experiment 1 (2 session lab) part 1. preparation and …chem125/w08/e1w08_1key.pdfu 238.07 93 np...
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Experiment 1 (2 session lab)Electrons and Solution Color
Session 1 One hour discussion (E2) Two hour lab (E1) Aim to complete Parts 1, 2, and 3 of E1.
Pre-lab Report, page 29
Part 1. Preparation and Color of Solutions
Goals Successfully prepare a solution of known Molarity. Determine if it possible to predict the color of asalt solution from the structure of the salt’s cation.
Part 1. Preparation and Color of Solutions
1H+
Hydrogen IIIA IVA VA
3Li+
Lithium
4Be
Beryllium
1 1Na+
Sodium
1 2Mg2+
Magnesium IIIB IVB VB VIB VIIB VIIIB ! VIIIB IB IIB
1 3Al3+
Aluminum
1 9K+
Potassium
2 0Ca2+Calcium
2 1Sc
Scandium
2 2Ti
Titanium
2 3V
Vanadium
2 4Cr3+
Chromium
2 5Mn
Manganese
2 6Fe3+Iron
2 7Co2+Cobalt
2 8Ni2+Nickel
2 9Cu2+Copper
3 0Zn2+Zinc
3 1GaGalium
3 2Ge
Germanium
3 7Rb
Rubidium
3 8Sr2+
Strontium
3 9YYitrium
4 0Zr
Zircon-ium
4 1Nb
Niobium
4 2Mo
Molyb-denum
4 3Tc
Technetium
4 4Ru
Ruthenium
4 5Rh
Rhodium
4 6Pd
Palladium
4 7Ag+Silver
4 8Cd2+
Cadmium
4 9InIridium
5 0
Sn2+
Tin
5 1Sb
Antimony
5 5Cs
Cesium
5 6Ba2+Barium
5 7La*
Lanthanum
7 2Hf
Hafnium
7 3Ta
Tantalum
7 4W
Tungsten
7 5Re
Rhenium
7 6Os
Osmium
7 7IrIridium
7 8PtPlatinum
7 9AuGold
8 0Hg2+
Mercury
8 1TlThallium
8 2
Pb2+Lead
8 3BiBismuth
Teams prepare solutions with different metal ionsRecord the color of solutions.
Solution Color Solutions with ions of the same metal element withdifferent ion charges may have different colors.
DEMO: V2+ vs. V3+ etc.
1H+
Hydrogen IIIA IVA VA
3Li+
Lithium
4Be
Beryllium
1 1Na+
Sodium
1 2Mg2+
Magnesium IIIB IVB VB VIB VIIB VIIIB ! VIIIB IB IIB
1 3Al3+
Aluminum
1 9K+
Potassium
2 0Ca2+Calcium
2 1Sc
Scandium
2 2Ti
Titanium
2 3V
Vanadium
2 4Cr3+
Chromium
2 5Mn
Manganese
2 6Fe3+Iron
2 7Co2+Cobalt
2 8Ni2+Nickel
2 9Cu2+Copper
3 0Zn2+Zinc
3 1GaGalium
3 2Ge
Germanium
3 7Rb
Rubidium
3 8Sr2+
Strontium
3 9YYitrium
4 0Zr
Zircon-ium
4 1NbNiobium
4 2Mo
Molyb-denum
4 3Tc
Technetium
4 4Ru
Ruthenium
4 5Rh
Rhodium
4 6Pd
Palladium
4 7Ag+Silver
4 8Cd2+
Cadmium
4 9InIridium
5 0Sn2+Tin
5 1Sb
Antimony
Periodic Table with common metal ion charges.
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Part 1 Data analysis and discussion.
Placement of the metal ion’s element in the periodic table? The metal ion’s radius? The metal ion’s electron configuration?
Questions 1-4, p.39
Is the presence or absence of saltsolution color predictable from ___?
Course theme
“There are structure andproperty and periodic tablerelationships”
Useful web sites are: www.merlot.org www.webelements.com
Discussion PreparationManipulate the class data.
You will NOT get points for justreproducing the class data.
Discussion Preparation Refer to the two grading rubricks, pages 37-38. - Note that one discussion grading rubrick refers toan exam question. Refer to the discussion information, page 222.
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Part 3. Solution Color and Light Interaction.
