1
CHEM*3440
FINAL EXAMINATION
Fall 2004
Duration: 2 hours
1. (8 points) A certain molecule was analyzed by electrospray ionization mass spectrometry. There are a lot
of overlapping bands, but a tetramer of the molecule appears with a residual charge of +7 in a region without
interference at an apparent m/z = 791.
(a) What is the mass of the monomer?
(b) At what value of m/z would the +6 charged tetramer appear?
2. (20 points) Describe the following instrumental components. Outline the physical principles behind
its operation and specify any limitations to its operation.
(a) A proportional counter detector of x-radiation.
(b) A PbS photoconducting transducer for the detection of near-IR radiation.
(c) The He:Ne laser source used in a Michelson Interferometer.
(d) Time-of-flight mass analyzer.
3. (10 points) Sucrose can be enzymatically broken down into fructose and glucose by the enzyme invertase.
sucrose
invertase
glucose + fructose
The following procedure was used to determine the sucrose content of sweet potatoes. A 82.3 g sample of
sweet potato was chopped into a juice. The juice was carefully collected (lots of washings) along with a buffer
solution to form a final solution volume of 500.0 mL. Several 3.00 mL aliquots were taken and analyzed by
a spectroscopic technique for glucose. Samples without any added enzyme determined the free glucose
concentration. Three trials of this gave the following concentrations 212 mg/dL, 204 mg/dL, and 207 mg/dL.
Invertase was added to three more aliquots and after 20 minutes, they were also analyzed for glucose, giving
the following results: 462 mg/dL, 457 mg/dL, and 454 mg/dL.
(a) In units of mg/dL, what is the concentration of glucose arising from the decomposition of sucrose in sweet
potato?
(b) Given the molar mass of glucose (180.16 g/mol) and sucrose (342.3 g/mol), what is the percent sucrose
(w/w) in the sweet potato?
4. (8 points) In hexane, acetone has an absorption maximum at 270 nm, where its molar absorptivity has
a value of 12 M-1 cm-1. Our spectrometer can reliably measure transmittance between 10% and 90%. What
is the range of acetone concentration that can be measured in a 1.00 cm cell under these circumstances?
2
5. (15 points) An ore is analyzed for its platinum content and nickel is used as an internal standard. X-ray
fluorescence is employed as a measurement technique and the count rate (in counts per second) were recorded
for the Kα emission lines for both atoms.
Wt% Pt
Ni (cps)
Pt (cps)
0.00
217
17
2.00
219
222
4.00
216
425
6.00
215
620
8.00
219
827
The calibration data are given here. A linear least squares fit to the data produced the following parameters.
m = 0.4633
sm = 0.0034
b = 0.09075
sb = 0.01635
! x = 20
!x = 120
!y = 9.720
2
i
i
sy = 0.02111
i
!y = 27.481
2
i
An unknown ore was measured four times and the average measured Pt/Ni x-ray ratio was 2.656.
(a) What is the weight% Pt in the unknown ore?
(b) What is the standard deviation of this unknown concentration?
(c) What is the platinum concentration within 95% confidence limits?
6. (6 points) This is a schematic of an electron impact ionization source for a mass spectrometer. Assign an
appropriate voltage to the six specified elements so that a 100 eV electron beam is produced which will ionize
entering gas molecules into cations and then these cations will be extracted for injection into the mass analyzer.
4
Molecules
Drift In
Ions Pass
Into Mass
Analyzer
1
Electron
Beam
5
2
3
6
1. Repeller Plate
2. Wehnelt (ionizer housing)
3. E-beam Filament
4. E-Beam Target Dump
5. Ion Lens 1
6. Ion Lens 2
+1030 V
+1010 V
+1000 V
+930 V
+700 V
0V
3
7. (8 points) Provide the best term that describes the following phrases:
(a) Incident radiation is reemitted without a change in frequency.
(b)The ratio of the power of the beam that emerges from the sample to the power of the beam that impinges
upon the sample.
(c) In a certain solvent, a molecule’s absorbance maximum shifts to shorter wavelengths.
(d) A high -energy gamma ray is absorbed by a nucleus and an electron and a positron are simultaneously
emitted.
