Separation of a Dye Mixture Using Chromatography
Inquiry Kit for AP® Chemistry
Materials Included In Kit
Food Dye FDC Blue No. 1, 2 g
Food Dye FDC Blue No. 2, 2 g
Food Dye FDC Green No. 3, 1 g
Food Dye FDC Red No. 3, 2 g
Food Dye FDC Red No. 40, 2 g
Food Dye FDC Yellow No. 5, 2 g
Food Dye FDC Yellow No. 6, 2 g
Isopropyl alcohol, CH3CHOHCH3, 70%, 250 mL
Sodium chloride solution, NaCl, 20%, 500 mL
Chromatography paper strips, 200
Additional Materials Required
Water, distilled or deionized*†
Balance, 0.01-g precision†
Beakers, 100-mL, 2*
Beakers, 250-mL, 7†
Erlenmeyer flasks, 250-mL, 2*
Graduated cylinder, 25-mL*
Graduated cylinders, 10-, 25- and 100-mL†
Stirring rods, 7†
Volumetric flask, 500-mL†
Watch glasses, 2*
*for each lab group
†for Prelab Preparation
- To prepare 100 mL of individual dye solutions, add 0.5 g of each FD&C solid dye to a separate beaker with 100 mL of distilled or deionized water. Mix thoroughly.
- To prepare the dye mixture for the Introductory Activity, combine 10 mL of Red No. 40, Blue No. 1 and Yellow No. 5 dye solutions in one beaker.
- To prepare 500 mL of 2% sodium chloride solution, fill a 500-mL volumetric flask one-third to one-half full with distilled or deionized water. Add 50 mL of 20% NaCl solution and dilute with distilled or deionized water to the mark. Mix thoroughly.
- To prepare 500 mL of 2% isopropyl alcohol solution, measure 14.3 mL of 70% isopropyl alcohol solution. Pour this into a 500-mL volumetric flask and dilute with distilled or deionized water to the line. Cover and mix thoroughly before dispensing.
- See the Lab Hints section for suggestions for unknown mixtures of dyes.
Isopropyl alcohol is a moderate fire risk and is slightly toxic by ingestion or inhalation. Use proper exhaust ventilation to keep airborne concentrations low. The FD&C dyes are slightly hazardous by ingestion, inhalation, and eye or skin contact. Red No. 40 may be absorbed through skin and Yellow No. 5 may be a skin sensitizer. All dyes are irritating to skin and eyes. Avoid contact with eyes, skin and clothing. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron. Remind students to wash their hands thoroughly with soap and water before leaving the laboratory. Please review current Safety Data Sheets for additional safety, handling and disposal information.
Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. Excess dye solutions and sodium chloride solution may be stored for future use or rinsed down the drain with excess water according to Flinn Suggested Disposal Method #26b. Small quantities of excess isopropyl alcohol solutions may be rinsed down the drain with excess water according to Flinn Suggested Disposal Method #18a.
- This laboratory activity can be completed in two 50-minute class periods. It is important to allow time between the Introductory Activity and the Guided-Inquiry Activity for students to discuss and design the guided-inquiry procedures. All student-designed procedures must be approved for safety before students are allowed to implement them in the lab. Prelab Questions may be completed before lab begins the first day, and the data compilation and calculations may be completed after the lab or as homework.
- Enough chromatography paper strips are included for 12 groups of students to develop 16 chromatograms each. Extra strips are provided in case of mistakes.
- Students should avoid over-handling the chromatography strips. Oil from the skin can interfere with the capillary action that draws water through the paper.
- In the Introductory Activity portion of the lab, it is recommended that half of the groups use 2% sodium chloride solution as the developing solvent and the other half use 2% isopropyl alcohol as the developing solvent. This split will allow for adequate testing of both solvents, providing enough data for students to compare benefits and drawbacks of both solvents.
- Good technique is important to achieve clean separations in paper chromatography. Common sources of student error include “overloading” the paper by placing too much dye on the initial spot and the band broadening that occurs because the initial spot is too large.
- Suggestions for unknown dye mixtures are below. The blue, red and yellow mixtures are suggested for identification of dyes found in commercial food products. Mix equal volumes of each dye desired to create a mixture.
Purple: Combine Blue No. 1 and Red No. 3, or Blue No. 2 and Red No. 3, or Blue No. 1 and Red No. 40.
Orange: Combine Red No. 40 and Yellow No. 6.
