Last Update: October 2, 2009
We are trying to find the percent Al in our alum sample by mass. To calculate this we need the mass of Al in the sample and the sample's total mass. If our alum sample was pure then this would be trivial because we know the chemical formula of alum. But, our sample is not pure and so we must find the mass of Al in our alum sample. We do this by diluting our original alum sample, adding a dye that will bind with Al3+ and measuring the absorbance of the dye solution. Using the absorbance of the diluted alum solution and the equation of the calibration curve, we can find the [Al3+] in the diluted alum solution and then by working backwards find the mass of Al in the alum sample. This is shown in Scheme 1 below.
Scheme 1. Steps for determining the mass of Al in the alum sample. Items in boxes are quantities that are either measured or calculated and items to the right of an arrow are things that get us from one box to another.
None of the steps in this scheme are individually difficult, but when taken together they can be intimidating. To help you better understand the method, we will work an example where we start from a calibration curve and guide you step by step through the calculations to the final % Al by mass in the sample. At each step you can click on the "Help Me" link to see the equation needed for that step and also the answer. However, it is in your best interest to try to set up the equation and perform the calculation before clicking on the "Help Me" link.
Concentration of Al3+ in the Dilute Solution The first step in determining the % Al by mass in alum is to extract the [Al3+] in the diluted alum solution from the absorbance reading of the diluted solution and your calibration curve. Your calibration curve should look like the one shown in Fig. 1. Note that your absorbance and [Al3+] values may differ slightly from those shown below, but your absorbance values should be less than about 1.1 and the [Al3+] values should be on the order of 10-5 M. To find the [Al3+] from a measured absorbance, one simply substitutes the absorbance of the solution (y) into the equation for the best fit line and solves for the concentration (x).
Figure 1. Typical calibration curve obtained for the colorimetric determination of Al3+.
If we assume that the final diluted alum solution had an absorbance of 0.713, what is the the [Al3+]? The answer is that the [Al3+] = 2.9428 x 10-5 M (note that we can only keep three significant figures and that we will retain two extra digits, shown as subscripts, to prevent rounding error).
Concentration of Al3+ in the Original Alum Solution We have just found that the [Al3+] in the final diluted solution is 2.9428 x 10-5 M. The next step is to determine the number of moles of Al3+ in the final diluted solution using the measured concentration and volume (50.00 mL). You should have calculated that there were 1.4714x10-6 moles Al3+ in that solution (Help Me).
We know that all of this Al3+ came from the 3.00 mL of the second solution that we made. From the moles of Al3+ above and the volume, you should find that the [Al3+] in the second solution was 4.9046 x 10-4 M (Help Me).
Since all of the Al3+ in the second solution came from the 3.00 mL that we took from the original alum solution, all we need do is repeat what we just did to find the [Al3+] in the original alum solution, which is 4.0872 x 10-3 M (Help Me).
There is a shortcut for this process. Remember that we performed a serial dilution of the initial solution, first by taking 3.00 mL of it and diluting it to 25.00 mL and then taking 3.00 mL of this new solution and diluting it to 50.00 mL. Instead of knowing the concentration of the original solution and finding the dilute solution, here we know the concentration of the dilute solution and need to find the original concentration (Help Me).
Mass of Aluminum in the Alum Sample With the [Al3+] in the original alum solution we can now calculate the mass of Al in original alum sample. The procedure is simple. Since we know the [Al3+] in the original alum solution and we know its volume (100.00 mL), we can calculate moles of Al3+ present. There is a one-to-one relationship between moles of Al3+ and moles of Al and we know the molar mass of Al. So, we can calculate the mass of Al present in our original alum sample, which is 0.011028 g (Help Me).
% Aluminum by Mass of Aluminum in Alum Now we have the mass of Al in the alum sample. Let's assume that we used 0.200 g of alum to prepare the original solution. This means that there is 0.011028 g Al in 0.200 g alum, and that alum is 5.51% Al by mass (Help Me).