Unformatted text preview: CHEM*3440
Instrumental Methods
MidTerm Examination
October 28, 2003
Duration: Two Hours
1. (5 points) A real battery actually behaves more like a voltage source in series with a
resistance, called its internal resistance. (This arises from the fact that the chemical
reactions have finite rate constants and can produce electrons only at a finite rate.) An
older 1.5 V dry cell battery, when shorted out with a wire of insignificant resistance, is
able to deliver only 5.0 mA of current. What is the internal resistance of the battery? +
Rint V
–
2. (10 points) You are analyzing a 2.0 g biological tissue sample, trying to determine if Ni is present. Your
experiment is sensitive down to the 0.1 ppb level. The tissue was obtained by slicing it with a stainless steel
scalpel. The stainless steel is 10% Ni.
(a) What mass of the scalpel rubbing off would contribute a sufficient quantity of Ni to give a positive (0.1
ppb) test result?
(b) Assume that the density of the steel is 7.0 g/cm3 and that the scalpel lost the material in a single, perfectly
cubic chip. What is the length of the edge of that cube? (For comparison, a red blood cell has a diameter of
about 10 µm.)
10 µF
3. (15 points) Analyze the accompanying op amp circuit.
100 kW
There are two input signals: VDC is a constant 1.00 V
V
and VAC is a sine wave of amplitude ±1V and a frequency DC
of 10 Hz. The op amps are being powered by ±10 V.
Carefully graph a voltagetime plot at the three indicated
locations, A, B, and C. Show all calculations you used
to arrive at any time or voltage relationships. Draw your VAC
curves carefully and emphasize the distinguishing features. –
+ 1 kW
B 1 kW
–
+ A 100 kW 4. (5 points) A spectrometric assay is designed to work with a final concentration of nicotine in the range
between 0.005 and 0.020 mg/L. If a 500.0 mL sample of urine is collected and all of the nicotine present is
concentrated during sample preparation into a final solution volume of 10.00 mL, what is the minimum
concentration of nicotine that can be measured by the assay? (Report your results in µg/L). C 5. (10 points) Here is an extension of our class
discussion of the op amp circuit that represents
A
a simplified threeelectrode electrochemical
cell. Two additional circuit elements have been
added (one is NOT the booster amplifier we
discussed). Bear in mind that the reference
electrode should draw negligible current, the
counter electrode is only present to carry the
2
–
main current, and that the working electrode
–
3
1
Vinput
+
is held at virtual ground.
+
(a) Identify the three electrodes (1, 2, and 3)
by name.
cell
(b) Give a general name for the role played
by the two new circuit elements labelled A and B and briefly (a sentence) describe what advantage they
bring to the circuit. 2 B 6. (16 points: 3 x 5 points + 1 x 1 point) We are passing a 1 A DC current through a 10 kW resistor at 20 ˚C.
It is amplified by an amplifier with a bandwidth of 20 kHz.
(a) What is the rms thermal noise voltage produced?
(b) What is the rms shot noise current produced?
(c) In this resistor, what is the rms shot noise voltage produced by the shot noise current?
(d) Which noise source dominates under these conditions, thermal or shot?
7. (12 points) Describe the differences in function and application of each pair of components in the following
list. You only need to discuss the basic differences and how different applications can benefit from those
differences; don’t go through and explain the complete function of all items. (You should only need two or
three sentences for each pair).
(a) filters or monochromators as wavelength selectors.
(b) phototubes or photomultiplier tubes as detectors.
(c) spectrophotometers or photometers as spectral systems.
(d) a tungsten lamp or a halogen lamp as a visible light source.
