Calculating Absorbance of Iron (III) Solution at 580nm for 2.50-cm Cell

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SUMMARY

The discussion focuses on calculating the absorbance of a diluted Iron (III) solution at 580 nm using a 2.50-cm cell. The user starts with a 5.00 mL aliquot of a 5.94 ppm Iron (III) solution, which is converted to a molarity of 0.000106 M. The user also notes the need for the molar absorptivity (ε) of FeSCN, which is essential for completing the absorbance calculation. The conversion from ppm to molarity is clarified, emphasizing that 1 ppm equals 1 mg/L.

PREREQUISITES
  • Understanding of ppm (parts per million) and its conversion to molarity.
  • Familiarity with Beer-Lambert Law, specifically the relationship between absorbance, molar absorptivity, path length, and concentration.
  • Knowledge of how to calculate molarity from mass concentration.
  • Basic understanding of spectrophotometry and absorbance measurements.
NEXT STEPS
  • Research the Beer-Lambert Law and its application in spectrophotometry.
  • Find the molar absorptivity (ε) of FeSCN at 580 nm from reliable sources.
  • Practice converting ppm to molarity with different solutes and volumes.
  • Explore the impact of path length on absorbance in spectrophotometric measurements.
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Chemistry students, laboratory technicians, and anyone involved in analytical chemistry or spectrophotometric analysis of solutions.

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Homework Statement



2. A 5.00 mL aliquot of a solution that contains 5.94 ppm iron (III) is treated with an appropriate excess of KSCN and is diluted to 50.0 mL.What is the absorbance of the resulting solution at 580 nm in 2.50-cm cell?

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The Attempt at a Solution



I'm having trouble finding the concentration of the solution. I'm not really good at using ppm. How do I convert this to concentration? If I have the concentration, I can finish the problem by plugging in ε,b and c.
 
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Not sure if correct but here's what I did to find the concentration.

5.49 ppm = 5.49 micrograms/mL = 0.00549 grams/L

0.00549 grams/L /(55.845 grams/mole) = 0.000106 M

0.000106 M * 5 mL / 50 mL = 0.0000106 M

EDIT : Also I just realized that the molar absorptivity of FeSCN is not given in the book, so am I doing the problem wrong or do I need to find this online?
 
Last edited:
Shortcut - ppm usually means mg per L.

1.06e-4 M is OK.
 

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