What is the relationship between the visiblecolor of a solution and its absorption andtransmission of visible light wavelengths?
Discussion question 5
Light source Diffractiongrating
Samplesolution
Detector
Use a spectrophotometer to examine the relationshipbetween solution color and absorption andtransmission of visible light.
Spectrophotometer
1. Record visible wavelength colors (Part 2)
Spectrophotometer
2. Record sample absorbance across the visiblespectrum to produce an absorption spectrum.
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A plot of absorbance versus wavelength
Absorption Spectrum
Absorbance λmax
Light Absorbance vs. Transmission
0% 10% T 100%
∞ 2 1 A 0
Abs = 1 10% light transmitted
Abs = 2 1% light transmitted
ABSORBANCE = -LOG TRANSMITTANCE
Solution preparation
Calibration line
Volumetric flasks
Chem.125/126 Mascot
__________________________________________________________
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Moles and the Periodic Table?
?
National Mole Day: October 23. Why?
1
H1.008
3
Li6.940
11
Na22.991
19
K39.100
37
Rb85.48
55
Cs132.91
87
Fr(223)
4
Be9.013
12
Mg24.32
20
Ca40.08
38
Sr87.63
56
Ba137.36
88
Ra226.05
58
Ce140.13
5
B10.82
13
Al26.98
31
Ga69.72
49
In114.82
81
Tl204.39
90
Th232.05
6
C12.011
14
Si28.09
32
Ge72.60
50
Sn118.70
82
Pb207.21
7
N14.008
15
P30.975
33
As74.91
51
Sb121.76
83
Bi208.9
8
O15.999
16
S32.06
34
Se78.96
52
Te127.61
84
Po(209)
9
F19.00
17
Cl35.457
35
Br79.916
53
I126.91
85
At(210)
10
Ne20.183
18
Ar39.944
36
Kr83.80
54
Xe131.30
86
Rn(222)
2
He4.003
21
Sc44.96
22
Ti47.90
23
V50.95
24
Cr52.01
25
Mn54.94
26
Fe55.85
27
Co58.94
28
Ni58.71
29
Cu63.54
30
Zn65.38
39
Y88.92
40
Zr91.22
41
Nb92.91
42
Mo95.95
43
Tc(99)
44
Ru101.1
45
Rh102.9
46
Pd106.4
47
Ag107.88
48
Cd112.41
57†
La138.92
72
Hf178.50
73
Ta180.95
74
W183.86
75
Re186.22
76
Os190.2
77
Ir192.2
78
Pt195.09
79
Au197.0
80
Hg200.61
89††
Ac(227)
104
Rf(261)
105
Ha(262)
106
--(263)
59
Pr140.92
60
Nd144.27
61
Pm(145)
62
Sm150.35
63
Eu152.35
64
Gd157.26
91
Pa(231)
92
U238.07
93
Np(237)
94
Pu(242)
95
Am(243)
96
Cm(245)
65
Tb158.93
97
Bk(249)
66
Dy162.51
67
Ho164.94
68
Er167.2
69
Tm168.94
70
Yb173.04
71
Lu174.99
98
Cf(251)
99
Es(254)
100
Fm(255)
101
Md(256)
102
No(254)
103
Lr(257)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Mole
= mol = 6.02 x 1023 item
= atomic wt (g) = formula wt (g)
1
H1.008
3
Li6.940
11
Na22.991
19
K39.100
37
Rb85.48
55
Cs132.91
87
Fr(223)
4
Be9.013
12
Mg24.32
20
Ca40.08
38
Sr87.63
56
Ba137.36
88
Ra226.05
58
Ce140.13
5
B10.82
13
Al26.98
31
Ga69.72
49
In114.82
81
Tl204.39
90
Th232.05
6
C12.011
14
Si28.09
32
Ge72.60
50
Sn118.70
82
Pb207.21
7
N14.008
15
P30.975
33
As74.91
51
Sb121.76
83
Bi208.9
8
O15.999
16
S32.06
34
Se78.96
52
Te127.61
84
Po(209)
9
F19.00
17
Cl35.457
35
Br79.916
53
I126.91
85
At(210)
10
Ne20.183
18
Ar39.944
36
Kr83.80
54
Xe131.30
86
Rn(222)
2
He4.003
21
Sc44.96
22
Ti47.90
23
V50.95
24
Cr52.01
25
Mn54.94
26
Fe55.85
27
Co58.94
28
Ni58.71
29
Cu63.54
30
Zn65.38
39
Y88.92
40
Zr91.22
41
Nb92.91
42
Mo95.95
43
Tc(99)
44
Ru101.1
45
Rh102.9
46
Pd106.4
47
Ag107.88
48
Cd112.41
57†
La138.92
72
Hf178.50
73
Ta180.95
74
W183.86
75
Re186.22
76
Os190.2
77
Ir192.2
78
Pt195.09
79
Au197.0
80
Hg200.61
89††
Ac(227)
104
Rf(261)
105
Ha(262)
106
--(263)
59
Pr140.92
60
Nd144.27
61
Pm(145)
62
Sm150.35
63
Eu152.35
64
Gd157.26
91
Pa(231)
92
U238.07
93
Np(237)
94
Pu(242)
95
Am(243)
96
Cm(245)
65
Tb158.93
97
Bk(249)
66
Dy162.51
67
Ho164.94
68
Er167.2
69
Tm168.94
70
Yb173.04
71
Lu174.99
98
Cf(251)
99
Es(254)
100
Fm(255)
101
Md(256)
102
No(254)
103
Lr(257)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Mole of H2O = _____grams?