8. (10 points) The two dotted curves are absorbance
spectra for Co and Ni EDTA complexes. From these
spectra, we can derive the molar absorptivities at any
wavelength. The following data have been determined:
e(Ni, 500 nm) = 1.0 M1 cm1
e(Ni, 590 nm) = 8.7 M1 cm1
e(Co, 500 nm) = 13 M1 cm1
e(Co, 590 nm) = 2.7 M1 cm1
The solid curve is that measured from a mixture of the
two ions and EDTA in a 1.00 cm cell. What is the
concentration of Ni and Co in the mixture? 1.1
0.9
0.7
A
0.5
0.3
0.1
400 500
600
Wavelength (nm) 700 3 Statistics = x Â xi = Â ( xi  x ) N
2 s2 Electricity
LOD = 3 sbl
LOQ = 10 sbl
Analytical sensitivity g = m/s
Confidence Limits = tsm N 1 E = hn = h c/l = h c n
1 eV = 8065.6 cm1 = 1.602 x 1019 J = 241.8 THz Noise
1
N RMS =
N
2 2 p p
S (dB) = 10 log Psignal = 20 log Vsignal = 20 log Isignal
10
10
10
N
Pnoise
Vnoise
Inoise
Thermal noise Vnoise,rms = 4 kB T R B Shot noise Inoise,rms = q = 1.602 x 1019 C Twopoint cx = kB = 1.38 x 1023 J/K S1 Vs
c
( S2  S1 ) Vx s f Nyquist 1
=
2D 0.2548
Multipoint cx = b cs =
8.7 = 12.7 ppm
( 0.035)(5.00 )
m Vx
Grating Equations
sin a + sin b = k N l db k N
=
dl cos b dl
cos b
=
dx k N Lexit l
= k N Wgrating
dl A =log10 T = log10 (Po/Pi)
A=ebc Current Amplifier Vo = ii R f
Voltage Amplifier Vo =  2 q I dc B Standard Addition R= P = I2 R = I V = V2/R Photometric Accuracy
1
1
Dc =
DA = ( log10 e)
DT
eb
e bT = s2
s
RSD =
x
s
sm =
N
s Ohm’s Law V = I R Rf
V
Ri i Integrator Vo ( t ) =  1
Ú Vi (t ) dt
Ri C f Differentiator Vo ( t ) =  R f Ci dVi (t )
dt 4
CHEM*3440
Instrumental Methods
MidTerm Examination
October 28, 2003 ANSWERS
1. Short circuiting a battery is just completing the circuit. It is just a trivial
circuit to analyze. The internal resistance is found from Ohm’s Law. V = IR R= V
1.5 V
=
= 300 W
I 0.005 A +
Rint V
– 2. (a) We first determine how much scalpel is needed to generate the minimum 0.1 ppb signal. We need to
know how much nickel is needed. Recall that a “parts per” measure refers to a ratio. Often that ratio is between
two masses – a mass ratio. Another common ratio is masstovolume ratio, such as in mg/ml. But that is
really just a masstomass ratio, based upon the observation that the density of water is almost exactly 1 g/ml.
When water is the solvent and we are not worried about 5 decimal place accuracy, then a masstovolume ratio
is the same as a masstomass. In this case, we have a mass ratio, since the stainless steel is not water and, as
given in the problem, its density is not 1.
By the definition of a billion, we have that this mass ratio is 0.1 ppb = 0.1¥ 10 9 = 1.0 ¥ 1010
This must be ratio of Ni when compared to the total mass of the biological sample, which is 2.0 g. The required
amount of Ni is m( Ni )
m ( Ni )
=
= 1.0 ¥10 10
m( sample) 2.0 g
m( Ni) = 2.0 g ¥ 1.0 ¥10 10 = 2.0 ¥10 10 g
This Ni arrives in the biological sample by a piece of stainless steel chipping of the scalpel. Since the stainless
is only 10% Ni, to get that much Ni, the stainless must have a mass given by m( Ni )
= 10% = 0.10
m( scalpel )
m( scalpel ) = m( Ni)
= 2.0 ¥ 10 10 ¥ 10 = 2.0 ¥ 10 9 g = 2.0 ng
0.1 5
2. (b) Given this mass of stainless steel, we can find the volume of such a chunk from its density. We have m 2.0 ¥ 10 9 g
V= =
= 2.86 ¥ 10 10 cm 3
r
7.0 g 3
cm
That volume, assumed to be a cube, has its sides equal in length to the cube root of the volume. 3 a = 3 V = 2.86 ¥ 10 10 3 cm = 6.6 ¥ 10 4 cm = 6.6 mm This is smaller than a red blood cell. Contaminating the sample with something that you would think is
completely inert is a common worry. Cleaning a piece of silicon in a glass beaker that had previously been
used to clean stainless steel tools often transfers enough Ni to destroy the surface order of the Si wafer.