(e) An FTIR experiment has an enhanced S/N over a dispersive instrument because no slits are required to
select a narrow band of light in order to achieve a high frequency resolution.
(f) A charged particle emits electromagnetic radiation when it is slowed down by collisions with a solid
target.
(g) Deactivation of an electronically excited state via emission of a photon following excitation and intersystem
crossing to a triplet state.
(h) A technique whereby the temperature difference between an unknown and a reference sample is kept at
0 while their temperatures are ramped up by controlling the amount of heat energy introduced to both.
8. (20 points) Explain the following physical phenomena:
(a) The Raman effect.
(b) Self-absorption in fluorescence.
(c) Ion evapouration process in electrospray ionization mass spectrometry.
(d) Quantum yield.
9. (5 points) An FTIR signal, once it has been modulated by the interferometer, consists of three frequency
components: 100 Hz (amplitude = 1), 250 Hz (amplitude = 2), and 475 Hz (amplitude = 3). The operator
chooses to digitize the signal by sampling at a rate of 400 Hz. Sketch the apparent spectrum that would
be observed, clearly indicating the apparent frequency and relative amplitude at which each spectral element
would appear.
4
Student’s t-Table
t factor for various confidence levels
Degree of Freedom
1
2
3
4
5
6
7
8
80%
3.08
1.89
1.64
1.53
1.48
1.44
1.42
1.40
90%
6.31
2.92
2.35
2.13
2.02
1.94
1.90
1.86
95%
12.7
4.30
3.18
2.78
2.57
2.45
2.36
2.31
99%
63.7
9.92
5.84
4.60
4.03
3.71
3.50
3.36
Mass Spectrometry
R=
m
Δm
Time of Flight
X-ray Beer’s Law
Magnetic Sector
n λ = 2 d sin θ
Bragg’s Law for Diffraction
Molecular Vibrational Energy Levels
Interferometer
1
1
Ev =  v +  hω =  v +  hν


2
2
h
2π
ω = 2π ν
ν=
δ = 2 (dmoving - dfixed)
λ
1
2π
k
µ
Fluorescence
φ=
kf
k f + kic + kec + kisc + kd + k pd
Electricity
P = I2 R = I V = V2/R
Ohm’s Law V = I R
X L = ~L
1 2 vm
2
f
=
=
τ=
τ
λ
vm
2v
2v ν
f = m = 2 vm ν = m
λ
c
Photometric Accuracy
1
1
Δc =
ΔA = −( log10 e)
ΔT
εb
ε bT
F = 2.3 K' P0 ε bc = K c
1
X C =- ~C
m B r e
=
z
2V
FTICR
A = ln(P0/P) = µm ρ x
h=
2 2
1z  1 2 1  B z e r  2 e B z
2V
e

ν cyclotron =
Ek == zLe V = m v = m
⇒
ttransit
2  m  2π m
 m2 
Z 2 = X 2C + X 2L + R 2
5
E = hν = h c/λ = h c ν
1 eV = 8065.6 cm-1 = 1.602 x 10-19 J = 241.8 THz
Statistics
!x
x=
N
RSD = s
A =-log10 T = log10 (Po/PT)
A= ε b c
i
s=
!^x -xh
s mean = s
x
S xx = ! x 2i -
` ! x ij
2
i
N -1
N
2
N
2
1 + 1 + ` y unk -y calibj
M
N
m 2 S xx
Confidence Limits: ! s calc t
N
LOD = 3 sbl
Analytical Sensitivity: c = m
s
sr
s calc = m
Noise
1
N RMS =
N p− p
2 2
S (dB) = 10 log Psignal = 20 log Vsignal = 20 log Isignal
10
10
10
N
Pnoise
Vnoise
Inoise
Thermal noise
Shot noise
q = 1.602 x
LOQ = 10 sbl
Vnoise,rms = 4 kB T R B
Current Amplifier Vo = −ii R f
Inoise,rms = 2 q I dc B
10-19
C
kB = 1.38 x
10-23
Voltage Amplifier Vo = −
J/K
Rf
Ri
Integrator Vo ( t ) = −
Grating Equations
sin α + sin β = k N λ
dβ k N
=
dλ cos β
R=
f Nyquist
dλ
cos β
=
dx k N Lexit
λ
= k N Wgrating
dλ
Vi
1
∫ Vi (t ) dt
Ri C f
Differentiator Vo ( t ) = − R f Ci
1
=
2Δ
dVi (t )
dt
Standard Addition
Two-point cx =
S1
Vs
c
( S2 − S1 ) Vx s
0.2548
Multi-point cx = b cs =
8.7 = 12.7 ppm
( 0.035)(5.00 )
m Vx
sb  2  s m  2

s u = cu
+
 b
 m
More Statistics
S xx = ∑ x i2 −
(∑ x i )
