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Exam 3 Material Type: Exam; Class: Fundamentals of Analytical Chemistry; Subject: Chemistry ; University: Illinois State University; Term: Summer 2007;
Typology: Exams
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Chemistry 215 – Fundamentals of Analytical Chemistry
July 10, 2007 100 POINTS Name __________________________ SEE EQUATIONS AND DATA ON LAST PAGE
PART I. DEFINITIONS (12 Points) Define the following terms as they relate to the topics covered in this course.
The energy per second per area of light
Any energy state in an atom or molecule that is higher in energy than the ground state
The distance between any two, adjacent identical points on a wave.
The emission of light that results from the absorption of energy by an atom or molecule to a singlet excited state (S 1 ), followed by some energy loss via vibrational relaxation (R) and finally the emission process where there is relaxation to a singlet ground state (S (^) o ).
PART II. CALCULATIONS (64 Points) Show all calculations in order to earn partial credit.
a. Calculate the wavelength, λ. (6 Points)
λ = c/ν = (3.0x10 8 m/s) / (2.4x10 9 s-1) = 0.125 m
b. Calculate the wavenumber, ν (in cm-1^ ). (6 Points)
ν = 1/λ = (12.5 cm) -1^ = 0.0800 cm-
c. Calculate the energy, E, of a single photon. (6 Points)
E = hν = (6.63x10 -34^ J.^ s) (2.4x10 9 s-1) = 1.64x10 -24^ J
d. What is the energy of a mole of photons? (6 Points)
(1.6x10 -24^ J) (6.02x10 23 photons) = 0.963 J/ mole (1 photon) (1 mole)
A = ε bc C 1 V 1 = C 2 V (^2)
c = A / ε b = (0.275) / (703 M -1cm -1) (1.00cm) C 2 = (100.00 mL)(3.9x10 -4^ M) (10.00 mL) c = 3.91x10 -4^ M C 2 = 3.91x10 -3^ M
(3.91x10 -3^ mol) (0.25000 L) (270.0 g) (1000 mg) (L) (mol) (g) = 264 mg
A = ε bc
c = A / ε b = (0.296) / (2.24x10 3 M-1cm -1) (1.00cm)
c = 1.32x10 -4^ M
(1.32x10 -3^ mol) (0.250 L) (54.398 g) (L) (mol) = 1.81x10 -3^ g
1.81x10 -3^ g x 100 wt% = 0.325 g
wt% = 0.557%
Concentration in Aliquot Amount in 100.0 mL
From calibration curve…. (3.85 μg) (100.0 mL) = 385 μg y = mx + b (mL) y ≡ Absorbance (A) x ≡ micrograms (μg)
Thus, Wt % (A – b) (0.230 – 0.105) μg = m = (0.0162) = 7.71 μg 385 μg / 2.10x10 3 μg x 100
7.71 μg / 2.00 mL = 3.85 μg/mL = 18.3%
PART III. SHORT ANSWER (12 Points)
See text for single-beam block diagram and components.
a. Make a sketch of the general experiment utilized by Einstein to illustrate the Photoelectric Effect. (4 Points)
See lecture notes for sketch.
b. What were the two significant outcomes of Einstein’s experiment, i.e., explain how the results of the experiment changed our concept of light. (4 Points)
E (^) photon = hνphoton
hc E = h = =
A log log 0
I = kP 0 c e EkT g
g N
0
0
−Δ ⎟⎟ ⎠
c = 2.998 × 108 m s-1^ h = 6.626 × 10 -34^ J s N = 6.02 × 1023 mol-