3. The opamp at point A is operating in the openloop configuration so that it is functioning as a comparator.
Since its input is a sine wave, it will produce a square wave of the same frequency but with an amplitude
defined by the operating voltages of the device, namely ± 10 V. At 10 Hz, each complete period will take 0.1
s. The correct graph would be
10 V –10 V
0.1 s
0.2 s
At point B, the op amp is set up so as to function as an integrator. Since the input is a DC voltage, the output
will be a linear ramp whose slope will be determined by the value of RC. Also, since the input is in the
inverting inputs and the input voltage is positive, the slope will negative. To determine the magnitude of the
slope, we find the value of RC.
Since RC = 1 s, and since
the input voltage is 1 V, the
ramp slope is 1 V/s.
time (s)
0.1 VB =  1
Vin
Vin
Ú Vin dt =  RC Ú dt =  RC t
RC ( RC = (100,000 W) 10 ¥ 10 6 ) 0.1
0.3 F =1 s
0.5 0.3 0.5 6
The other op amp is a summing amplifier. Recall that the amplification ratio is Rfeedback/Rinput. With these
resistors, it is clear that the ramp voltage experiences a unity gain amplification while the square wave
experiences a 100fold attenuation. Since the square wave has an amplitude of 20 V (± 10 V), the attentuated
square wave has an amplitude of 0.2 V (±0.1 V). The input is into the inverting input so the signal polarity
is reversed. The final waveform has the form
0.5 Note how the square wave steps 0.2 V up and down and goes
through a complete period in 0.1 s. During that time, the ramp
has increased its potential by 0.1 V. Note that the slope affects
all aspects of the square wave (i.e. it is not flat on top and
bottom, but is slanted at the same slope.) 0.3
0.1
0.1 0.3
0.5
time (s) 4. Here you need to recognize that the detection of 0.005 mg/L constitutes the quantitation limit of the
technique. Hence the final analyte sample must have at least this concentration of nicotine. However, our
workup procedure concentrates the nicotine from 500 mL to 10 mL. By this volume ratio, we can lower the
detection limit in the original sample. The concentration of the original sample at the quantitation limit is mg
mg
= 5.0
L
L
mg 10.00 mL
mg
5.0
¥
= 0.1
L 500.0 mL
L
0.005 5. (a) Electrode 3 is at the input to an op amp whose other terminal is at ground. We know that the op amp
works to keep its inputs at the same potential so this electrode is at virtual ground. We should recall that the
working electrode is at virtual ground. Electrode 2 passes through a voltage follower whose task is to draw
as little current from that electrode as possible. We should recall that the reference electrode draws minimal
current. The other electrode passes the current for the experiment through it. It is the counter electrode. Hence
we have
Electrode 1: Counter electrode
Electrode 2: Reference electrode
Electrode 3: Working electrode
(b) Component A: This op amp functions as a voltage follower. It is there to buffer the reference electrode
to ensure that it draws negligible current.
Component B: This op amp is a current to voltage converter. It takes the current at the working electrode and
converts it to a voltage for measurement. An IV curve is a plot of this potential against the potential at the
reference electrode. 7
6. (a) Use the equations for thermal noise. The original exam had an incorrect value for Boltzmann’s constant.