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Molar Reflections* Yesterday I took a drink of water And something strange happened in my head I made an actual connectionWith something that my chemistry teacher saidI could visualize lots and lots of moleculesMovin in the water to and fro…and then I thoughtIf I just drank 18 grams of waterI just drank a
Molar reflections, my mind is running freeMaking connections between the world and chemistryA mole is more than a number, more than just a wordIt’s an amount that’s equal to
*Mike Offutt©2002
6.02 times ten to the twenty third!
mole of H2O!
Moles and solution concentration
#M = Molarity of Solution
# = moles per 1000mL of solutionor mmoles per mL of solution
Solution preparationWhat is the formula weight of copper sulfate? CuSO4• 5 H2O CuSO4• 2.5 H2O CuSO4
ExampleCuSO4• 5 H2O = Cu + S + 4 ( O ) + 5 (2 x H + O) = 63.55 + 32.07 + 4(16) + 5 (2 + 16) = 249.62
DEMO
= Hydrated copper sulfate
= anhydrous copper sufate
Solution preparation1. You prepare 100 mL of 0.10 M copper sulfate
using CuSO4• 5H2O. How many grams of CuSO4•5H2O (FW = 249.68) do you weigh out?
249.62 g x 0.10 mol x 100 mL = 2.50 g1 mol 1000 mL
DEMO
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Solution preparation
Calibration line
Volumetric flasks
0.1 M Copper sulfateTransfer 2.50 g of CuSO4·5H2O to a100 mL Vol flask and add water tothe 100 mL calibration line
Solution Preparation
22.99 + 35.46 = 58.45 g NaCl
2. Your teammate adds 1 liter (1000 mL) of water to58.45 g of NaCl to prepare 1.0 M NaCl. Theresulting solution was too dilute ( < 1.0 M ). Why?
Solution dilution
• One buret with water; one buret with sample to be diluted• Deliver the volume (mL) of sample to be diluted into a flask.• Deliver the needed volume (mL) of water into the flask.
Burets clamped to ring stand
Solution Dilution
Initial mols or mmoles = Final mmol or mmoles
Add H2O
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Dilution of Solutions
If V= milliliters: M x V = mmol x mL = mmol
mL
Initial mol or mmol = Final mol or mmol M1V1 = M2V2
If V= liters: M x V = mol x L = mol L
Solution Dilution
0.10 M x 10 mL →
add 10 mL H2O
initial mmoles = final mmoles
Add H2O
1 mmol = 1 mmol
.05 M x 20 mL
Q. What volume of 0.20 M Ni(NO3)2 and water do youuse to prepare 10 mL of 0.16 M Ni(NO3)2?
M1V1 = M2V2
___ mL 0.20 M Ni(NO3)2+ ___ mL H2O ?
Q. What volume of 0.20 M Ni(NO3)2 and water do youuse to prepare 10 mL of 0.16 M Ni(NO3)2?
0.20 M x ? mL = 0.16 M x 10 ml
M1V1 = M2V2
8 mL 0.20 M Ni(NO3)2 +2 mL H2O
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Any Questions?Contact [email protected]