Question

How do I find the weight precent iron (II)?

We measured 1.0014 g of unknown iron sample and added 100.0 mL water. Then pipetted 10.00 mL of the first solution into a beaker and added 90.00 mL of water. Then added 1,000 microliters of the second solution to a plastic container along with the reagents needed to analyze it with a spectrophotmeter. This solutions total volume was 10.00 mL. I then measured the absorbance of this solution and got .019 A. I then used the equation form my calibration curve and found the concentration of the solution to be 0.18 ppm. Now, how do I find what the weight percent of iron (II) in my unknown is?

Answer 1

0.18 ppm = 0.18 mg/L

That means 1000 mL contains 0.18 mg

Therefore, 10 mL contains = 0.18 x 10/1000 = 0.0018 mg

This means 0.0018 mg of iron(II) is there 1000 microliter or 1 mL.

Thus 1 mL contains 0.0018 mg.

This 1 mL solution is from the 100 mL stock solution.

Therefore 100 mL contains = 0.0018 x 100 = 0.18 mg

Now, this 100 mL is made by diluting 10 mL from the original stock solution.

Thus 10 mL original stock solution contains = 0.18 mg iron(II)

Therefore 100 mL original stock solution contains = (0.18 x 100)/10 = 1.8 mg

Weight of the unknown iron sample = 1.0014 g = 1.0014 x 1000 = 1001.4 mg

Therefore weight percent of iron(II) = (1.8 x 100)/ 1001.4 = 0.18 %