Ruth Zalalem (1) - Analysis of Food Dyes in Beverages Lab - Checked on 1_4

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Feb 20, 2024

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Analysis of Food Dyes in Beverages Background nnn The color of a solution is an important tool used by scientists to gain information about the composition of the solution. Color is a physical property that is useful for both qualitative and quantitative analysis. A qualitative method yields information about the nature or type of compound in a sample, whereas a quantitative method provides numerical data for the amount of a compound in a sample. Spectroscopy is the study of the interaction of light and matter. A spectrophotometer is an instrument that uses electromagnetic radiation from a selected region of the electromagnetic spectrum, such as ultraviolet, visible, or infrared light, to analyze the absorption or transmission of radiation by a sample. The basic function of a spectrophotometer is shown in Figure 1. The electromagnetic spectrum (see Figure 2) is the entire range of possible wavelengths or frequencies of electromagnetic radiation. In this investigation a visible spectrophotometer will be used – it scans the visible region of the electromagnetic spectrum, from 380 nm to 750 nm. Typical light sources for visible spectrophotometers include xenon and tungsten lamps. Figure 1 Figure 2 Plastic cuvettes may be used as sample cells for visible spectrophotometers. More specialized spectrophotometers require quartz cells, which are “invisible” to and do not absorb ultraviolet radiation. In addition to the energy sources used in spectrophotometers, a diffraction grating called a monochromator is also incorporated. The monochromator spreads the beam of light into the light’s component wavelengths. The desired wavelength is then focused onto the sample cell to detect any absorption or emission of light by a substance in a sample. Spectrophotometry is an analytical procedure that uses electromagnetic radiation to measure the concentration of a substance. The success of a spectrophotometric technique requires that the absorption of light by the substances being analyzed must be distinct or different from that of other chemical species in solution. How do scientists select the desired wavelength for spectrophotometry? The absorption of visible light by a substance results from electron transitions, that is, the promotion of a ground state electron to a higher energy atomic or molecular orbital. Both light energy and electron energy levels are quantized, so that the specific wavelength of light absorbed by a substance depends on the energy difference between two electron energy levels. The optimum wavelength for spectrophotometric analysis of a substance is selected by measuring the visible spectrum of the substance, corresponding to a plot of absorbance ( A ) versus wavelength (λ, “lambda”). Just seven unique dyes are approved by the Food and Drug Administration for use in foods, drugs and cosmetics. These seven FD&C dyes give rise to the entire palette of artificial food colors. One of these dyes, FD&C Blue 1, is discussed in this inquiry lab for the analysis of sports drinks and other beverages. The structure of FD&C Blue 1 is shown in Figure 3. Notice the extensive series of alternating single and double bonds (also called conjugated double bonds) in the center of the structure. This feature is characteristic of intensely colored organic dyes and pigments. Every double bond added to the system Figure 3

reduces the energy difference between the bonding and nonbonding molecular orbitals so that the resulting energy gap corresponds to visible light. A solution containing FD&C Blue 1 appears blue under normal white light – blue is the color of light transmitted by the solution. The colors or wavelengths of light that are absorbed by this solution are complementary to the transmitted color. A color wheel (Figure 4) provides a useful tool for identifying the colors or wavelengths of light absorbed by a substance. The blue solution absorbs orange light and we would expect the visible spectrum of FD&C Blue 1 to contain a peak in the 600-640 nm region. The optimum wavelength for spectrophotometric analysis of a dye solution is generally determined from the wavelength of maximum absorbance (abbreviated λ max or “lambda max”). The value of lambda max for FD&C Blue 1 is 630 nm. Figure 4 The wavelength of light absorbed by a substance is characteristic of its molecular or electronic structure. The intensity of light absorbed depends on the amount of the substance in solution. Generally, the more concentrated the solution, the more intense the color will be, and the greater the intensity of light the solution absorbs. A digital spectrophotometer measures both the percent transmittance of light and the absorbance. When light is absorbed, the radiant power ( P ) of the light beam decreases. Transmittance ( T ) is the fraction of incident light ( P/P o ) that passes through the sample (see Figure 5). The relationships between transmittance and percent transmittance ( % T ) and between transmittance and absorbance ( A ) are given in Equations 1 and 2, respectively. % T = T x 100 = P/P o x 100% Equation 1 A = absorbance = -log T Equation 2 Figure 5 The amount of light absorbed by a solution depends on its concentration ( c ) as well as the path length of the sample cell ( b ) through which the light must travel. See Equation 3, which is known as Beer’s law. The constant a in the equation is a characteristic of a substance and is known as the molar absorptivity coefficient. A = abc Equation 3 Pre-Lab The visible absorption spectrum for FD&C Blue #1 is shown in Figure 6. The estimated concentration of the dye was 7.0E10 -6 M. (FYI: The estimated concentration of FD&C Red 40 is 3.2E-5 M. The estimated concentration of FD&C Yellow 5 is Figure 6

