You can do this mathematically but it is easier, I think, to draw a graph for the calibration curve.
On the Y axis you place the absorbance which I assume is the "signal" in your post. If not, it might be %T; whatever it is plot it (Absorbance or whatever) on the Y axis and mg Fe on the X axis. You have the A units in the post. Calculate the mg Fe for the X axis this way.
You have 400 mg Fe in 250 mL. So in the 5.00 mL sample you have
400 mg x 5/250 = ?mg Fe and that is diluted to 50.
In the 10.0 mL sample you have
400 mg x (10/250) = ? mg Fe and that is diluted to 50 mL.
Likewise for the other volumes.
You can see the dots representing the points, draw the best line through those, then read the mg Fe for the sample; i.e., for 0.310 A. You can just look at the data, as I've just done, and the sample is approximately whatever mg Fe is for about half way between the Fe you found for 10 mL and the Fe you found for the 20 mL. (Again, I'm assume the "signal" is the absorbance.) Anyway, that gives you the mg Fe in the sample and since the calibration curve was made with a 50 mL flask and the sample was in the 50 mL flask, that is mg Fe in that 1.00 grams sample of spinach.
Then % w/w = [(GRAMS Fe--not mg)/mass spinach sample)]*100 = ?
If you want to do it mathematically, you use A = kc. Plug in the A value and the concn(mg Fe) and calculate k. I would do all of them, then average the k values.
Then A for sample = k(just found) x c and solve for c. Then place that c in the % formula and solve.
I am having major issues with this problem and I no clue where to start. It also doesn't help that the professor never taught us how to calculate mass % from data like this...
Students in the analytical chemistry class were given a sample of spinach to determine its iron content. 1.00 g of this sample was ashed and dissolved in 5.0 mL of nitric acid. Resulting solution was added to 50.0 mL volumetric flask and diluted to the mark with DI water. This sample was analyzed with atomic absorption spectroscopy. Standard stock solution was prepared by dissolving 400 mg of iron in 5 mL nitric acid and diluting the resulting solution to 250 mL in a volumetric flask. A calibration curve was obtained by taking following volumes from the standard stock solution and diluting them to 50.0 mL with DI water. The Table shows the results of the AAS analysis. Calculate the mass % of the iron in spinach sample.
Volume of the standard Signal
5.0 mL - 0.123
10.0 mL - 0.256
20.0 mL - 0.410
30.0 mL - 0.610
Sample - 0.310
Please and Thank you!!
3 answers
@ Dr.Bob222
Thank you for the help! it is greatly appreciated!
However - Do I graph the last number saying Sample and 0.310?
Or is that not necessary?
Thank you for the help! it is greatly appreciated!
However - Do I graph the last number saying Sample and 0.310?
Or is that not necessary?
No. You have no concn for that so how would you put that on the graph? You look at the A value on the graph, read the concn in mg Fe from the curve, and use that in the final % w/w calculation.
1 answer