Light Spectrometry of Food Dye

Light Spectrometry of Food Dye Abstract The food dye experiment was done to determine the varieties of food dye present in an assigned solution. To reach this part first the absorptions and wavelengths were recorded through the light spectrometer for the basic color dye (red, blue, yellow) and later Beers law plot was created for each of these colors. The given solution (Gatorade) was also examined through the light spectrometer and it gave a graph of absorption for red, blue and yellow color. Plugging the absorption values for that respective color’s wavelength into the beers law plot, the molarity for each color in the given solution was found. The absorption and molarity for the given solution were: Red (0.156 abs, 1.06x10-5 M), Blue (0.260 abs, 1.420x10-6 M), Yellow (0 abs,0 M). Then using the formula M1V1 = M2V2 the new volume was found which was red (13.166 mL), blue (5.635 ML), yellow (0ml). Through mixing this combination of a specific volume of liquid, the given solution could be determined. Introduction We taste food by our eyes, that’s the power of food color. One of the significant roles that play in food is food dye, and food dye is the quality to make food look more attractive (1). Sometimes natural food colors like carotenoids, chlorophyll, anthocyanin, and turmeric are used in food (2). Using those natural food dyes could be way expensive, so different food colors are prepared in the lab by mixing basic colors (2). There is a limitation for types of natural color could be prepared but for lab-made food colors, there is no limitation (2). Nowadays most food contains the synthesis of food dye; food dye is a quality to make food look more natural and eyecatching (1). Since dyes contain the color substance, they absorb light from the visible spectrum. To find what food dye is present, a UV-vis spectrometer is used. UV-vis spectrometer uses white light for the absorption, and white light is the composition of many different wavelengths. If the solution absorbs light, then less intense light will be received by the receiver (3). The given solution was a purple liquid (Gatorade Grape), using beers law plot the given liquid is replicated. Beers law plot provides a result with for concentration versus abortion which is: A = εbc. Experimental First the basic color dyes were diluted with DI water for specific volumes: 4mL (Red/Blue/Yellow Dye) and 1 mL DI water, 3mL (Red/Blue/Yellow Dye) and 2 mL DI water, 2 mL (Red/Blue/Yellow Dye) and 3 mL DI water, and 1 mL (Red/Blue/Yellow Dye) and 4 mL DI water. Each of these diluted liquids was contained in different cuvettes and was put into the light spectrometer to find the absorbance for a constant wavelength of each respective colors. Therefore, M1 represents the initial molarity of the specific dye (Red=4.03 x 10-5, Blue=1.26 x 10-5, and Yellow=9.35 x 10-5). V1 describes the amount of dye present without DI water (i.e., 4mL, 3mL, 2mL, or 1 mL). V2 represents 5 mL because that was the constant amount of liquid present in each test tube once the specific dyes and DI water were added to each test tube. Using the dilution formula M1V1 = M2V2; M2 is discovered, which was the molarity of each dilution that was created with each specific dye. Putting absorption on the y-axis and molarity on the xaxis beers law plot was created. By adding a trendline on the graphs it gave the equation for each graph. Later the given liquid (Gatorade) was tested in the light spectrometer, and a graph was generated which gave the peaks for red, blue, yellow absorption. For the color’s previous respective wavelength, the absorption for the given solution was: Red (0.156 abs), Blue (0.260 abs), Yellow (-0.070 abs). Examining the results, it can be observed that there is no presence of yellow dye in the solution since the absorption is negative. Using the blue beer’s law plots equation y=10139 x 0.1152 and substituting 0.260 in for y, the molarity of blue dye in the purple liquid is found 1.42x10-6 M. Then again Using the equation of the beer’s plot for the red 0.156 and substituting 0.156 in for y we get the new molarity of the red dye in the purple liquid which is 1.0612x10-5 M. Again, using the equation M1V1 = M2V2 the new volume (V2) could be found for each color. For the blue dye, the equation is (1.26x10-5 M) (V1) = (1.42x10-6 M) (50mL), after calculating correctly the answer gave V1=5.635 mL. For the red, the equation is (4.03x10-5 M) (V1) = (1.0612x10-5 M) (50mL), after calculating correctly the answer gave V1=13.166 mL. Mixing 5.635mL of blue and 13.166mL of red dyes would give the same exact color as the given solution. Results Table-1: The below data shows the absorption at the constant wavelength for each of the diluted red, blue and yellow dyes. Table-2: The below data shows the concentration for the diluted dyes for red, blue and yellow. Figure-1: The figure shows the absorption against concentration graph for the red dye. Figure-2: The figure shows the absorption against concentration graph for the blue dye. Figure-3: The figure shows the absorption against concentration graph for the yellow dye. Table-3: The below data shows the absorption of the given purple dye (Gatorade) of the respective wavelength for each color. Using the blue beer’s law plots equation y=10139 x 0.1152 and substituting 0.260 in for y, the molarity of blue dye in the purple liquid is found 1.42x10-6 M. Then again Using the equation of the beer’s plot for the red 0.156 and substituting 0.156 in for y we get the new molarity of the red dye in the purple liquid which is 1.0612x10-5 M. Using M1V1 = M2V2 the molarity was calculated. For the blue dye, the equation is (1.26x10-5 M) (V1) = (1.42x10-6 M) (50mL), after calculating correctly the answer gave V1=5.635 mL. For the red, the equation is (4.03x10-5 M) (V1) = (1.0612x10-5 M) (50mL), after calculating correctly the answer gave V1=13.166 mL. Table-4: The below data shows the absorption of the solution made (mixing 5.635mL of blue and 13.166mL of red dyes) of the respective wavelength for each color. Discussion After adding 5.635mL of blue and 13.166mL of red dyes, it gave a color like the given solution. So as the liquid was put in the light spectrometer it generated a graph with peaks, for the mixed dyes as shown in table Table-4. So, comparing the absorbance from Table-3 (given solution) with Table-4 (made solution) it could be seen that absorption for red and blue dye of the made solution was more than it should be. The surplus absorbance was 0.406 abs for red and 0.130 abs for blue, which means red and blue dyes were added more. So the beers law plot was not plotted correctly which led to these wrong results. There might be an accuracy problem while diluting the dyes or maybe a calculation error for calculating the molarity, either if these two reasons could lead the plotting of the wrong beers law plot. Conclusion Determining the beers law plots for all the basic colors the absorbance for the given solution was plugged into the equations of the beers law plot. The absorption and molarity for the given solution were: Red (0.156 abs, 1.06x10-5 M), Blue (0.260 abs, 1.420x10-6 M), Yellow (0 abs,0 M). Then using M1V1 = M2V2 the final volume to be added was red (13.166 mL), blue (5.635 mL), yellow (0mL). After these volumes were added a similar color was observed, it was then checked in the light spectrometer and discovered that more red and blue dyes were added. It is clear that the beers law plot was not correctly plotted which led these errors. If the dilution and calculation were done more accurately a better result is being expected. In future accuracy in measurement and calculation should be more focused to get a more excellent result.