N
2
Syy = ∑ yi2 −
(∑ yi )2
N
S xy = ∑ x i yi −
∑ xi ∑ yi
N
Solutions
6
1. Recall that in electrospsray ionization, charge accumulates on the parent ion by the accretion of protons.
They not only add on the charge but they add a small amount of mass also.
(a) With a charge of +7, and an m/z = 791, the detected species must have a mass of 7 x 791 = 5537 amu.
However, 7 of those amu units are from the H+ bringing on the additional charge. Therefore, the mass of the
tetramer is 5537 - 7 = 5530 amu. The mass of the monomer must be around 5530/4 = 1382.5 amu.
(b) The tetramer mass is still 5530 amu. With 6 protons, its mass becomes 5536 amu. Its m/z value is therefore
5536/6 = 922.7
2. (a) A gas-filled tube with an electrode running down the central axis is the basic proportional counter. The
filling gas is usually Ar, Kr, or Xe. A potential is applied between the central anode and the outer cathode. An
X-ray photon strips electrons from gas atoms. The potential difference accelerates the electrons towards the
anode and the cations towards the cathode. The electron current is measured. In a proportional counter, the
volrtage is in the range of 600-800 V. The unit provides considerable gain in the electron cascade, thereby
improving sensitivity. The magnitude of the current is proprotional to the energy of the X-ray photon. hence,
one can do an energy analysis of the spectral dispersion. The bias voltage is not too high so as to remain oustide
the Gieger range where the current gain is high, but saturated.
(b) PbS is a material which is very resistive. When near-IR radiation impinges upon it, it generates e-h pairs
which increase the conductivity of the material. Photons must have sufficient energy to be detected so that it
cannot work in the mid-IR region.
(c) The purpose of the He:Ne is to provide a high frequency signal which can be used to time the data acquisition
triggering. Its wavelength in the red of the visible spectrum is much greater than any signal to be detected in
the infrared. Everytime the signal at the detector goes through a zero-crossing, we can use that to trigger a data
acquisition event. Counting several zero-crossings before initiating a data acquisition event allows one to change
the timing frequency and hence the spectral range to be covered.
(d) Time of flight. An ionizer produces a collection of ions. A gate pulse drives them into a field-free region
where they drift at a velocity that depends upon their mass. All the ions receive the same kinetic energy, so that
the velocity is smaller for heavier ions. A timer measures the time between the gate pulse (signaling the start
of their flight) and the pulse at the detector (signaling their arrival at the end). The length of the flight path and
the speed of the timing circuit determine the resolution.
7
3. Understand the problem. We first measure the amount of free glucose in the original sample. it constitutes
a background correction. The three samples average to a vlaue of 207.7 mg/dL. We then enzymatically convert
the sucrose to produce glucose and measure it again. Those three trials average a value of 457.7 mg/dL.
(a) The difference between these two is just the concentration arising from sucrose. 457.7-207.7 = 250 mg/dL.
(b) We note that the stoichiometry between the original sucrose and the product glucose is 1:1 We need only
find the moles of gluocse, which gives the moles of sucrose, to find the weight of sucrose and hence the weight
percent.
mg
mglucose = 250 dL # 500.0 mL # 1 dL = 1250 mg
100 mL
1.250 g
nglucose =
= 6.938 # 10- 3 mol = nsucrose
g
180.16 mol
g
msucrose = 6.938 # 10- 3 mol # 342.3 mol = 2.375 g
This mass of sucrose all came from the original 82.3 g sample. The %w/w is simply
% ^w/wh =
2.375 g
# 100 = 2.89%
82.3 g
4 Here we need to determine the concentration at the two transmittances from Beer’s Law. We obtain
P
T = T = 0.9 or 0.1
P0
A = − log10 T
A = ε bc
c1 =
∴ A1 = − log10 (0.9) = 0.04576
c=
A2 = − log10 (0.1) = 1.00
A
εb
A 0.04576
=
= 0.00381 M = 3.81 mM
(12)(1)
εb
The usable range is between 3.81 and 83.3 mM.
