EXERCISE SEVEN: SEEING THE LIGHT WORKSHEET
NEVAEH LINDSEY
Name
923 3/4/2026
Section Date
Part One. Standard Green Sample (6 points)
The undiluted yellow dye (Yellow 6) had a concentration of 73.260 μM and an Amax of 480 nm.
The undiluted blue dye (Blue 1) had a concentration of 15.385 μM and an Amax of 630 nm.
Use the information from Lab Exercise Six to fill in the blanks below for how you made your
Standard Green solution:
I made a dilution by putting __1__ mL of each dye (V1 - initial volume of
stock solution used) into a final volume of __4__ mL (V2 - final volume).
Use the dilution equation (below) to determine what the final concentration of each of the dyes
are in your Standard Green solution. Do two separate calculations, one for Yellow 6 and one for
Blue 1.
C1 (initial concentration) is the undiluted dye concentration from above. You are solving
for C2 (final concentration of your diluted dye sample).
(C1)(V1) = (C2)(V2)
My diluted dye sample should have a concentration of ____18.315_ μM for Yellow 6.
My calculation is
My diluted dye sample should have a concentration of ____3.846___ μM for Blue 1.
My calculation is
Convert the diluted concentrations from μM to M. The conversion factor is 1 M = 1x106 μM.
For Yellow 6, __18.315__ μM equals __0.000018315___ M (put this value in Table One.)
For Blue 1, __3.846____ μM equals __0.000003846___ M (put this value in Table One.)
Ex. 7 Seeing the Light - 1
, Using the Beer-Lambert Law (A = εbc) to calculate the predicted absorbance of each of your
dyes using Table One.
Now you have to measure your Green Standard solution with the spectrophotometer to get the
actual measured absorbance of Yellow 6 and Red 40. Follow the instructions in checklist for
how to measure the absorbance.
Table One. Dye Absorbance Predictions and Measurements
Absorbance = ε X b X c ε b
c (M) = Absorbance
(see checklist) (cm)
The predicted 0.000018
absorbance of my 25,900 1 315 = 0.4744
Yellow 6 at Amax is
My actual measured
absorbance for Yellow 0.58
6 was
The predicted 0.000003
absorbance of my Blue 130000 1 846 = 0.499
1 at Amax is
My actual measured
absorbance for Blue 1 0.59
was
Conclusions
Did your actual absorbance values match your prediction absorbance values?
For Yellow 6 it was ___close_______For Blue 1 it was _________close_
Put in very close (within 0.02 of the prediction), Close (withing 0.06 of the prediction) or not
close
As we discussed in Exercise Six, there are three types of error. For your measurements, rank the
possible sources by circling or highlighting below. There is only one primary cause, one
secondary cause, etc.
The primary (or major) source of error was _______ Systematic______.
The secondary source of error was _____Human_______.
The least source of error was _________Random__.
Use Random, Systematic or Human as answers.
What could you do to improve your results? proper instrument calibration, precise
dilutions, careful cuvette handling, and repeated measurements.
Ex. 7 Seeing the Light - 2
NEVAEH LINDSEY
Name
923 3/4/2026
Section Date
Part One. Standard Green Sample (6 points)
The undiluted yellow dye (Yellow 6) had a concentration of 73.260 μM and an Amax of 480 nm.
The undiluted blue dye (Blue 1) had a concentration of 15.385 μM and an Amax of 630 nm.
Use the information from Lab Exercise Six to fill in the blanks below for how you made your
Standard Green solution:
I made a dilution by putting __1__ mL of each dye (V1 - initial volume of
stock solution used) into a final volume of __4__ mL (V2 - final volume).
Use the dilution equation (below) to determine what the final concentration of each of the dyes
are in your Standard Green solution. Do two separate calculations, one for Yellow 6 and one for
Blue 1.
C1 (initial concentration) is the undiluted dye concentration from above. You are solving
for C2 (final concentration of your diluted dye sample).
(C1)(V1) = (C2)(V2)
My diluted dye sample should have a concentration of ____18.315_ μM for Yellow 6.
My calculation is
My diluted dye sample should have a concentration of ____3.846___ μM for Blue 1.
My calculation is
Convert the diluted concentrations from μM to M. The conversion factor is 1 M = 1x106 μM.
For Yellow 6, __18.315__ μM equals __0.000018315___ M (put this value in Table One.)
For Blue 1, __3.846____ μM equals __0.000003846___ M (put this value in Table One.)
Ex. 7 Seeing the Light - 1
, Using the Beer-Lambert Law (A = εbc) to calculate the predicted absorbance of each of your
dyes using Table One.
Now you have to measure your Green Standard solution with the spectrophotometer to get the
actual measured absorbance of Yellow 6 and Red 40. Follow the instructions in checklist for
how to measure the absorbance.
Table One. Dye Absorbance Predictions and Measurements
Absorbance = ε X b X c ε b
c (M) = Absorbance
(see checklist) (cm)
The predicted 0.000018
absorbance of my 25,900 1 315 = 0.4744
Yellow 6 at Amax is
My actual measured
absorbance for Yellow 0.58
6 was
The predicted 0.000003
absorbance of my Blue 130000 1 846 = 0.499
1 at Amax is
My actual measured
absorbance for Blue 1 0.59
was
Conclusions
Did your actual absorbance values match your prediction absorbance values?
For Yellow 6 it was ___close_______For Blue 1 it was _________close_
Put in very close (within 0.02 of the prediction), Close (withing 0.06 of the prediction) or not
close
As we discussed in Exercise Six, there are three types of error. For your measurements, rank the
possible sources by circling or highlighting below. There is only one primary cause, one
secondary cause, etc.
The primary (or major) source of error was _______ Systematic______.
The secondary source of error was _____Human_______.
The least source of error was _________Random__.
Use Random, Systematic or Human as answers.
What could you do to improve your results? proper instrument calibration, precise
dilutions, careful cuvette handling, and repeated measurements.
Ex. 7 Seeing the Light - 2
