Introduction to Chemistry I with Lab GT-SC1

Andria Patton July 16, 2013 Experiment #10 Beer’s Law and Colorimetry PURPOSE The purpose of this experiment is to construct a Beer’s Law plot and to use it to determine the concentration of an unknown. It is also to determine the concentration of FD & C Blue Dye #1 in commercial drinking using a colorimeter and visual colorimetry. DATA Concentration of Various Samples Sample Iden tific a tion Code Concentration of M (mol/L) % T A = 2 - log(%T) Q5000 4.00 x 10 -4 17.9 0.747 Q5001 3.20 x 10 -4 25.0 0.6021 Q5002 2.40 x 10 -4 35.7 0.4474 Q5003 1.60 x 10 -4 50.2 0.2993 Q5004 8.000 x 10 -5 70.8 0.1500 0 0.0001 0.0002 0.0003 0.0004 0.0005 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 f(x) = 1871 x + 0.00012 R² = 0.999928107622122 Series2 Linear (Series2) Concentration Absorbance Figure 1: Calculated Absorbance of M Figure 2: Graph of Absorbance vs Concentration Concentration of M Sample Iden tific a tion Code %T A = 2 - log(%T) A M Q021015-01 43.7 0.35951856 0.36 0.000192 Q021015-02 44.1 0.35556141 0.36 0.000189 Q021015-03 43.8 0.35852589 0.36 0.000192 Q021015-04 44.1 0.35556141 0.36 0.000189 Q021015-05 43.8 0.35852589 0.36 0.000192 Concentrations of Blue Dye Well # 1 2 3 4 5 6 7 8 9 10 Blue Dye Drops 1 2 3 4 5 6 7 8 9 10 Water Drops 9 8 7 6 5 4 3 2 1 0 Concentration 2.60E-05 5.20E-05 7.80E- 05 1.04E- 04 1.30E- 04 1.56E- 04 1.82E- 04 2.08E- 04 2.34E- 04 2.60E- 04 Data Table 2: Dilution Test tube Number Solutions Concentra tion (M) Concentra tion (ppm) 3 Starting dilution (100 mL water and drops of blue#1 dye) 6.5 x 10 -5 51.53 2 1:1 Dilution (1 part water - 1 part starting dye solution) 1.3 x 10 -2 10,306 1 2:1 Dilution (2 parts water - 1 part starting dye solution) 8.58 x 10 -3 6,802.6 0 Zero standard – distilled water 0 0 Data Table 3: Resistance vs. Concentration Zero Tube #0 2:1 Tube #1 1:1 Tube #2 Original Tube #3 Commercial drink; Tube 4 Concentration (M) 0 8.58 x 10 -3 1.3 x 10 -2 6.5 x 10 -5 1.56 x 10 -3 Concentration (ppm) 0 6,802.6 10,306 51.53 Figure 3: Calculated Concentrations of M Figure 3: Calculated concentrations of blue dye Figure 4: Dilution concentrations in M and ppm. Resistance, Ω 3.1 10.3 8.9 5.6 5.3 Resistance, Ω 3.0 4.2 3.3 4.1 3.9 Average Resistance, Ω 3.05 7.25 6.1 4.85 4.6 OBSERVATIONS In exercise 1 of this experiment, I observed the linear relationship between absorbance and concentration. After creating a plot that visually represented that relationship, I was able to use the equation of the line of best fit to then predict the unknown concentration of a solution M. In exercise 2, I observed the change in color of a diluted dye and how the concentration affects that color. The more concentrated solutions were a much darker blue than were the non- concentrated solutions. Also, I discovered a way to find the concentration of an unknown liquid by comparing the solutions with varying concentrations in front of a light. In exercise 3, I observed the resistance of solutions based on their concentrations. I also observed how placing the solution in the chamber in different directions can vary the results of resistance. Also, the reading of the resistance tends to waver and flex around a number which is why it is important to take the average of multiple readings. CALCULATIONS To calculate the absorbance for a given %T, I used the given equation A = 2 – log(%T) A = 2 – log(17.9) = 0.747 A = 2 – log(25) = 0.6021 A = 2 – log(35.7) = 0.4474 A = 2 – log(50.2) = 0.2993 A = 2 – log(70.8) = 0.1500 To calculate M, I used the line of best fit equation and plugged in my given values of A. To calculate the concentrations of the blue dye mixes, I multiplied the concentration of blue dye (6.5 x 10 -5 by the number of drops of blue dye I used. The amount of (10 – number of drops) was the number of water drops accounted for. Because the concentration of water is 0, we did not have to add the concentration. Then I divided the result by 10. For example, for 7 drops of blue dye and 3 drops of water, I computed (7 * 6.5 x 10 -5 )/10= 7.8 x 10 -5 . To calculate the dilutions of the solutions, I first calculated in molarity. For the 1:1 dilution, I knew the solution was half water and half dye. Therefore, I calculated 0.5(0.026) + 0.5(0) = 1.3 x 10 -2 . To then convert these concentrations from molarity to ppm, I multiplied the M concentration by 1000 and then multiplied by the molar mass, which in this case is 792.84. An example of this calculation is 0.013 M * 1000 * 792.84 = 10,306 ppm Figure 5: Resistance vs Concentration observations and calculations

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