1. Cu2+ + 4NH3 → Cu(NH3)42+ (deep blue)
? %
White KNO3
? %
Blue CuSO4.5H2O
Cu NH3 → Cu(NH3)42+ (deep blue)
EX7b-1 (of 16)

2. A more saturated color means a higher concentration of the colored component
EX7b-2 (of 16)

3. A light bulb emits white light A prism separates the colors of light
SPECTROPHOTOMETER – A device that measures the amount of light absorbed by a sample
A light bulb emits white light
A prism separates
the colors of light
Light passes through the sample
Light passes through a slit to form a narrow beam
Rotating the prism allows just one color to pass
through another slit
A detector measures the final amount of light
EX7b-3 (of 16)

4. Incident Light
100 photons
Transmitted Light
10 photons I0 It
TRANSMITTANCE (T) – the fraction of the incident light that passes through the sample
T = It / I0
T = photons = 0.1
________________
100 photons
EX7b-4 (of 16)

5. 100 photons
10 photons I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.1)
= 1
EX7b-5 (of 16)

6. 100 photons 1
photon I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.01)
= 2
EX7b-6 (of 16)

7. STEP 1 – DETERMINE THE WAVELENGTH OF LIGHT TO USE
ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths
EX7b-7(of 16)

8. STEP 1 – DETERMINE THE WAVELENGTH OF LIGHT TO USE
WAVELENGTH OF PEAK ABSORBANCE (λmax) – The wavelength of light most strongly absorbed
When measuring the absorbance of solutions, detection is most sensitive when the spectrometer is set to the wavelength of peak absorbance
EX7b-8 (of 16)

9. A solution will appear a certain color if it absorbs its complementary color from the color wheel
A solution appears blue if it strongly absorbing its complimentary color, orange
EX7b-9 (of 16)

10. a combination of all colors
White light is
a combination of all colors
Sample absorbs orange, but transmits all other colors
Your eye sees the
remaining combination
of colors as blue
EX7b-10 (of 16)

11. STEP 2 – MAKE A CALIBRATION LINE
BEER’S LAW (or BEER-LAMBERT LAW) – The mathematical relationship between concentration and absorbance
A = ɛ l c
A = absorbance
ɛ = extinction coefficient
(a constant for a given solute at a given wavelength)
l = path length
(width of the cuvet holding the sample: 1.00 cm)
c = concentration of the colored substance
(in our lab it’s in “percent CuSO4.5H2O”)
l = 1.00 cm
EX7b-11 (of 16)

12. This means a graph of A vs. c will produce a straight line
STEP 2 – MAKE A CALIBRATION LINE
A = ɛ l c
+ 0
constant
y = m x + b
This means a graph of A vs. c will produce a straight line
variable to plot on y-axis = A (absorbance)
slope = ɛ l
variable to plot on x-axis = c (concentration of colored substance)
y-intercept = 0
EX7b-12 (of 16)

13. STEP 2 – MAKE A CALIBRATION LINE
Measure the absorbance of CuSO4.5H2O solutions of known concentrations at the peak wavelength, and plot absorbance vs. concentration
Absorbance
C: % % % % A:
Concentration of CuSO4.5H2O
EX7b-13 (of 16)

14. This is called a CALIBRATION LINE
m (slope) =
b (Y-intercept) =
correlation =
A = ɛ l c
A = m c + b
A = (0.9612%-1)c
+ 0
This is called a CALIBRATION LINE
m = Δy
____ Δx
= Δ Absorbance
______________________
Δ Concentration
= no units
____________ %
= %-1
EX7b-14 (of 16)

15. STEP 3 – MEASURE ABSORBANCE OF YOUR SAMPLE
If your crystal solution has an absorbance of 0.351, find its concentration
A = (0.9612%-1)c
= (0.9612%-1)c
0.350 = (0.9612%-1)c
= c
____________
0.9612%-1
0.364% = c
EX7b-15 (of 16)

16. Calculate the extinction coefficient at this wavelength, with units
A = (0.9612%-1)c
A = ɛ l c + b m
m = ɛ l
m = ɛ
___ l
= %-1
_____________
1.00 cm
= %-1cm-1
EX7b-16 (of 16)