1

Experiment 1 (2 session lab)

Electrons and Solution Color

LAB Perform Parts 1, 2, 3, 4, and 5 B.

Work on discussion question (assigned) anddiscussion abstract. Work on the team report. Be sure to refer to“Team Report Tips”, page 30.

Pre-lab Report, page 29

Course theme

“There are structure andproperty and periodic tablerelationships”

Useful web sites are: www.merlot.org www.webelements.com

Part 1. Preparation and Color of Solutions

1H+

Hydrogen IIIA IVA VA3Li+

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 0Sn2+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 2Pb2+Lead

8 3BiBismuth

Teams prepare solutions with different metal ionsRecord the color of solutions.

Solution ColorSolutions with ions of the same metal element withdifferent ion charges may have different colors.

DEMO: V2+ vs. V3+ etc.

1H+

Hydrogen IIIA IVA VA3Li+

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 0Sn2+Tin

5 1Sb

Antimony

Periodic Table with common metal ion charges.

Part 1 Data analysis and discussion.

Is the presence or absence of saltsolution color predictable from ___? 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

Discussion PreparationManipulate the class data.

You will NOT get points for justreproducing the class data.

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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.

Part 3. Solution Color and Light Interaction.

What is the relationship between the visiblecolor of a solution and its absorption andtransmission of light?

Discussion question 5 p. 39

Light source Diffractiongrating

Samplesolution

Detector

Use a spectrophotometer to examine the relationshipbetween solution color and absorption andtransmission of visible light.

Spectrophotometer

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 A plot of absorbance versus wavelengths.

00.10.20.30.40.50.60.7

400 450 500 550 600 650 700

Abso

rban

ce

Wavelength λ (nm)

Viole

t

Blue

Yello

w

Gree

n

Oran

ge

Red

Absorbance λmax

Absorption Spectrum

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Can be used to identify unknown samples.

Absorption Spectrum Chem.125/126 Mascot

__________________________________________________________

Moles and the Periodic Table?

?

National Mole Day: October 23. Why?

1 H1.0083 Li6.94011 Na22.99119 K39.10037 Rb85.4855 Cs132.9187 Fr(223)

4 Be9.01312 Mg24.3220 Ca40.0838 Sr87.6356 Ba137.3688 Ra226.05

58 Ce140.13

5 B10.8213 Al26.9831 Ga69.7249 In114.8281 Tl204.39

90 Th232.05

6 C12.01114 Si28.0932 Ge72.6050 Sn118.7082 Pb207.21

7 N14.00815 P30.97533 As74.9151 Sb121.7683 Bi208.9

8 O15.99916 S32.0634 Se78.9652 Te127.6184 Po(209)

9 F19.0017 Cl35.45735 Br79.91653 I126.9185 At(210)

10 Ne20.18318 Ar39.94436 Kr83.8054 Xe131.3086 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 Tb158.9397 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 H1.0083 Li6.94011 Na22.99119 K39.10037 Rb85.4855 Cs132.9187 Fr(223)

4 Be9.01312 Mg24.3220 Ca40.0838 Sr87.6356 Ba137.3688 Ra226.05

58 Ce140.13

5 B10.8213 Al26.9831 Ga69.7249 In114.8281 Tl204.39

90 Th232.05

6 C12.01114 Si28.0932 Ge72.6050 Sn118.7082 Pb207.21

7 N14.00815 P30.97533 As74.9151 Sb121.7683 Bi208.9

8 O15.99916 S32.0634 Se78.9652 Te127.6184 Po(209)

9 F19.0017 Cl35.45735 Br79.91653 I126.9185 At(210)

10 Ne20.18318 Ar39.94436 Kr83.8054 Xe131.3086 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 Tb158.9397 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 preparation

Calibration line

Volumetric flasks

Solution preparation

What 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 preparation

1. You prepare 100 mL of 0.10 M copper sulfateusing CuSO4• 5H2O. How many grams ofCuSO4•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

Solution preparation

1. You prepare 100 mL of solution using CuSO4•5H2O. You weigh out 5.00 gram of CuSO4•5H2O(FW = 249.68). What is the M of the solution?

5.00 g x 1 mol x 1000 mL = 0.20 M 249.68 g 100 mL

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Solution Preparation

22.99 + 35.46 = 58.45 g NaCl

2. Your teammate adds 1 liter (1000 mL) of water to58.5 g of NaCl to prepare 1.0 M NaCl. Theresulting solution was too dilute ( < 1.0 M ). Why?

Solution Preparation and dilution

VolumetricFlask

Solution prep

BuretteSolutiondilution

Solution dilution

Burets clamped to ring stand Flask beneath buretOne with water; one with sample

Serial Dilution

Example:0.10 M → 0.05 M → 0.025 M → 0.0125 M

AddequalvolumeH2O

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

Dilution of Solutions

If V= milliliters:M x V = mmol/mL x mL = mmol

M1V1 = M2V2If V= liters:

M x V = mol/L x L = mol

Example: 15 mL of a 2 M solution contains 30 mmol.

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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

Any Questions?Contact nkerner@umich.edu