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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 nkerner@umich.edu