Name _________________________________

Analytical Chemistry 415
Hour Exam 2
March 4, 2004

ALL WORK MUST APPEAR ON TEST FOR ANY CREDIT

(10 pts.)  1. List and describe the five major components of all spectrometers.

source - source of the radiation
sample - material being inestigated
monochromator - selects one waelength of light to strike the detector at any one time
detector - monitors the radiation which has interacted with the sample and passed throught the monochromator
signal processor - takes electrical signal from the detector and transforms it into a form useful by the readout deice
readout deice - takes the signal from the signal processor and displays it to the operator

(3 pts.)    2. For an inexpensive photometer 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.

(3 pts.) 3.  Describe what is meant by spectral interferences in atomic absorption spectroscopy, AAS.

• caused by broadband absorption due to combustion products
• power can be reduced by matrix components

(4 pts.) 4. Give two reasons why the transmitted radiation, P, is always less than the radiation, P_o incident onto a cuvet, even if there is no absorption of radiation by the contents of the cuvet?

Any two of the following
• light can reflect from the front surface of the cuvet
• light can reflect from the inside front surface of the cuvet
• light can reflect from the inside back surface of the cuvet
• light can reflect from the back surface of the cuvet
• light can scatter from substances floating in the solution

(4 pts.)    5.  Compare the advantages and disadvantages of sequential inductively coupled plasma (ICP) and multi-channel with Rowland circle optics ICP instruments.

advantages
sequential inductively coupled plasma (ICP)
less complex, thus cheaper
multi-channel with Rowland circle optics ICP
can analyize several metals simultaneously
disadvantages
sequential inductively coupled plasma (ICP)
does each analysis in sequence, thus slower
multi-channel with Rowland circle optics ICP
more complex and close lines can not be separated

(3 pts.)    6. Describe one cause of deviations from Beer’s Law.

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

(8 pts.) 7. In spectrometry, the instrument response can be described by the equation:

G = KP + K'

Define each variable in this equation.

G => instrument response
K' => dark current, set at 0.00 with the beam blocked
Po => power of the radiation at detector with only solvent in the beam
K => slope of linear response of the detector

(6 pts.) 8. Give sources appropriate for each of the following regions of electromagnetic radiation:

(a.) ultraviolet region deuterium lamp

(b.) visible region tungsten lamp

(c.) infrared regionNernst glower or glowbar or incandescent wire

(2 pts.) 9. Give an example of a detector suitable for the UV-Visible region of the electromagnetic spectrum.

phototube
photomultiplier tube
photo-diode
diode-array

(4 pts.)    10. Give an example of one advantage and one disadvantage of a prism monochromator.

Any of the following:
Advantages
prism
• cheaper
Disadvantages
prism
• larger
• may be unstable to atmosphere of lab

(3 pts.)    11. What is the main advantage of Inductive Coupled Plasma (ICP) over Atomic Absorption Spectrometry (AAS)?

much higher temperatures than AA, thus higher number of atoms in excited state, thus stronger signal

(2 pts.)    12. What is the most common radiation source in an atomic absorption spectrometer?

hollow cathode lamp

(2 pts.)    13. 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.)    14. In multi-channel ICP, explain a limitation of the number of elements that can be analyzed at one time using photomultiplier tubes.

the physical size of the photomultiplier tubes causes a limit on how many elements can be analyzed at a gien time

(3 pts.)    15. Describe how a deuterium lamp can be employed to a provide a background correction for an atomic absorption spectrum.

A D2 lamp emits radiation from the far-UV region (less than 190nm through approximately 350nm). With our proprietary in-line system, the D2 emission corresponds precisely with the spectrum of the analyte. Using modulated signals, the absorbance of analyte and background interferences are ratioed resulting in a clean, unbiased absorbance signal.

Name _______________________________

Analytical Chemistry 415
Exam 2
Take Home Questions
Feb. 26, 2002
Due: 4:00 pm March 5, 2004

(10 pts.)  16. Compounds A and B absorb in the Ultraviolet region.  A exhibits an absorption maximum at 267nm (a = 157) and a tailing shoulder at 312nm (a=12.6).  B has an absorption maximum at 312nm (a=186) and does not absorb at 267nm.  A solution containing the two compounds exhibits absorbances (using a 1-cm cell) of 0.726 and 0.544 at 267 and 312nm, respectively.  What are the concentrations of A and B in mg/L?

A = abc

a_A,267nm = 157

a_A,312nm = 12.6

a_B,267nm = 0

a_B,312nm = 186

b - 1.00 cm

c_A = x

c_B = y

a_A,267nmbc_A + a_B,267nmbc_B = 0.726

157 *1 x + 0 *1 y = 0.726

157 x = 0.726

x = 0.00462 M

a_A,312nm b c_A + a_B,312nm b c_B = 0.544

12.6 * 1 x + 186 * 1 y = 0.544

12.6 * 1 * 0.00462 + 186 * 1 y = 0.544

0.058 + 186 y = 0.544

186 y = 0.486

y = 0.00261 M

(6 pts.)    17. Calculate the wavelength and the energy in joules associated with an NMR signal at 60Mhz.

l = c/n = 3.00x10⁸m/sec/60x10⁶6 /sec = 5.0 m
E = hn =(6.63x10^-34J sec)(60x10⁶ /sec) = 3.98x10^-25J

(2 pts.)    18. Express the absorbance, 0.567, in terms of percent transmittance?

%T = 10^-A-A*100 = 10^-0.261*100 = 27.1%

(2 pts.)    19. Convert the percent transmittance data, 64.5%,  to absorbance.

A = -log(%T/100) = - log(64.5/100) =0.190

(6 pts.)    20. For a grating, how many lines per millimeter would be required in order for the first-order diffraction line for wavelength of 585 nm to be observed at a reflection angle of 14 deg when the angle of incidence is 43 deg?

nl = d (sin r + sin i)

1*585x10^-9m = (d) (sin(14 degrees) + sin(43 degrees))

d = (1*585x10^-9)/(sin(14 degrees) + sin(43 degrees)

d = (1*500x10^-9m)/(0.242 + 0.682)

d = (1*585x10^-9m)/(0.924) = 5.41x10^-7m

d = 5.41x10^-4mm

1/d = 1.68x10³ lines/mm

(8 pts.)    21. The molar absorptivity of a particular solute is 2.1 x 10⁴. Calculate the transmittance through a cuvet with a 5.00-cm light path for a 2.00 x 10^-6M solution.

A = ebc

A = (2.1x104)(5.00)(2.00x10-6) = 0.21

T = 10-A = 10-0.21 = 0.62

(6 pts.)    22. What is the molar concentration of a compound in a solution that has an absorbance of 0.98 when measured in a 5.00-cm cell (Assume a molar absorptivity of 9.32 x 10³L cm^-1 mol^-1 at 470 nm)

c = (A/ab) = 0.98/((9.32 x 10³L cm^-1 mol^-1)(5.0 cm)) = 2.1x10^-5M

   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 this problem. Date ______________