1. What would an optimum wavelength for measuring the absorbance versus concentration of FD&C Blue #1 dye solution? Absorbance measurements are most accurate and sensitive in the range 0.2-1.0. 630 nm 2. To construct a calibration curve, a series of known concentration standards is prepared. Using the estimated concentration of the FD&C Blue #1 stock solution, determine the concentration of each of the following dilutions. Use: M 1/ V 1 = M 2/ V 2 Dye Stock Solution (A) B C D E F G H Concentration (μM) 7.00 5.60 4.20 2.80 2.10 1.40 0.700 0 Water (mL) 0 2 4 6 7 8 9 10 Stock Solution (mL) 10 8 6 4 3 2 1 0 3. Using the information provided in Question #1, predict the absorbance of each solution A-H at the optimum wavelength. Refer to Equation 3 in the Introduction . The value of a is 130,000 M -1 cm -1 and b = 1 cm. (Don’t forget to convert concentration from μM to M.) A= abc A- (130,000 M -1 cm -1 ) ( 1 cm) ( 7e-6) = 0.910 B- (130,000 M -1 cm -1 ) ( 1 cm) ( 5.6e-6)= 0.728 C- (130,000 M -1 cm -1 ) ( 1 cm) ( 4.2e-6)= 0. 546 D- (130,000 M -1 cm -1 ) ( 1 cm) ( 2.8e-6)=0.364 E-(130,000 M -1 cm -1 ) ( 1 cm) ( 2.1e-6)= 0.273 F-(130,000 M -1 cm -1 ) ( 1 cm) ( 1.4e-6)= 0.182 G-(130,000 M -1 cm -1 ) ( 1 cm) ( 0.7e-6)= 0.091 H-(130,000 M -1 cm -1 ) ( 1 cm) ( 0e-6)= 0 Materials (list materials below) -cuvettes -calorimeter -distilled water - stock solution (red food dye) - test tubes -plastic pipettes -pipette bulb - test tube racks - beverage containing dye Procedure 1. Calibrate the colorimeter using distilled water. Based on the maximum absorbance of the dye being tested at your station, select the appropriate wavelength on the colorimeter. 2. Using a graduated pipet for volume measurements, dilute the stock solution as indicated in the following table to prepare 5 mL of each of a series of standard solutions, A-H. Carefully mix each solution. To avoid contaminating the solutions, use a pipet for the stock solution and a separate pipet for the distilled water. Solution Stock (A) B C D E F G Water (mL) 0 1 2 3 3.5 4 4.5 Stock solution (mL) 5 4 3 2 1.5 1 0.5 3. Measure and record the percent transmittance (% T) of the stock solution and each standard solution (B-G) at the optimum wavelength. Remember to handle the cuvettes only at the top and polish the cuvette with lens tissue before inserting it into the colorimeter. Do not fill the cuvette more than ¾ full. Red Food Dye Sample T

