Spectrophotometry Lab
Spectrophotometry of Food Dyes

(Reference:  CH111 General Chemistry I Laboratory Manual, Department of Chemistry , Michigan Technological University, Wiley, 1998)

 
PREPARATION

As many spectrophotometers as you can get
1 Spec 20 cuvette for each group
Kimwipes
Distilled water, in bottles – 1 for each group
1 L Blue food dye- 20 drops food coloring (store kind) per 1 L of water.  Absorbance should be between 0.9 and 1.0 at 610 nm
1 L Yellow food dye – 8 drops food coloring (store kind) per 1 L of water.  Absorbance should be between 0.9 and 1.0 at 420 nm
250 mL Unknown – make up your own, something with less that 0.9 absorbance
5 Beakers for each group
1 10 or 25 mL graduated cylinder for each group
 

BACKGROUND

Chemicals absorb and transmit different wavelengths of light.  The light transmitted by a substance is the color which is seen by the human eye.  All other light, especially the complement of the color transmitted, is absorbed.  For example, a green plant is transmitting the color green, that you see, and absorbing all other colors, particularly red.  Red is said to be the complement of green.
An instrument called a spectrophotometer is used to determine how much light is absorbed by certain substances.  A blue substance, for example, ABSORBES yellow light.  If we shine yellow light on different concentrations of the blue substance, the higher concentrated blue substances will absorb more of the yellow light, and vice versa.  This is Beer’s Law.

A=abc

Where “A” is the ABSORBANCE value at a certain CONCENTRATION “c”, using a tube with constant PATH-LENGTH, “b”, and having a characteristic MOLAR ABSORPTIVITY constant “a”.  We will only be concerned about “A” and “c” today.

PROCEDURE

1. Make necessary solutions; you should have 6 total, ready to test when you get to the spectrophotometer.  These can be prepared in beakers or test tubes.
2. If you are testing BLUE set wavelength (l) = 610nm, if YELLOW l = 420nm
3. With nothing in the sample chamber, adjust spec to zero transmittance with left knob
4. With spec cell full of water, place in chamber, adjust spec to 100% transmittance with right knob
5. For each sample you should start with your spec cell rinsed, pour in about 2 mL of your solution to test, shake it around, discard, then fill up with the same fresh solution.
6. Wipe fingerprints from the cell and then test.

Blue

 
Measure Transmittance of:
  • 100% Blue Solution
  • 3/4 Blue Solution, 1/4 Water Solution
  • 1/2 Blue Solution, 1/2 Water Solution
  • Trans. 100% Solution ____________________
  • Trans. 75% Solution   ____________________
  • Trans. 50% Solution   ____________________

  •  
    Yellow
     
    Measure Transmittance of:
  • 100% Yellow Solution
  • 3/4 Yellow Solution, 1/4 Water Solution
  • 1/2 Yellow Solution, 1/2 Water Solution
  • Trans. 100% Solution ____________________
  • Trans. 75% Solution   ____________________
  • Trans. 50% Solution   ____________________
  •  
    7. Now you need to convert from Transmittance to Absorbance

    Absorbance = -log (% transmittance / 100)

    Show a sample calculation:
     
     

    Fill in table:

     
    0 Absorbance % Transmittance
    100% Blue Solution 00 0
    75% Blue Solution 0 0
    50% Blue Solution 0 0
    Unknown Blue Solution 0 0
    100% Yellow Solution 0 0
    75% Yellow Solution 0 0
    50% Yellow Solution 0 0
    Unknown Yellow Solution 0 0
     
    8. Graph your points, EXCEPT UNKNOWN, for Blue; then make the same kind of graph for yellow.

     
     
    9. Connect the points on your graph, with the best-fit STRAIGHT line, through ZERO.

    10. Find where the unknown falls on each graph.

    11. What is the approximate value of the CONCENTRATION for your unknowns:

    BLUE Unknown_________________  YELLOW Unknown _______________
     

    The concentrations you calculated are probably very close to being correct because you could use a best-fit straight line through your data points.  The reason for this is because of Beer’s Law having a linear relationship between CONCENTRATION and ABSORBANCE:
     

    A=(ab)c + 0
    is like
    y=mx + b

    where y=mx + b is just the equation of a straight line.


     
    QUESTIONS

    1. If your dilution were 1/3 water 2/3 blue solution, what would be the theoretical Absorbance?  Explain.
     
     
     
     
     
     
     
     

    2. Why must each group zero the machine for their individual spec cell?
     
     
     
     
     
     
     
     
     

    DISPOSAL

     · Extra solutions can be saved
     · Student solutions can go down drain
     

    Ginger Chateauneuf, 2000.
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