c1 =
A
1.00
=
= 0.0833 M = 83.3 mM
ε b (12)(1)
8
5. All the hard work has been done already. We just have to solve some of the easy stuff.
(a)
Ratio = 0.4633 # Wt% + 0.0908
2.656 - 0.0908 = Wt% ^ Pt h = 5.54%
0.4633
(b) This uses the equation for a calibration curve.
sr
s=m
1 + 1 +
M
N
` yunk - ycalibj
2
m2 Sxx
M=5, the number of points in the calibration curve, N = 4 the number of unkown measurements, sr is given
in the statistics, m is the slope of the curve, yunk is the average result found above, ycalib is the average of
the y terms in the calibration curve. We need also to find Sxx. We get
ycalib =
^0.0783 + 1.014 + 1.968 + 2.884 + 3.776h
S xx = ! x2i -
5
` ! xi j
N
= 1.944
2
2
= 120 - 20
5 = 40
Put it altogether to find the standard deviation of the answer.
2
1 + 1 + ^2.656 - 1.944h = 0.0325
s = 0.02111
2
0.4633 4 5
^0.4633h ^40h
(c) Now we invoke Student’s t-table to find the confidence limits. We have 4-1 = 3 degrees of freedom so
that for 95% confidence, we have t = 3.18. This gives a confidence limit of
C.L. ^95%h = s t = 0.0325 # 3.18 = 0.052
N
4
The final answer is 5.54 ± 0.05% Pt.
6. We must account for the charged particle’s polarity and graduate the potentials to drive the electrons and
ions in the correct direction. To do that, we must make the following assignments.
1. +1010 V
2. + 1000 V
3. +1030 V
4. +930 V
5. +700 V
6. 0 V
9
7 (a) Rayleigh scattering
(b) Transmittance
(c) Hypsochromic
(d) Pair Formation
(e) Jacquinot or Throughput Advantage
(f) Bremsstralung radiation.
(g) Phosphorescence.
(h) Power Compensated Differential Scanning Calorimetry (DSC)
8 (a) The Raman effect occurs when light is inelastically scattered, losing or gaining energy from the
vibrational motion of the molecular scatterer. When it loses energy – downshifted – it is a Stokes shifted
peak, leaving energy behind in a molecular vibration. Gaining energy corresponds to an Anti-Stokes shifted
peak, which takes energy out of a molecular vibration. The Raman process is coherent where the photon
interaction is through a virtual state of the photon field – molecule complex.
(b) When a fluorescing molecule’s concentration is high enough, the emitted light can be absorbed by the
other molecules (if the shift between the fluorescence band and the absorption band is small enough). This
will decrease the apparent fluorescence intensity at the detector. A calibration curve will start to curve over
at this point and can actually start to go down at high enough concentration.
(c) Charged droplets quickly evapourate their solvent as they flow towards the capillary. The surface
charge density increases as the droplet size decreases. Finally the surface charge density forces exceed
the surface tension of the droplet and it breaks up into many smaller droplets. The process repeats itself
until small molecular fragments with residual charge enter the mass analyzer.
(d) An absorbed photon can lead to fluorescence if the excited state is not quenched by other, non-radiative
processes such as internal conversion, external conversion, dissociation, pre-dissociation, or intersystem
crossing. The ratio of the fluorescence rates to all possible decay process rates is the quantum yield.
9. Here we have to consider aliasing of the signals. Since the data is acquired at 400 Hz, the Nyquist
frequency is at 1/2 that frequency so that 200 Hz is the highest frequency that can be correctly measured.
The spectrum above 200 Hz is folded over like a z-fold paper. The peak at 250 Hz is 50 Hz above the
Nyquist frequency so it shows up 50 Hz below that level, or at 150 Hz. The 475 Hz signal is 75 Hz above
400 Hz which is fold ed down to 0 so it appears at 75 Hz. The spectrum would look like
0
100
200