Different amounts of FD&C Blue I were diluted with water to make eight differently concentrated 10 mL solutions. Samples were placed in the spectrophotometer to determine the percent transmittance of FD&C Blue 1. All of the data was summarized in graphs to predict the concentration of FD&C Blue I in a sample of Gatorade.
Materials:
FD&C Blue 1 stock solution, 50-mL
Sample of consumer sports drink, 10-mL
Water, distilled or deionized
Beakers, 50-mL, 2-3
Cuvets or test tubes, 13 x 100 mm, 3-8
Kimwipes or lens tissues
Pipet, serological, 10-mL
Pipet bulb or pipet filler
Spectrophotometer or colorimeter
Test tube rack
Procedure:
1) Turn the spectrophotometer on allow to warm up for 15-20 minutes. 2) Based on maximum absorbance of the dye tested, select the appropriate wavelength on spectrometer. 3) Read the entire procedure. Construct an appropriate data table to record measurements and the results of calculations. Note: As part of a cooperative lab activity, your instructor may assign different groups to prepare and analyze different solutions, and to graph the results. Each group will need analyze the results for all solutions in order to complete the guided-inquiry activity. 4) Obtain approximately 50 mL of stock solution containing FD&C Blue 1 dye. 5) Using a serological pipet for accurate volume measurements, dilute the stock solution as indicated in the following table to prepare 10 mL each of a series of standard solutions, B-H.
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Thoroughly mix each solution. Hint: To avoid contaminating the stock solution, and then measure and add the required amount of stock solution to each test tube. 6) Measure and record the percent transmittance (%T) of the stock solution and each standard solution (B-H) at the optimum wavelength. 7) Convert %T to transmittance (T) for each measurement, and calculate the appropriate values of both log T and –log T. Record all results. 8) Use Beer’s law to calculate the precise concentration of FD&C Blue 1 in the stock solution. The molar absorptivity (ε) of FD&C Blue 1 is 130,000 M-1 cm-1 at 630 nm and the path length (b) is 1cm. Record the micromolar (µm) concentration (1 µm= 1 x 10-6 M) in your data table. 9) Prepare separate graphs of (a) %T, (b) T, (c) 1/T, (d) log T, (e) –log T (on the y-axis) versus dye concentration (on the x-axis) for each solution.
Data Table
Results:
The concentration for solution A is 6.29 x 10 -6 M, the percentage of transmittance is 15.2%, the transmittance is 0.152 , the inverse transmittance is 6.58, the log of transmittance is -0.818, and the negative log transmittance is 0.818. The concentration for solution B is 5.05 x 10-6M, the percentage of transmittance is 22.0%, the transmittance is 0.220, the inverse transmittance is 4.55, the log of transmittance is -0.657, and the negative log transmittance is 0.657. The concentration for solution C is 3.85 x 10-6M, the percentage of transmittance is 31.6%, the transmittance is 0.316, the inverse transmittance is 3.16, the log of transmittance is -0.500, and the negative log transmittance is 0.500. The concentration for solution D is 2.88 x 10-6M, the percentage of transmittance is 42.2%, the transmittance is 0.422, the inverse transmittance is 2.37, the log of transmittance is -0.374, and the negative log transmittance is 0.374.
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The concentration for solution E is 1.85 x 10-6M, the percentage of transmittance is 57.4%, the transmittance is 0.574, the inverse transmittance is 1.74, the log of transmittance is -0.241, and the negative log transmittance is 0.241. The concentration for solution F is 1.12 x 10-6M, the percentage of transmittance is 71.4%, the transmittance is 0.714, the inverse transmittance is 1.40, the log of transmittance is -0.146 and the negative log transmittance is 0.146 The concentration for solution G is 5.35 x 10-7M, the percentage of transmittance is 85.2%, the transmittance is 0.852, the inverse transmittance is 1.17, the log of transmittance is -0.0695 and the negative log transmittance is 0.0695. The concentration for solution H is 0, the percentage of transmittance is 100%, the transmittance is 1.00, the inverse transmittance is 1.00, the log of transmittance is 0, and the negative log transmittance is 0.
Sample Calculations:
15.2%T/100= 0.152 T
-log(.152 T)= 0.818
Log(.152 T)= -0.818
1/0.152= 6.58
0.818 = (130,000 M-1cm-1)(1 cm) (c)
0.818 / 130,000 M-1cm-1= 6.29 x 10-6 M
y = 15.2x + 96.182
Figure 1 Displays the relationships between percent transmittance and concentration.
Figure 2 Displays the relationship between transmittance and concentration.
y = .-818 + 0.0339
Figure 3 Displays the relationship between log T and concentration.
y = .818 – 0.0339
Figure 4 Displays the relationship –log T and concentration.
y = 6.58x + 96.182
Figure 5 Displays the relationship of 1/transmittance and the concentration.
Conclusion:
The –log T graph has a positive slope and shows that the values of the concentration can’t be negative. The concentration of blue dye in Gatorade is 1.3×10-6M. The percent transmittance of Gatorade 67.4% was divided by 100 to find T as 0.674. The –log of T was found to be 0.17, and divided by 130,000 M-1cm-1 to make the concentration in M. 1.3×10-6 mol/L x 793 g/1 mol x 1000 mg/1 g = 1.03 mg FD&C Blue 1
Errors that happened, could be caused by using a pipet since amount might be inaccurate when making the solutions since there could be inaccurate values of water and stock solution in each serial solution. Using a measuring device that would be easier to see the exact measurement of solution for example using a beaker or a syringe.
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