Green: Combine Blue No. 2 and Yellow No. 6; or Blue No. 1 and Yellow No. 6; or Blue No. 1, Yellow No. 5 and Green No. 3; or Blue No. 2, Yellow No. 6 and Green No. 3.
Blue: Combine Blue No. 1 and Blue No. 2.
Red: Combine Red No. 3 and Red No. 40.
Yellow: Combine Yellow No. 5 and Yellow No. 6.
- It is critical to allow enough time for the development of the chromatography paper. The chromatography paper must be left in the chromatography chamber long enough for the solvent to be drawn up near the top of the strip. Do not stop the development until the solvent front nears the top of the strip. Underdevelopment will lead to incomplete separation. Do not allow the solvent front to move off the paper, however.
- The developing solvent and dyes will continue to move even after the paper strip is removed from the solvent. It is necessary for students to mark the solvent front and positions of the dye bands or spots immediately after the strip is removed from the flask.
- As students plan their investigation in the inquiry portion, they must remember to run a control or baseline trial. The 2% sodium chloride and 2% isopropyl alcohol solutions are convenient baseline runs because those were used in the Introductory Activity.
Opportunities for Inquiry
As noted in the Background section, FD&C food dyes are used in a wide range of food products, most notably the outer shells of candies. Candy may be placed in 5–6 drops of water. Stir the candy until the color dissolves. Repeat with two more candies. This is the color sample. Design an experiment to determine the composition of the dye mixture in the candy shell.
Alignment to the Curriculum Framework for AP® Chemistry
Enduring Understandings and Essential Knowledge
Matter can be described by its physical properties. The physical properties of a substance generally depend on the spacing between the particles (atoms, molecules, ions) that make up the substance and the forces of attraction among them. (2A)
2A3: Solutions are homogenous mixtures in which the physical properties are dependent on the concentration of the solute and the strengths of all interactions among the particles of the solutes and solvent.
Forces of attraction between particles (including the noble gases and also different parts of some large molecules) are important in determining many macroscopic properties of a substance, including how the observable physical state changes with temperature. (2B)
2B2: Dipole forces result from the attraction among the positive ends and negative ends of polar molecules. Hydrogen bonding is a strong type of dipole-dipole force.
2B3: Intermolecular forces play a key role in determining the properties of substances, including biological structures and interactions.
2.7 The student is able to explain how solutes can be separated by chromatography based on intermolecular attractions.
2.10 The student can design and/or interpret the results of a separation experiment (filtration, paper chromatography, column chromatography, or distillation) in terms of relative strength of interactions among and between the components.
2.13 The student is able to describe the relationships between the structural features of polar molecules and the forces of attraction between particles.
1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively.
4.2 The student can design a plan for collecting data to answer a particular scientific question.
4.3 The student can collect data to answer a particular scientific question.
5.1 The student can analyze data to identify patterns or relationships.
5.2 The student can refine observations and measurements based on data analysis.
5.3 The student can evaluate the evidence provided by data sets in relation to a particular scientific question.
6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices.
6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
Answers to Prelab Questions
- Figure 1 is a sample paper chromatogram for three samples: A, B and C. Label the drawing with the following items: the stationary phase, the mobile phase and the solvent front.
- Calculate the Rf value for the spot in sample B using sample A as an example.
- Sample C gave two spots on the paper chromatogram. What does this tell you about the composition of the sample?
Sample C is a mixture of at least two components, A and B.
- Based on the Rf values of samples A and B, what can you conclude about the intermolecular attractions both samples have for the eluent and the paper?
The Rf value of sample A is larger than sample B. Sample A has a stronger affinity (attraction) for the eluent (solvent) than the paper, so it traveled farther with the solvent. Sample B has a stronger attraction for the paper than the solvent, so it traveled a shorter distance.
Observations for 2% Sodium Chloride Solution
Total time for chromatograms to develop ranged from 25–30 minutes. Three separate dye spots were visible: blue on top, yellow in the middle, and red on the bottom. Each spot was fairly spread out, with the largest band being the Blue No. 1 dye. All three dyes traveled the same direction in a straight line (not curved).
