Problem Set 2

spectroscopy theory

 

1.         Give the energy of the following things in J, kJ/mol and eV (1 eV = 96.485 kJ/mol)

 

a.         uv photon at 230 nm 

b.         C=O vibration at 1630 cm-1

c.         visible photon at 5400 Å

d.         microwave photon at 175 GHz

 

8.64E-19

520.5443

5.39508

3.24E-20

19.51521

0.202262

3.68E-19

221.7133

2.297905

1.16E-22

0.06984

0.000724

 

 

2.  The emission spectrum of hydrogen gas has many lines that vary in wavelength from the uv to the ir. The lines in the visible portion spectrum form the Balmer series and arise from atomic transitions that originate in upper levels and terminate in the 2nd level (in the Bohr sense).  Balmer showed that the lines could be fit to the form

 

1/l = RH (¼ – 1/n2) where n is the principle quantum number and RH is the Rydberg Constant (109677 cm–1). 

a.         calculate the positions of the first 4 Balmer lines in nm.

656

486

434

410

 

b.         describe the color of each of the four lines

c.         determine the energy of these transitions in both of the commonly used

spectroscopic units eV and cm-1.  1 eV = 96.485 kJ/mol.

15233

1.89

20564

2.55

23032

2.86

24373

3.02

 

3.         The colorless, water-soluble ligand 1,10-phenanthroline (phen) forms a chelate complex with Fe2+.  The wavelength maximum is 510 nm.  This ligand can be used to determine the concentration of unknown Fe2+. The experimental procedure involves treating an iron containing solution with a reducing agent to ensure than all iron is in the +2 oxidation state, then adding excess phen.

 

 

 

 

An 8.00 ml aliquot of a solution containing 15.00 mg of iron is added to a 100 ml flask and diluted to volume with distilled water.  The absorption at 510 nm is recorded as 0.397 using a standard 1 cm cell.  In a separate experiment, a 5.00 ml aliquot of an unknown iron solution was also diluted to 100 ml, but the 1 cm cell was being used by some pesky graduate student.  Instead, a longer 20 cm cell was used and the absorbance at 510 nm was 0.142.  How many grams of Fe2+ were in the aliquot?  What color is the chelate complex?

15ug / 55.85 g/mol = 2.68 x 10-7 mol in 100 ml = 2.68 x 10-6 M = c

A = ebc = 0.397 =e(1)(2.68 x 10-6)   e = 1.48 x 105

 

0.142 = 1.48 x 105 (20) c      c = 4.79 x 10-8 M = 4.79 x 10-9 moles in 100 ml =  0.268 ug

 

4.         A hyperactive analytical chemistry professor assigns his students the problem of determining the amount of copper in his Diet Coke® (the copper would be introduced in the bottling process from the copper pipes used to transfer the carbonated water to the storage tank). 5 ml samples were made available to the students.

a.         Diligent student A decides to make a calibration curve using a 50 ppm Cu standard that she finds lying around in the lab.  She adds the following amounts of the 50 ppm standard to a 100 ml volumetric and dilutes with distilled water:

0.635 ml  :  2.54 ml  :  5.64 ml  :  7.62 ml  :  9.53 ml (the standard is cheap)

the following %T readings were taken

87.1  :  53.7  :  33.1  :  17.8  :  14.1

She then transfers the entire 5 ml sample of Diet Coke® to a 100 ml volumetric flask and measures the %T of 43 %.

What did researcher A find the [Cu] in the diet coke to be?  A standard can of Diet Coke® contains 355 ml.  How many grams of copper are in the can.  Assume the density of all solutions is 1.0 g/ml

 

50 ppm = 50ug/ml = 7.87 x 10-7 mol/ml diluted to 100 ml.

43% = 0.3175  = 11414x + 0.0163   x= 3.1 x 10-5 M = 3.1 x 10-5 mole in 1 L, 3.1 x 10-6 moles in 100 ml all of which came for 0.005 L , [Cu] = 6.2 x 10-4 M * 0.355 L = 2.2 x 10-4 moles in can * 63.5 g/mol,   13.9 mg in can.

 

b.         Diligent but lazy student B figures out that he can figure out the Cu concentration

by only making 2 measurements using an internal standard.  He takes the 5 ml sample and divides it into 2 equal parts.  He puts one 2.5 ml sample in a 100 ml volumetric flask and dilutes to volume.  This sample has a %T of 66%.  Into a second 100 ml flask, he adds the 2.5 ml sample and 4.5 ml of the 50 ppm Cu standard he found lying on student A’s desk.  He dilutes with water to the mark and measures a %T of 24.8%.

What did researcher B find the [Cu] to be?  in the total can?

 

C0  = (C1A0)/(A1 – A0)

 

66% = .180 = A0

24.8% = .606 = A1

c1 = 7.87 x10-7 moles/ ml * 4.5 = 3.54x 10-6 mol diluted to 100 ml = 3.54 x 10-5  M

 

co = 3.54 x 10-5 M (0.180) / (.426) =  1.5 x 10-5 M in 100 ml = 1.5 x 10-6 mols in sample. all of which came from 0.25 ml , [Cu] in can or 6.0 x 10-4M.