three hour lab - university of michiganchem125/f08/lec04f08e2wk2.pdf · 1 chem.125-126: sept.24 -...
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Chem.125-126: Sept.24 - 30Experiment 2 Session 2
Preparation• Pre-lab prep and reading for E2, Parts 3-5
Experiment 2 Session 2Electrons and Solution Color
Three hour lab• Complete E2 (Parts 1 - 5)
• Prepare discussion presentation• Prepare team report.• Give team report to GSI at the end of labor turn in by grace deadline to GSI’s atriumlevel mailbox.
Electrons will move from a lower to an availablehigher energy level if the provided energy = exactlythat needed for a possible energy level transition
DEMO
Background Information: Energy and electrons
If electrons move from a higher to a lower energylevel, the difference in energy will be released.
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λ 400 Violet - Blue - Green - Yellow - Orange - Red λ 800
Visible Light and Energy
•The shorter the wavelength (λ), the greater is its energy•The shorter the wavelength (λ), the higher is its frequency
Light is a form of energy
Light source →
Wavelength and Energy
Balloon containing H2 and Cl2
DEMO1. Expose a balloon containing H2 and Cl2 to red light (≅ λ 650).2. Expose a balloon containing H2 and Cl2 to blue light (≅ λ
450).
_____________________________________________________________________________________________________________________________________________________________________
• The shorter the wavelength, the greater its energy
Part 3. Solution Color and Light Interaction
• Determine if salt solution color is predictable based onthe cations interaction with visible light versus:
• Plot an absorption spectrum for each teamassigned salt solution
- placement of the cation’s element in the periodic table?
- the cation’s electron configuration?
- the cation’s ionic radius?
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1A VIIIA1H1s1 IIA IIIA IVA VA VIA VIIA
2He1s2
3Li2s1
4Be2s2
5B
2s22p1
6C
2s22p2
7N
2s22p3
8O
2s22p4
9F
2s22p5
1 0Ne
2s22p6
1 1Na3s1
1 2Mg3s2 IIIB IVB VB VIB VIIB VIIIB ⇔ VIIIB IB IIB
1 3Al
3s23p1
1 4Si
3s23p2
1 5P
3s23p3
1 6S
3s23p4
1 7Cl
3s23p5
1 8Ar
3s23p6
1 9K4s1
2 0Ca4s2
2 1Sc
3d14s2
2 2Ti
3d24s2
2 3V
3d34s2
2 4Cr
3d54s1
2 5Mn
3d54s2
2 6Fe
3d64s2
2 7Co
3d74s2
2 8Ni
3d84s2
2 9Cu
3d1 04s1
3 0Zn
3d1 04s2
3 1Ga
4s24p1
3 2Ge
4s24p2
3 3As
4s24p3
3 4Se
4s24p4
3 5Br
4s24p5
3 6Kr
4s24p6
3 7Rb5s1
3 8Sr5s2
3 9Y
4d15s2
4 0Zr
4d25s2
4 1Nb
4d35s2
4 2Mo
4d55s1
4 3Tc
4d55s2
4 4Ru
4d75s1
4 5Rh
4d85s1
4 6Pd
4d10
4 7Ag
4d1 05s1
4 8Cd
4d1 05s2
4 9In
5s25p1
5 0Sn
5s25p2
5 1Sb
5s25p3
5 2Te
5s25p4
5 3I
5s25p5
5 4Xe
5s25p6
5 5Cs6s1
5 6Ba6s2
5 7La*
5d16s2
7 2Hf
5d26s2
7 3Ta
5d36s2
7 4W
5d46s2
7 5Re
5d56s2
7 6Os
5d66s2
7 7Ir
5d76s2
7 8Pt
5d96s1
7 9Au
5d1 06s1
8 0Hg
5d1 06s2
8 1Tl
6s26p1
8 2Pb
6s26p2
8 3Bi
6s26p3
8 4Po
6s26p4
8 5At
6s26p5
8 6Rn
6s26p6
8 7Fr7s1
8 8Ra7s2
8 9Ac#
6d17s2
1 0 4 +
6d27s2
1 0 5 +6d37s2
1 0 6 +6d47s2
1 0 7 +6d57s2
1 0 8 +6d67s2
1 0 9 +6d77s2
+ Element synthesized,but no official name assigned
Pre-transition.
Color versus Periodic Table Position
Transition Post-transition
• Compare the color data of salt solutions containing cationsfrom transition and pre and post transition families
Beer-Lambert Law A λ = ε c l
absorptivity factor • concentration • path length
Absorbance at λ =
• Concentration and path length are held constantwhile taking the sample’s absorption spectrum(Part 3)
Recording a spectrum
• Calibrate (0 absorbance and 100% transmission) thespectrophotomer with the blank every time you changethe λ before taking the absorbance of your sample
• Use the same sample and holder (cuvette) so thatsample concentration and path length are constant
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• Record the solution color and the wavelength andcolor of the wavelengths of absorption andtransmission max
Absorption Spectrum (Part 3)
Transmission λmax
Absorptionλmax
Light Absorbance vs. Transmission
Abs = 0 100% light transmitted
Abs = 1 10% light transmitted
ABSORBANCE = -LOG TRANSMITTANCE
• Absorbance reading values = 0 to 1.0 for minimal error
Absorbance vs. Transmission Spectrum
The identity of a solution can be determined from itsabsorption (or transmission) spectrum.
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Solution Color and Light Interaction
DEMO
Q. What wavelength colors will a solution of NiSO4transmit?
Absorbance differences across wavelengths are due to?1. Differences in the absorptivity coefficient ( ε )2. Differences in the concentration of the sample.3. Differences in the path length of the sample holder.4. All the above.
