Name ________________________

Analytical Chemistry 415
Exam 2
Feb. 23, 2006

All work must appear on the exam for any credit!

Under Construction

(3 pts.) 1.  For an inexpensive photometer with an analog meter that displays both absorbance and %T scales, why is it advisable to take the readings in %T and convert to absorbance?

The %T scale is linearwhie the absorbance scale is exponential. Therefore, it is very hard to interpolate values on the absorbance scale compared to the %T scale.

(6 pts.) 2. Sketch a typical line spectrum and a continuum spectrum and describe how they differ.

As can be seen, a line spectra has a very narrow band width of wavelengths whereas a continuum spectra show a broad range of absorbed wavelengths.

(4 pts.) 3. Compare a flame photometer and a flame atomic absorption spectrometer as to similarities and differences.

In both the flame breaks down the sample into atoms of the analyte and causes influences the distribution of the atoms between the ground and excited state. In a flame photometer the flame causes the atoms in the sample to exict light. In the flame atomic absorption light is emitted by a separate lamp and the sample absorbs some of that light depending on the concentration of the analyte.

(2 pts.) 4. What material must be used for the cuvet while working in the UV spectral range?

Quartz

(2 pts.) 5. Give an example of a detector that can be used in the UV-Visible spectral range?

phototube
photomultiplier tube
photo-diode
diode-array

(2 pts.) 6. What is used as the source of radiation in atomic absorption spectroscopy?

HCL - hollow cathode lamp

(2 pts.) 7. Give an example of a source that can be used in the UV-Visible spectral range?

ultraviolet region - deuterium lamp

visible region - tungsten lamp

(8 pts.) 8. State Beer’s Law and define each parameter.

Beer's Law: A = abc

Where A => absorbance, a measure of the amount of radiation that is absorbed by the sample
a => absorptivity, depends on the substance being analyzed and the analytical wavelength
b => path length, how much radiation is absorbed depends on how much of the solution the radiation passes through
c => concentration, the greater the concentration, the greater the amount of radiation is absorbed

(8 pts.) 9. Describe the advantages and disadvantages of a prism versus diffraction grating monochromator.

Advantages
Gratings
• dispersion nearly constant w/ wavelength
• simplier monochromator
• better dispersion for same size
Prisms
• cheaper in the past
Disadvantages
Gratings
• stray radiation
• higher order spectra
Prisms
• larger
• may be unstable to atmosphere of lab

(3 pts.) 10. Give one advantage of a diode-array spectrometer over a conventional spectrometer.

A diode-array spectrometer analyzes all wavelengths at the same time, thus it is faster.

(3 pts.) 11. Describe one source of reflection or scattering loss in the use of a cuvet.

Any one of the following:
• reflection off front surface outside interface
• reflection off front surface inside interface
• reflection off back surface inside interface
• reflection off back surface outside interface
• scattering losses in solution

(3 pts.) 12. Why do optical instruments use front surface mirrors?

This avoids absorption of some of the radiation from passing the radiation through an additional medium.

(3 pts.) 13. Describe a cause for deviation in Beer's Law.

• chemical interactions
• high concentration causing some of the molecules to not see the beam of the radiation
• non-monochromatic radiation
• stray light

Name ________________________

Analytical Chemistry 415
Exam 2
Take Home Questions
Feb. 22, 2006
Due:  4:00 PM Tuesday Feb. 28, 2006

All work must appear on the exam for any credit!

(8 pts.) 14. Compare the resolution obtainable with a 15.2-cm grating of 1180 grooves/mm to another grating of the same size but with 295 grooves/mm when each grating is used in any of the first four orders.

R1/R2 = (152mm*1180grooves/mm)/(152mm*295grooves/mm) = 4

(8 pts.) 15. A Duboscq colorimeter has a 50.0 mm scale for measuring the depth of each solution.  The colors for a known and an unknown matched when the known solution was set at 40.0 mm and that for the unknown at 33.0 mm.  Calculate how much iron must be in 50.0 mL of unknown solution if the known solution contained 2.00-mg of iron per 100. mL of solution.

Beer's law states that A = abc. The operation of the Duboscq colorimeter assumes that the value of A is the same on both sides of the instrument. Since the same analyte is being investigated on both sides of the instrument, we will assume that a is constant. That menas that b1c1 = b2c2 where b1 = 40.0 mm and c1 = 2.00 mg Fe/100mL and b2 = 33.0 mm and c2 = ? c2 = (b1c1)/b2 = (2..00*40.0)/33.0 mg/100mL = 2.42 mg/100mL #mg Fe = (2.42 mg/100mL)(50.0mL) = 1.21 mg

(6 pts.) 16. For Na⁺ ions, what is the ratio of the number of particles in the 3p excited state to the number in the ground state at 2500^oC.

Pj/Po = 3

1 erg = 6.24e11ev

Ej = 2.50 ev

k = 1.38e-23 J/K


E_j/kT = (2.5/6.24e11*1.00e7)/(1.38e23*2773) = 1.05e1

exp(-(E_j/kT) = 2.84e-5

N_j/N_o = P_j/P_oexp(-(E_j/kT) = 3*2.84e-5 = 8.52e-5

(8 pts.) 17. For a grating, how many lines per millimeter would be required in order for the third-order diffraction line for wavelength of 567 nm to be observed at a reflection angle of 27 deg when the angle of incidence is 61 deg?

nl = d (sin r + sin i)

3*567x10^-9m = (d) (sin(27 degrees) + sin(61 degrees))

d = (3*567x10^-9m)/(sin(27 degrees) + sin(61 degrees)

d = (3*567x10^-9m)/(0.454 + 0.875)

d = (3*567x10^-9m)/(1.33) = 1.28^-6m

d = 1.28x10^-3mm

1/d = 7.81x10⁵ lines/mm

(10 pts.) 18. The molar absorptivity of a particular solute is 3.6 x 10³. Calculate the percent transmittance (%T) through a cuvet with a 4.00-cm light path for a 1.23 x 10^-4M solution.

A = abc = (3.6 x 10³)(4.00)(1.23 x 10^-4) = 1.77

%T = 10^(-T)x100 = 1.69E-02 x100 = 1.69%

(6 pts.) 19. Calculate the wavelength and the energy in joules associated with a proton NMR signal at 60 Mhz.

l = c/n = (2.998x10¹⁰mm/s)/(60x10⁶ /s) = 500 mm

E = (6.626x10^-34J*s/photon)(60x10⁶ /s)(6.02x10²³photons/mol) = 2.39x10^-2J/mol

(2 pts.) 20. Convert the transmittance of 0.261 to absorbance.

A = -log(T) = -log(0.261) = 0.583

(2 pts.) 21. Express the absorbance, 0.645, in terms of transmittance?

T = 10^-A = 10^(-0.645) = 0.226

I, ________________________________, did not receive nor did I give aid on the preceding questions. Nor did I see others give or receive help on the preceding questions. I understand that I may use my notes, text, and other materials to work these problems.

Date ______________