The Report
In the results section, show two figures. The first will be your Excel plot of absorbance v. wavelength for solution 10; this plot allowed you to find lmax, the wavelength of maximum absorbance. Be sure to state what lmax value you used to make the rest of your measurements!
Next, you'll need to prepare a plot of absorbance v. concentration (measured at lmax) for your four standard solutions. You're measuring the absorbances using a spectrophotometer, but you need to calculate the concentrations of the standards before you can make the plot. Here's an outline of the calculations:
Preparation of stock solution
Suppose that your sample of Red #40 had a mass of 46.92 mg (0.04692 g). You added this sample to a 500 mL volumetric flask and diluted to 500 mL; so the concentration of the stock solution = 0.04692 mg/0.500L = 93.84 mg/L.
To make up standard solution 1, you took a 1.00 mL sample of the stock solution, added it to a 100 mL volumetric flask, and diluted to 100 mL. We need to calculate the concentration of the new solution - we do this using the solution dilution formula,
Here, C1 and C2 are the concentrations before and after dilution, respectively. V1 is the volume of the stock solution we're taking out to dilute, and V2 is the final volume that we're diluting to. So, here....
C1 = 93.84 mg/L (calculated above)
V1 = 0.001 L (1 mL, expressed in L)
V2 = 0.1000 L (100 mL, in L)
C2 is the final concentration we're trying to find. Here,
C2 = (C1V1)/(V2) (93.84 mg/L)(0.001 L) / (0.100 L) = 0.9384 mg/L
Now, to make up solution 5, we took 5.00 mL of stock and diluted it to 100 mL. To calculate the concentration, we use the solution dilution formula and just set
V1 = 0.005 L (we took 5.00 mL of stock)
so for solution 5, C2 = (93.84 mg/L)(0.005 L) / (0.100 L) = 4.692 mg/L
For solutions 10 and 15, we took 10.00 and 15.00 mL of stock, respectively, and diluted again to 100 mL. We change V1 to 0.010 L for solution 10, and to 0.015 L for solution 15. Check the math - the concentrations of these solutions are 9.384 mg/L and 14.076 mg/L.
Now we plot absorbance v. concentration for the standard solutions, and fit a line to the data....
Now, suppose that you measured an absorbance of 0.123 for your diluted Kool aid sample. Using the equation of the best-fit line on the absorbance v. concentration plot, we can calcuate the concentration (in mg/L) of Red #40 in the Kool aid. Absorbance is the y-variable; substitute your Kool-aid absorbance into the equation of the line on your plot, and find x (the concentration.) In this example,
x = (0.123 - 0.0088) / 0.0419 = 2.73 mg/L
Don't forget that the Kool aid sample was diluted by a factor of 10!!! Multiply x 10 to get the real concentration of Red #40 in the sample. Here, the true concentration of Red 40 = 27.3 mg/L.
Calcuation of Acceptable Dietary Intake (ADI)
Here, you'll use the ADI value for Red #40 to calculate how many liters of Kool aid you could drink without exceeding the ADI. The assumption you're making is that the Red 40 concentration in Kool aid is the value you calculated above (here, 27.3 mg/L.)
The ADI for Red 40 is 7 mg dye/kg body mass. For a 165 lb person (use your body weight here),
(7 mg dye/kg)(1 kg /1000g)(453.6 g/1 lb)(165 lb) = 523.9 mg dye.
So, our 165 lb person can ingest 523.9 mg of Red 40 without appreciable risk.
Now, we found that Kool aid contains 27.3 mg Red 40 /L. What volume (in L) of Kool aid must a 165 lb person drink to ingest 523.9 mg of dye? Look at the units...
523.9 mg / (27.3 mg/L) = 19.2 L Kool aid (with Red 40 = 27.3 mg/L).
Now repeat the calculation using the ADI for Red dye #2; it's ADI is 0.8 mg dye/kg body mass. Assume that there's the same concentration of Red #2 as Red 40 (27.3 mg/L, in this case.)
For comparison purposes, the mass of Red 2 which can be consumed is 59.87 mg; the volume of liquid (at 27.3 mg/L Red #2) is 59.87 mg/ (27.3 mg/L) = 2.19 L.