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A 0.986 B 0.961 C 0.934 D 0.761 E 0.729 F 0.512 G 0.414 Steps 4-6 are to be completed outside of class. 4. Convert % T to transmittance ( T ) for each measurement, and calculate the appropriate values of both log T and –log T. Record all results in a table. Sample T 1/T %T log T -log T micromolar (μM) concentration A 0.986 1.01 98.6 % -0.00612 0.00612 2.87 e-7 B 0.961 1.04 96.1 % -0.0173 0.0173 C 0.934 1.07 93.4 % -0.0297 0.0297 D 0.761 1.31 76.1 % -0.119 0.119 E 0.729 1.37 72.9 % -0.137 0.137 F 0.512 1.95 51.2 % -0.290 0.290 G 0.414 2.42 41.4 % -0.383 0.383 5. Use Beer’s law to calculate the precise concentration of the stock solution (A). The molar absorptivity ( a ) of FD&C Blue #1 is 1.3E5 M -1 cm -1 at 630 nm. The molar absorptivity ( a ) of FD&C Red #40 is 2.13E4 M -1 cm -1 at 565 nm . The molar absorptivity ( a ) of FD&C Yellow is 2.43E4 M -1 cm -1 at 430 nm. The path length ( b ) is 1 cm. Record the micromolar (μM) concentration (1 μM = 1 x 10 -6 M) in your data table. A= -log T A= -log(.986) A= 0.00612 c= A/ab c= (0.00612)/(2.13E4 M -1 cm -1 )( 1 cm) c= 2.87e-7 6. Prepare 5 separate graphs that compare (a) % T, (b) T, (c) 1/T, (d) log T, and (e) –log T (on the y-axis) versus dye concentration (x-axis) for your solution. Dye concentrations were calculated in the Pre-Lab Questions using the estimated concentration of the stock solution. a. % T

b c. d. e. -Log T Concent ration l o g T Concentration 1 / T Concentrati on T Concentration

7. Measure and record the percent transmittance of the beverage containing the dye. Recall that absorbance measurements are most accurate in the range of 0.2 to 1.0. A= - logT 0.367= - log T T= 0.430 8. Determine the concentration (micromolar, μM) of the dye in the beverage and calculate the amount (mass) of dye in milligrams per liter of the beverage. The molar mass of FD&C Blue #1 dye is 793 g/mol. The molar mass of FD&C Red 40 is 496.42 g/mol. The molar mass of FD&C Yellow 5 is 534.3 g/mol. c= A/ ab c= (0.367)/ (2.13E4 M -1 cm -1 )( 1 cm) c= 1.72e-5 μM (1.72e-5 μM)(e-6)= 1.72e-11 M M= mol/ L 1.72e-11 M= ( Analysis and Conclusions 1. Calculate the value of %T for an absorbance value A = 1.5. Use these results, explain why absorbance measurements that are greater than one may not be accurate. A= -log T 1.5 = -log T T=0.316 %T= 3. 16 Absorbance measurements greater than 1 may not be accurate because it is overly concentrated and the percent transmission will get closer to 100 % . 2. Spectrophotometric studies can be conducted on any colored compound. The transition metal group of the periodic table exhibits a wide array of different colored compounds. The complex ion tetraaminecopper (II) contains four ammonia molecules covalently bonded to a copper (II) ion. In aqueous solutions, Cu 2+ ions will bond to four water molecules in a square planar geometry. The solution is a light blue color. The water molecules can be displaced by ammonia molecules, which form more stable complex bases than water. The appearance of the intense dark blue- violet color of the [Cu(NH 3 ) 4 ] 2+ ion is often used as a positive test to verify the presence of Cu 2+ ions. a. Write a balanced chemical equation for the reaction of copper (II) sulfate and concentrated ammonia to produce tetraaminecopper (II) sulfate. CuSO4(aq) + 4NH3(aq) + H2O [Cu(NH3)4H2O]S04(aq) → b. [Cu(NH 3 ) 4 ] 2+ solutions exhibit a deep blue-violet color. How can you use spectrophotometry to confirm this reaction has occurred and that the product formed is in fact tetraaminecopper (II) sulfate? Would you expect the wavelength of maximum absorbance (λ max ) for Cu(NH 3 ) 4 2+ to be greater than or less than λ max for Cu(H 2 O) 6 2+ ? Explain. You can use spectroscopy to confirm the reaction by the difference in wavelengths of the maximum absorbance. If [Cu(NH 3 ) 4 ] 2+ has a different wavelength it indicates that the absorbance has changed, therefore a product. I would expect the wavelength of maximum absorbance of Cu(NH 3 ) 4 2+ to be less than Cu(H 2 O) 6 2+ . This is because the absorbance is less, therefore it is a shorter wavelength. In addition to this, it absorbs similar to the wavelength of the yellow lights, then the solution looks like a deep-violet color. Concentration

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Ruth Zalalem (1) - Analysis of Food Dyes in Beverages Lab - Checked on 1_4

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