Observations for 2% Isopropyl Alcohol Solution
Total time for chromatograms to develop ranged from 25–30 minutes. Two separate dye spots were visible: blue and yellow on top, and red on the bottom. The Blue No. 1 and Yellow No. 5 colors overlapped resulting in a blue front followed by a green line then a yellow tail end. The blue-yellow color band traveled straight for a portion of the time, and then the edges began trailing while the middle continued at a “faster” rate. The shape of the blue-yellow color band was a crescent. The top of the Red No. 40 color spot was near the middle of the crescent.
Comparison of Solvent Concentration on Dye Mixture Separation
Based on the data and observations of the Introductory Activity, the sodium chloride solvent was chosen for further investigation. The concentration of the solvent was increased by a factor of four to 8% and decreased by a factor of four to 0.5%. The data table below summarizes the findings.
*Values are averaged from two trials each.
†Rf values are copied from Introductory Activity here for comparison.
In the 8% sodium chloride solvent, total time for chromatograms to develop ranged from 20–30 minutes. All three dyes were visible: blue on top, yellow in the middle, and red on the bottom. Blue No. 1 was the largest band. The Yellow No. 5 band overlapped with roughly half of the Red No. 40 band. These two bands were much closer than in the 2% sodium chloride solvent. All three dyes traveled in the same direction in a straight line (not curved).
In the 0.50% sodium chloride solvent, the total time for chromatograms to develop ranged from 20–30 minutes. The three dyes were visible: blue on top, yellow in the middle, and red on the bottom. Red No. 40 was the largest band. The Yellow No. 5 band overlapped with the very top of the Red No. 40 band. These two bands were much closer than in the Introductory Activity. All three dyes traveled in the same direction in a straight line (not curved).
Based on the data collected for the 8% and 0.50% sodium chloride solutions, the more dilute solvent separated the mixture of dyes better. The three dyes were more distinguishable from one another in the 0.50% solution than in the 8% solution. This was evident with the overlapping of the Yellow 5 and Red 40 dyes with the 8% solvent. The 0.50% sodium chloride solution was the more optimal solvent, compared to the 2% and 8% solutions, because the dyes traveled farther on the paper and there were greater distances separating the dye bands.Effect of Solvent Concentration on FD&C Food Dyes
The following chart is a comparison of the seven FD&C food dyes at two different concentrations of sodium chloride solutions: 2% and 0.10%.** The dyes were run simultaneously on the same piece of chromatography paper. Green No. 3 and Blue No. 1 have very similar Rf
values in both 2% and 0.10% sodium chloride solutions. Similarly, Red No. 40 and Blue No. 2 have similar Rf
values in the two sodium chloride solutions. These two pairs of dyes will be the most difficult to separate.
** The 0.10% concentration was chosen based on the work of Peter Markow (see the References section).
Answers to Questions
- Examine the structures of the FD&C Red No. 40, Blue No. 1 and Yellow No. 5 dyes. What are the similarities and differences in the structures of the three dyes?
All three dyes, Red No. 40, Blue No. 1 and Yellow No. 5, have sulfonate (—SO3¯) functional groups. However, Blue No. 1 has the most sulfonate groups (three); Red No. 40 and Yellow No. 5 both have two. Blue No. 1 also has a positively charged nitrogen atom and Yellow No. 5 has a carboxylate group (—CO2–). Red No. 40 and Yellow No. 5 both have double bonded nitrogen atoms near the middle of the structures and a single —OH group. Red No. 40 has an —OCH3 group and a methyl group on the leftmost benzene ring. Blue No. 1 is the largest molecule of the three dye molecules.
- In the Introductory Activity, the developing solvents were 2% sodium chloride aqueous solution and 2% isopropyl alcohol aqueous solution. Draw separate molecular diagrams of how sodium chloride and isopropyl alcohol would interact in water. Identify the types of intermolecular attractions within each diagram.
- Based on the diagrams and intermolecular attractions identified in Question 2, predict and compare the nature of intermolecular attractions experienced by the FD&C Red No. 40, Blue No. 1 and Yellow No. 5 dyes with the two solvents.
In the sodium chloride solution, all three dye molecules would experience ion–dipole interactions due to the charged sulfonate groups, although the overall strength of these interactions will vary due to the number of groups on each molecule. Similarly, each dye molecule will experience hydrogen bonding between water and the functional groups. Blue No. 1 would experience the strongest interactions because it has the most charged side groups—three SO3– groups and a positively charged nitrogen atom. Red No. 40 and Yellow No. 5 would experience weaker ion–dipole and hydrogen bonding interactions with the sodium chloride solution because the molecules have fewer charged side groups. Yellow No. 5 experienced a stronger interaction with the sodium chloride solution because it has a charged carboxylate group in addition to two charged sulfonate groups. Red No. 40 only has two charged sulfonate groups.
