Students in the analytical chemistry class were given a sample of spinach to det
ID: 1018094 • Letter: S
Question
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
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
Explanation / Answer
Concentration of standard Fe stock solution = 0.4 g/55.845 g/mol x 0.25 L = 0.028 M
Using, M1V1 = M2V2 we determine concentration of each standard diluted solution,
Volume of conc. of Fe Signal
standard(ml) (M)
5 0.0028 0.123
10 0.0056 0.256
20 0.0112 0.410
30 0.0168 0.610
sample 0.310
concentration of Fe in sample solution = 0.310 x 0.0168/0.610 = 0.00854 M
mass of Fe in sample = 0.00854 M x 0.05 L x 55.845 g/mol = 0.024 g
Mass% of iron in the spinach sample = 0.024 x 100/1 = 2.4%