This document has corrected that and I use the correct value here. The question was marked based upon the
constant you used. You need to find the resistance of the resistor. Its power is given as 10 kW. Since this is
being produced when a 1 A current is flowing, we know from P = IV that V = P/I = 10 kW/1 A = 10 kW. If
you just used the number 10k, you still got the same result. ( ) ( )( Vthermal = 4 k B T R B = 4 1.38 ¥ 10 23 ( 298) 10 4 2 ¥10 4 ) = 3.10 ¥ 10 12 = 1.7 ¥ 10 6 V = 1.7mV
(b) The shot noise current is ( ) ( Ishot = 2 q IDC B = 2 1.609 ¥ 10 19 (1) 2 ¥ 10 4 ) = 6.405 ¥10 15 = 8.0 ¥10 8 A = 80nA
(c) The shot noise current is passing through this resistor. If we take the current and resistor and Ohm’s Law,
we can find the voltage that will result from this current and then can compare it to the voltage noise arising
from thermal sources. V = IR = 80 ¥ 10 9 4 A ¥ 1 ¥10 W = 8.0 ¥10 4 V = 800 mV (d) Clearly, the 800 µV coming from shot noise swamps the 1.7 µV coming from thermal noise. Shot noise
dominates.
7. (a) Filters are lower cost and have a wider bandwidth. They are more useful for portable measurements.
The monochromator can measure over a wide spectrum and has an adjustable bandwidth. The device is more
costly and cumbersome.
(b) Phototubes are low cost and low sensitivity light detectors. They are acceptable when light levels are high
and when portability is important. The photomultiplier tube can measure much lower light levels by providing
electron gain through a series of dynodes which are secondary electron emitters. They are more expensive.
(c) A photometer only measures total light. Coupled with a filter, it can make an effective portable measuring
system for a specific, fixed task. A spectrophotometer is able to tune to any wavelength and can indeed measure
a spectrum. While it is more versatile, it is also more expensive.
(d) Both light sources derive their spectrum from a hot tungsten filament. The lifetime and operating temperature
of the filament is increased when a halogen gas, such as Br2 is added to the atmosphere around the filament.
It reacts with sublimating W to form a volatile tungsten bromide which decomposes and redeposits the W on
the filament. 8
8. Normally you would have to measure the pure compounds in solution and calculate their molar absorptivities,
but I did that for you in this case. The dotted spectra then were superfluous – you should not have had to
refer to them at all. When you have two unknowns (the concentrations of Ni and Co), then making two
measurements (the absorbance at 500 and 590 nm) you can solve for the two unknowns. By inspection of
the solid curve, the absorbance at 500 nm is (by my eye) 0.86 and at 590 nm it is 0.62. Your eye may have
measured differently and you were graded only on your calculations unless your measure was way off,
suggesting that you were not looking at the right curve.
First we solve for the concentration of Ni in terms of the Co concentration. We choose to do this at 500 nm.
The specific choice is not important; you should get the same answer going from a different starting point. Ni
Co
A500 = e 500b c Ni + e 500 bc Co 0.86 = 1.0 ¥ 1.0 ¥ c Ni +13 ¥1.0 ¥ c Co
c Ni = 0.86 13 ¥ cCo
Now set up the same equation at 590 nm and then substitute for the concentration of Ni. The equation then
only has 1 unknown (Co concentration) and is readily solved. Ni
Co
A590 = e 590b c Ni + e 590 bc Co 0.62 = 8.7 ¥ 1.0 ¥ c Ni + 2.7 ¥1.0 ¥ c Co [ = 8.7 0.86 13c Co + 2.7c Co
= 7.5 113.1c Co + 2.7cCo = 7.5  110.4 c Co
6.88 = 110.4 c Co
6.88
c Co =
= 0.062 M
110.4
Substitute this value for the Co concentration back into the first equation to solve for the Ni concentration. c Ni = 0.86 13 cCo = 0.86  13( 0.062 ) = 0.054 M ...
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This note was uploaded on 05/15/2010 for the course CHEM 3440 taught by Professor Danthomas during the Fall '06 term at University of Guelph.
 Fall '06
 DanThomas

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