words) Published: December 17, 2012
Practical 3
Absorption Spectra and the Beer-Lambert Law

INTRODUCTION
Electromagnetic radiation can be described in terms of frequency (V) and wavelength (λ). Absorbance is the capacity of a substance to absorb radiation and transmittance (the fraction radiation at a specified wavelength that passes through a sample) is physical properties that all molecules have. The purpose of experiment one was to find the maximum absorbance (λmax) occurred for both bromophenol blue and methyl orange by using spectrophotometer. Spectrophotometer is a device that measures the absorption of radiation at a particular wavelength. In experiment two, the purpose was to investigate the relation between concentration of the solution and its absorbance at the (λmax) obtained from experiment one for each dye.

Beer's Law states that concentration of a substance is directly proportional to its amount of light absorption (Department of Biology, 2008). The number of molecules of the solute is related exponentially to the amount of light that is absorbed while passing through the solute, called the solute concentration. Using Beer's Law with known absorptivity at the absorption maximum a substance's concentration within a solute can be measured. Since spectrophotometers are constructed to give absorbance, concentration can be figured out through their relationship to each other (Jones, A, et al, 2007).

Methods
All the steps were carried out as accurately as possible as stated in lab manual provided. The only problems encountered were the quick change in value of the sample so reading the absorbance from the spectrophotometer had to be done quickly so that accurate reading could be noted down.

Results
Experiment one

Absorbance at all the different wavelengths tested for each dye: Bromophenol Blue|
Wavelength (nm)| Absorbance| Further readings to determine ‘λmax’| 400| 0.096| Wavelength (nm)| Absorbance|
420| 0.046| 590| 0.895|
440| 0.042| 610| 0.690|
460| 0.054| 595| 0.951|
480| 0.074| 585| 0.794|
500| 0.109| | |
520| 0.177| | |
540| 0.300| | |
560| 0.461| | |
580| 0.706| | |
600| 0.948| | |
620| 0.399| | |
640| 0.008| | |
660| 0.000| | |
680| 0.000| | |
700| 0.000| | λmax = 595nm
λmax = 595nm
|

Methyl orange |
Wavelength (nm)| Absorbance| Further readings to determine ‘λmax’| 400| 0.401| Wavelength (nm)| Absorbance|
420| 0.530| 450| 0.708|
440| 0.644| 470| 0.747|
460| 0.739| 465| 0.746|
480| 0.716| 475| 0.733|
500| 0.592| | |
520| 0.340| | |
540| 0.136| | |
560| 0.033| | |
580| 0.006| | |
600| 0.004| | |
620| 0.001| | |
640| 0.000| | |
660| 0.000| | |
680| 0.000| | λmax = 470nm
λmax = 470nm
|
700| 0.000| | |

Experiment 2:
* Concentration found by using following equation:
C1V1 = C2V2
Where C1 = 10 mg/l
C2 = the unknown
V1 = the volume of stock solution used
V2 =the total volume of stock solution and distilled water

For example,
Tube 1,
The above equation can be rearranged to give:
C1V1
V2
C1V1
V2

Tube 2,
C2 = = 2

Tube 2,
C2 = = 2

C2 =
10 x 0.6
3
10 x 0.6
3

10 x 0.3
3
10 x 0.3
3
Therefore,
C2 = = 1

* To find concentration (%):
% concentration = Volume of stock solution used X 100 Total volume of solution

Tube 4,
% = 1.2 x 100
3
= 40%

= 50%

Tube 4,
% = 1.2 x 100
3
= 40%

= 50%

For example,
Tube 3,
0.9
3
0.9
3

% = x 100

= 30%

* Table showing the components and concentrations of the different tubes: Tube no.| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10|
Stock solution (10mg/l) (ml)| 0.3| 0.6| 0.9| 1.2| 1.5| 1.8| 2.1| 2.4| 2.7| 3.0| Distilled water (ml)| 2.7|...