In the isopropyl alcohol solution, all three dye molecules would experience ion–dipole interactions due to the charged functional groups and polar alcohol group in isopropyl alcohol. In addition to their changed side chains, all of the food dyes are large organic molecules with significant nonpolar rings and groups. These nonpolar regions interact with relatively nonpolar isopropyl alcohol molecules and thus have a greater affinity for this solvent than for either the NaCl solution or the hydrophilic paper substrate.
- Chromatography paper, and paper in general, is highly hydrophilic. Paper is made from a natural polymer called cellulose, which is a long chain of glucose molecules. Glucose is a cyclic structure with a number of —OH groups around the ring.
- Predict and explain the types of intermolecular forces that would occur between paper and water. How do these interactions account for the hydrophilic nature of paper?
The —OH groups around the glucose rings are sites for ion–dipole interactions and hydrogen bonding with water. The hydrogen bonding interaction between the paper and water will be strong because the —OH groups will be able to interact strongly with the hydrogens and oxygen in water due to the large dipole moments. Paper has a strong affinity for water and draws water up by capillary action.
- Explain the types of intermolecular interactions that would occur between the FD&C Red No. 40, Blue No. 1 and Yellow No. 5 food dyes and the paper.
The —OH groups around the glucose rings are sites for hydrogen bonding with the charged functional groups on the dye molecules. In addition, Red No. 40 would experience hydrogen bonding at the —OH group and —OCH3 group. Yellow No. 5 would have hydrogen bonding at the —OH group, as well. In both Red No. 40 and Yellow No. 5, hydrogen bonding would occur with the double-bonded nitrogen atoms and —OH groups around the glucose rings.
Answers to Review Questions for AP® Chemistry
- Hydrocarbons are nonpolar compounds containing carbon and hydrogen atoms. The properties of three hydrocarbons are summarized.
- How do the attractive forces between molecules change in the transition from the gas to the liquid to the solid state?
Attractive forces between molecules increase in the order gas << liquid < solid. Molecules in the gas state are very far apart—there are almost no attractive forces between the molecules. As gases condense into liquids and then solidify, the molecules get closer together and the strength of attractive forces between molecules increases. Attractive forces between molecules are strongest in the solid state, because the molecules are locked into fixed positions.
- Based on its properties, which compound has the strongest attractive forces? The weakest attractive forces?
Eicosane, a hydrocarbon with 20 carbon atoms, has stronger intermolecular attractive forces than octane or methane, which contain eight carbon atoms and one carbon atom, respectively. Methane has the weakest attractive forces.
- Write a general statement describing how the size of a molecule influences the strength of London dispersion forces between molecules.
The strength of London dispersion forces between molecules increases as the size of the molecules increases (all other factors being equal).
- Dyes are organic compounds that can be used to impart bright, permanent colors to fabrics. The affinity of a dye for a fabric depends on the chemical structures of the dye and fabric molecules and also on the interactions between them. Three common fabrics are wool, cotton and nylon. Wool is a protein, a naturally occurring polymer made up of amino acids with ionized (charged) side chains. Cotton is a naturally occurring polymer made up of glucose units with hydrophilic groups surrounding each glucose unit. Nylon is a synthetic polymer made of hydrocarbon repeating chains joined together by highly polar amide (–CONH–) functional groups.
- The chemical structure of methyl orange is drawn below. Identify the groups in the dye that will bind to ionic and polar sites in a fabric.
- Complete the following “If/then” hypothesis to explain how the structure of a fabric will influence the relative color intensity produced by methyl orange.
“If a fabric contains more ionic and polar groups in its structure, then the intensity of the dye color due to methyl orange should increase, because there would be more sites on the fabric for the dye molecules to bind.”
- Using this hypothesis, predict the relative color intensity that would be produced by methyl orange on cotton, nylon and wool. Rank the fabrics from 1 = lightest color to 3 = darkest color.
1 = nylon, 2 = cotton, 3 = wool
AP® Chemistry Guided-Inquiry Experiments: Applying the Science Practices; The College Board: New York, NY, 2013.
Markow, P. G. The Ideal Solvent for Paper Chromatography of Food Dyes. J. Chem. Ed. 1988, 65, 10, pp 899–900.