Since ancient times, organic dyes have been used for coloring cloth and leather. Numerous species of — Organic Chemistry Chemistry Question
Separation of a dye mixture using thin-layer chromatography (TLC)
Since ancient times, organic dyes have been used for coloring cloth and leather. Numerous species of plants and animals have been used as sources of natural dyes. The extraction and purification of dyes as well as the dyeing process itself are sophisticated chemical processes. The first human-made (synthetic) organic dye, mauveine, was discovered as late as the 19th century. Since then, however, thousands of synthetic dyes have been used for various purposes extending beyond coloring, including indispensable uses for digital photo-recording media such as compact discs (CDs) and digital versatile discs (DVDs). The apparent color of a dye solution comes from the absorption of light preferred by the dye molecule. When a dye solution in a transparent vessel is seen against a white background, the complementary color of its absorption can be recognized.
In this experiment, you will learn the basic principles and procedures for separating and distinguishing individual dyes from their mixture.
[VISUAL]
Chemicals
* organic dyes (respective names of dyes intentionally hidden)
* methanol
* developer (mixture of methanol and water (90/10 = v/v))
Apparatuses and glassware
* glass capillary
* wide-mouth bottle with cap (developing chamber) (× 3)
* TLC plates
Note:
If P-2 and P-3 plates are not available, prepare them from modified silica gel and calcium sulphate (CaSO4·1/2 H2O; binder) (you can use silica gel with calcium sulphate instead of pure calcium sulphate). Typically, slurry modified silica gel and calcium sulphate in a methanol/water mixture (2/18) and homogenize the slurry using an electric blender. Spread the slurry on a glass plate. Dry and then activate it at 110 – 130 °C.
Procedures
Using a pencil, draw a starting line approximately 10 mm above the shorter edge of a silica gel plate.
1. Draw cross marks on the line as chromatography starting points.
2. Use a glass capillary to collect some of the sample solution, spot the solution lightly on one of the starting points, and dry the spot with a dryer, if necessary. Repeat this operation a few times to concentrate the sample in a small spot, measuring less than 2 mm in diameter.
3. Pour developing solvent into the respective wide-mouth bottles about 5 mm in height.
4. Close the caps and wait a few minutes until the bottles are saturated with solvent vapour.
5. Open the cap of one bottle and grip the upper edge of a TLC plate with tweezers. Place the TLC plate in the bottle so that the bottom of the plate is immersed in the solvent and the top of the plate is leaning against the wall of the bottle. The solvent should be drawn straight up.
6. Finish the development when the solvent front reaches about 10 mm below the upper edge of the TLC plate.
7. Take the TLC plate out and immediately mark the front line of the developing solvent with a pencil.
8. Record the shapes and colors of the spots.
9. Use the same steps to develop the other TLC plates.
Code Stationary phase
P-1 Silica modified by octadecylsilyl ligands
P-2 Silica modified by anion-exchange ligand
P-3 Silica modified by cation-exchange ligand
[VISUAL]
[VISUAL]
Questions
I. TLC results
From the spots recorded on the TLC plates, calculate the Rf value of each dye on each plate.
Rf = a / b
a = distance from the starting point to the center of gravity of the sample spot.
b = distance from the starting point to the front of the developing solvent.
Determine the color of dyes A, B, and C by considering the nature of the surface of the TLC plate and the properties of the molecules (acidic or basic and hydrophilic or hydrophobic) anticipated by the structural formulae.
Model Answer
Dye A (rhodamine B) is red, dye B (brilliant blue) is blue and dye C (berberine chloride) is yellow.
Explain concisely how you reached your conclusion.
Model Answer
Judging from their structural formulae, dyes A and B have relatively larger numbers of dissociable sites than dye C. Obviously, dye A comprises cationic molecules and dye B anionic molecules, on the whole. Dye C also comprises cationic molecules but it should have weaker hydrophilicity than the others. The TLC results indicate that (1) the red dye is strongly retained on the cation-exchange plate, (2) the blue dye is strongly retained on the anion-exchange plate, and (3) the yellow dye is strongly retained on both the cation-exchange and ODS-modified plates. The conclusion entered in answer 2 was reached based on these observations.
II. Absorption spectra
The apparent color of a dye solution comes from the light absorption preferred by the dye molecule. We can obtain more in-depth information on dyes from their optical absorption spectra. The figure shows the absorption spectrum obtained by measuring the 3.30×10–6 mol dm–3 solution of one of dyes A – C using a cuvette with a 10-mm optical path length. Maximum absorbance (0.380) is observed at 545 nm, which corresponds to the wavelength of yellow-green light.
[VISUAL]
The following are questions concerning the phenomena of light absorption and the Beer–Lambert law.
Calculate the molar absorption coefficient of the dye at 545 nm.
Model Answer
According to the Beer–Lambert law, the absorbance, A, is given by following formula: A = ε × c × l . The molar absorption coefficient of the dye at 545 nm is: ε = 0.380 / (3.30 × 10–6 mol dm–3 × 1.0 cm) = 1.15×105
Calculate the % transmittance of the dye solution at 545 and 503 nm (the absorbance is 0.100 at 503 nm). Then calculate the % transmittance that will be measured at each wavelength when the dye solution is diluted by 50%. By comparing these results, estimate which wavelength of the light source results in a more sensitive change in transmittance when the concentration of the dye solution is modified.
Model Answer
According to Beer–Lambert law, the relationship between % transmittance and absorbance is given as follows: %T = 10–A × 100. Hence, the %T for the original dye solution, is 41.7 % and 79.4% at 545 and 503 nm, respectively. When the concentration of dye is decreased to 50%, on the other hand, the absorbance will be decreased to 50%, since absorbance is proportional to the concentration of the species which absorbs the light. Hence, the absorbance will be 0.190 and 0.0500 at 545 and 503 nm respectively, and the %T calculated from the absorbance will be 64.6 % and 89.1 % at 545 and 503 nm, respectively. Thus, by diluting the dye solution by 50%, the intensity of the transmitted light is increased by 1.55-fold and 1.12-fold at 545 nm and 503 nm, respectively. The above results suggest that wavelengths that show larger absorptivity are more sensitive to changes in the concentrations of species that absorb the light.
Calculate both the absorbance and % transmittance obtained for the original dye solution at 545 nm using a cuvette with a 30-mm path length.
Model Answer
Since absorbance is proportional to the absorption pass length, when the pass length is increased 3-fold, the absorbance will be increased 3-fold. Hence, the absorbance will be 1.14, and the %T calculated from the absorbance will be 7.24 %.