Spectrum of 0.10 M _________
Which statement below is correct?1. Color of Abs λ max = blue-purple.2. The sample is green.3. ε is greater at λ 500 than at λ 400.
00.10.20.30.40.50.60.7
400 450 500 550 600 650 700
Absor
bance
Wavelength λ (nm)
Violet
Blue
Yello
w
Green Orang
e
Red
00.10.20.30.40.50.60.7
400 450 500 550 600 650 700
Absor
bance
Wavelength λ (nm)
Violet
Blue
Yello
w
Green Orang
e
Red
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Beer-Lambert Law and Path Length
Path length and light absorbance are directlyproportional at a fixed wavelength and concentration.
• Different spectrophotometers have different pathlengths! Don’t change spectrophotometers in themiddle of an analysis (Parts 4 and 5)!
DEMO
Path length changes will result in a proportional change inabsorbance values if sample concentration is constant.
Absorbance readings will alter proportionately across allwavelengths; the spectrum pattern will not alter.
Path Length and Absorption Spectra
Fig. Absorption spectra at different sample path lengths
Sample concentration changes will result in a proportionalchange in absorbance values if path length is constant.
Absorbance readings will alter proportionately across allwavelengths; the spectrum pattern will not alter.
Concentration and Absorption Spectra
Fig. Absorption spectra at different sample concentrations
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Part 4. Concentration and Light Absorbance
- Write a mathematical expression to express thepattern between the concentrations and theabsorbance values of your team assigned sample
• Plot a calibration curve for your team assigned sample
• Successfully use the calibration curve to determine theunknown concentration of your team assigned sample (Part 5)
Beer-Lambert Law A λ = ε c l
absorptivity factor • concentration • path length
Absorbance at λ =
• Wavelength and path length are held constantwhen producing a calibration curve (Part 4)
Beer-Lambert Law and Concentration
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abso
rban
ce
Concentration (mM)
Absorbance is proportional to concentration at aconstant wavelength (λ )and constant path length
DEMO
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Absorbance and Path Length
= 1/2pathlength
Absorbance readings for a calibration curve will alter ifthe path length is altered
0.50.40.30.20.10.00.0
0.5
1.0
1.5
2.0
2.5
[Plastocyanin], mM
Abso
rban
ce
Absorbance and Path Length
Answer:________________________________________________________________________________________________________________________
Path length must be fixed for a calibration curve orabsorbance readings will be in error.
Q. How do you hold the path length constant?
Preparation of Calibration Curve
Prepare a set of solutions of known and accurateconcentration by diluting the team prepared andassigned 0.1 M solution
M1V1 = M2V2
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The wavelength of max absorbance is typically chosen. Why?
Preparation of Calibration Curve
• Refer to the samples absorption spectrum to choose awavelength for the calibration curve
_____________________________________________________________________________________________________.
1.20
0.80
0.40
0.00
Abs
orba
nce
Wavelength (nm)250 350 450 550 650 750 0.50.40.30.20.10.0
0.0
0.5
1.0
1.5
2.0
2.5
[Plastocyanin], mM
Absorb
ance
Spectrum of 0.16 mM Plastocyanin Calibration curve at 600nm
Calibration Curve Wavelength
Q. Will the slope of the linear line of a calibration curve produced at550 nm be greater or less than the slope of the line of the abovecalibration curve produced at 600 nm? _______________
Wavelength of Calibration Graph?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Abs
orpt
ion
0.8
0 0.1 0.2 0.3 0.4 0.5[M+] (Molar)
0
0.2
0.4
0.6
0.8
1
1.2
Abs
orpt
ion
400 450 500 550 600 650 700
Purp
le
Blue
λ (nm)
Gre
en
Yell
ow
Ora
nge
Red
Q. A 0.4 M solution of M+ has the absorption spectrum onthe left. Circle the wavelength of its calibration graph: 425 500 550 600 650
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Preparation of Calibration Curve
Calculate the slope of the linear line of yourcalibration curve
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abso
rban
ce
Concentration (mM)
_________________________
Calibration Curve Slope
Q. What does the slope of the calibration curverepresent in the Beer-Lambert expression Aλ = εlc?
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abso
rban
ce
Concentration (mM)
Answer: _____________________________________
Unknown concentration determination
Q. A sample of unknown concentration of XY(aq) has Abs >1.2 at a λ of 600nm, What is its concentration (mM)?
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abso
rban
ce
Concentration (mM)
Fig. Calibration Curve of XY(aq) at 600 nm
Slope = 0.15Abs/mM
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Beer-Lambert and Calibration Curve
Fig. Calibration Curve Deviation
• The Beer-Lambert law only applies at low concentrations!• Do NOT extrapolate the linear line of a calibration curve!
• Determine the concentration of a diluted sample of theunknown if the unknown’s absorbance reads above anAbs = 1 or outside the linear line of the calibration curve
“Eyeball” the graph ONLY for an approximateconcentration.
Unknown concentration determination
• Use the slope and Beer-Lambert law to determinean exact concentration
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abs
orba
nce
Concentration (mM)
What is the unknown concentration?The diluted sample has an absorbance at λ of 600nm = 0.57.
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
Abso
rban
ce
Concentration (mM)
Fig. Calibration Curve of XY(aq) at 600 nm
Slope = 0.15Abs/mM
_________ = c
Abs λ600 = elc
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What is the unknown concentration?Q. A diluted sample of unknown concentration = 3.8 mM. Youprepared the diluted sample by adding 6.0 mL of water to 2.0mL of the sample of unknown concentration. What is theunknown concentration (mM)?
a) 3.8 mMb) 7.2 mMc) 11.4 mMd) 15.2 mM
Questions?Contact [email protected]