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Determination of Fe Lab

By urk1012 Jul 30, 2013 278 Words
CHM 3120L
ANALYTICAL CHEMISTRY I
LABORATORY REPORT

EXPERIMENT: SPECTROPHOTOMETRIC DETERMINATION OF IRON IN DRINKING WATER

Name: Steven Adrien
Section: 3
Date Experiment Completed: Wednesday, July 17, 2013

1. Complete the following table
Fe(II) stock solution |
mass, g| 0.1756|
volume, mL| 500.0|
MW(Fe(NH4)2(SO4)2 x 6H2O), g/mol| 392.14|
AW(Fe), g/mol| 55.85|
conc. Fe(II), ppm| 50.0|

Use Equation Editor to show how you calculated the concentration of Fe(II) in the stock solution. grams of Ferrous ammonium sulfate *1392.14gm*55.851gm*1000mg1g*10.500L=ppm

0.1756 *1392.14*55.851*10001*10.500=50.019

2. Report concentrations and absorbance of standards and samples. Solution # | Stock
volume, mL | Conc. Fe(II), ppm | Absorbance at 510 nm|
| | | 1| 2| 3| Average|
 1| 1|  0.500|  0.082| 0.083| 0.082| 0.082|
2| 2| 1.00| 0.182| 0.181| 0.182| 0.182|
3| 5| 2.50| 0.476| 0.476| 0.479| 0.477|
4| 10| 5.00| 1.002| 1.002| 1.001| 1.002|
Unknown| -| -| 0.571| 0.575| 0.574| 0.573|

3. Paste Excel plot of your calibration data

4. Enter the equation for your calibration line:
y = 0.3055x - 0.328

5. Report the concentration of Fe (in ppm) in your sample. Remember to account for the dilution of the sample before measurements. 5.90 ppm

Use MS Word equation editor to show how you calculated the concentration of Fe in your unknown. M1V1=M2V2
M150.0mL=M2100mL
M2=0.573+0.3280.3055=2.95
M150.0mL=2.95ppm100mL
M1=2.95ppm100mL50.0mL=5.90 ppm

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