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Chemistry 215 – Fundamentals of Analytical Chemistry
Exam 3
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.
- Irradiance
The energy per second per area of light
- Excited State
Any energy state in an atom or molecule that is higher in energy than the ground state
- Wavelength
The distance between any two, adjacent identical points on a wave.
- Fluorescence
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.
- Wireless networking (WiFi) works with radio waves at 2.4 GHz (G = giga, or 10^9 ). Answer the following questions for electromagnetic radiation of this frequency.
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)
- The drug tolbutamine (MW = 270.0 g mol-1^ ), a treatment for type 2 diabetes, has a molar absorptivity of 703 M-1^ cm-1^ at 262 nm. One tablet was dissolved in 250.00 mL of water. A 10.00 mL aliquot of this solution was diluted to 100.00 mL in a volumetric flask. This diluted solution exhibited an absorbance 0.275 at 262 nm in a 1.00 cm cell. Calculate the mass (in mg) of tolbutamine in the tablet. (8 points)
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 0.325 g sample of steel is analyzed for its manganese content by dissolving the sample in nitric acid, oxidizing the manganese to the intensely purple permanganate ion (MnO 4 -^ ), and then diluting the solution with water to 250 mL in a volumetric flask. The absorbance at 525 nm in a 1.00 cm cell is 0.296. The molar absorptivity of MnO 4 - at 525 nm is 2.24 × 103 M-1^ cm-1^. Calculate the wt % of Mn in the steel. (AWMn = 54. g mol-1^ .) (8 Points)
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%
- A tissue sample is analyzed for its protein content by the Lowry method. First, a series of standards were prepared containing 5 to 20 micrograms of a pure protein. After following the Lowry procedure, the absorbance of each standard was measured at 750 nm in a 1.00 cm cell. The data were then subjected to a linear least square analysis resulting in the following parameters: m = 0.0162, b = 0.105. Next, the tissue sample weighing 2.10 mg was digested to release the protein then diluted to 100.0 mL. A 2. mL aliquot of this solution was analyzed by the Lowry procedure in the same manner as the standards resulting in an absorbance of 0.230 in a 1.00 cm cell. Calculate the % protein in the sample. (8 Points) Given: 1 mg = 1000 μg.
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)
- Draw a block diagram of a typical single-beam spectrophotometer, labeling each component. (4 Points)
See text for single-beam block diagram and components.
- Einstein’s Photoelectric Effect
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)
- The energy content of light is proportional to its corresponding frequency (ν).
E (^) photon = hνphoton
- In theory, a single “particle” (i.e, photon) of light could eject a single electron from the surface of a metal. This lead to the particle-wave duality of light.
Equations and Data
c = λν ν
ν hc
hc E = h = =
P 0
P
T = % T = T × 100 T
P
P
A log log 0
A = ε bc C 1 V 1 = C 2 V 2 n λ = d (sin θ−sin φ)
I = kP 0 c e EkT g
g N
N /
0
0
−Δ ⎟⎟ ⎠
c = 2.998 × 108 m s-1^ h = 6.626 × 10 -34^ J s N = 6.02 × 1023 mol-
