Chemical Instrumentation CHEM*3440 Mid-Term Examination Fall 2005 TUESDAY, OCTOBER 25, 2005

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1 Chemical Instrumentation CHEM*3440 Mid-Term Examination Fall 2005 TUESDAY, OCTOBER 25, 2005 Duration: 2 hours. You may use a calculator. No additional aids will be necessary as a series of data and equation sheets are included as part of the examination paper. 1. (8 points) Polycyclic Aromatic Hydrocarbons (PAH) is a class of organic compounds that have important toxicological effects. Their detection and measurement in environmental conditions is a vital health concern. A method was developed to measure these compounds from waste water samples. The first step was the extraction of the PAH from the water into methylene chloride. The efficiency of this step was tested by injecting known amounts of various PAH substances into an aqueous sample and then quantifying the extracted amount. The following table reports the results of several trials for four different compounds. Note the different added amounts in each case. The mean recovered values and the standard deviation of those means are shown at the bottom. Chrysene Benzofluoranthenes Benzo(a)pyrene Dibenz(a,h)anthracene Amount added 1.0 µg/m µg/m 3 10 µg/m 3 50 µg/m Mean St. Dev (a) Calculate the relative standard deviation for each substance. (b) Which substance is most completely recovered? (c) Which substance is most unreliably recovered? (d) What are the 95% confidence limits for the recovery of benzo(a)pyrene?

2 2. (6 points) An experiment provided a rapidly and chaotically time varying signal. On average, however, the input signal current was 1.5 µa. It was decided that the signal could be smoothed by integrating it for a fixed length of time. The op amps are driven by a ± 10.0 V source. Here is a basic schematic of the arrangement. R = kω C = 0.1 µf I in = 1.5 µa - A R = 100 kω - B (a) What will be the average voltage at point A? (b) What will be the maximum useful integration time for the circuit? 3. (4 points) A spectrophotometric assay was undertaken with a substance with a molar absorptivity of 4000 M -1 cm -1. A stock solution is available whose concentration is M. The instrumental path length is cm. A final absorbance in the range between is desired to calibrate in the range of the unknown solutions. You are to make up.0 ml of the calibration solution in a volumetric flask. Do you need to pipette 1.00, 2.00, 3.00, 4.00 or 5.00 ml of the stock solution into the.0 ml volumetric flask to provide a solution in this absorbance range? Show your work, 4. (6 points) Look at the six spectra on the next page. The background has already been corrected for all spectra. Read the scales and the titles very carefully. The unknown sample contains a mixture of Cu and Zn ions. What is the concentration of each ion in the unknown? 5. (4 points) Your instrument is using a 12-bit analog-to-digital converter (ADC) to digitize an input signal. The range of this device is 0 to 10 V. (a) What is the voltage resolution of this device? (b) What decimal number would be produced if an input voltage of 2.50 V was converted? 6. (12 points) Describe the following terms in a couple of sentences each. When appropriate, give both and equation and an explanation. (a) Limit of quantitation. (b) Dynamic range. (c) Impedance. (d) 1/f noise. (e) Alias (in the context of digital signal acquisition). (f) Auxochrome.

3 7. (8 points) The following data were obtained for forming a calibration curve in a spectroscopic experiment. The concentrations are in ppm and the instrument signal is in mv. The calibration curve is given in the following graph, along with the parameters of the fit. Concentration (ppm) Average Signal (mv) Calibration Curve Signal Concentration (ppm) The fit parameters are: m = mv/ppm b = mv s r = mv s m = mv/ppm s b = mv An unknown sample was measured 5 times, giving the following results: , , , (a) What is the concentration of the unknown sample? (b) What are the 95% confidence limits on that concentration?

4 CHEM*3440 Mid-Term Examination Fall 2005 ANSWERS 1. (a) The relative standard deviation is just the ratio of the standard deviation to the mean (divide the standard deviation by the mean). We find Chrysene Benzofluoranthenes Benzo(a)pyrene Dibenz(a,h)anthracene (b) The highest recovery rate comes from the ones whose mean is the largest fraction of the originally added amount. We can easily see that chrysene is recovered at a rate of about 70.4% and is the largest of the four. (c) The one which is most unreliably recovered is the one with the largest relative standard deviation. We can compare the standard deviation directly because that depends upon how much was recovered. In fact, benzofluoranthenes is the most unreliably recovered material with its relative standard deviation of 49.1%. (d) The equation needs a value for Student s t from the tables. To find that, we need the number of degrees of freedom, which is N-1. There are 14 measurements, so dof = 13. From the table, t = for 95% confidence. The result then is ts ^2.160h^1.419h = = N ! ng m3 at the 95% confidence level. 2. (a) First element is just a current-to-voltage amplifier. We simply have VA =- ^1.5 # 10-6 Ah^ # 10 3 Xh = V (b) This 0.75 V will be integrated at a rate given by the RC time constant of the integrator until the integrator is saturated (reaches its maximum or minimum output). The time required to do this is the maximum useful integration time. x = RC = ^100 kxh_ 0.1 nfi = 0.01s Voutput =- RC 1 Vmax = 10.0 V 0 t # Vinput dt =- ^-0.75 V h t 0.01s tmax = ^10.0 Vh 0.01s = s =133 ms 0.75 V 3. This is a common activity one would need to do: form a calibration solution whose absorbance falls within a particular range. There are two approaches. You could solve Beer s Law to find the volumes that would give the two ends of the desired absorbance range; they would then bracket the desired volume. Or you could solve Beer s Law for each of the possible volumes and find which one falls within the desired range. I ll so the work for the last procedure.

5 It is obviously central to the problem to consider the dilution of the concentration. This is easily accounted for by scaling the original concentration by the ratio of the added volume to the final volume (.0 ml). The final concentration in each of the five cases is then C1 V1= C2 V2 V1 C1 V2 = C2 Z 1 = 3.90 # 10-5 M 2 = 7.80 # 10-5 M M # [ 3 = 1.17 # 10-4 M 4 = 1.56 # 10-4 M 5 = 1.95 # 10-4 M \ Now we use these five concentrations in Beer s Law to find the expected absorbance in each case. A = fbc Z 3.90 # 10-5 M = # 10-5 M = M - 1 cm - 1 # cm # [ 1.17 # 10-4 M = # 10-4 M = \ 1.95 # 10-4 M = Clearly the third entry falls within the desired absorbance range of We need to add 3.00 ml of the stock solution to make the desired calibration standard solution. 4. Note the spectra are for three different solutions a Zn 2 calibration standard, a Cu 2 calibration standard, and an unknown containing both ions. The spectra are taken at two different wavelength ranges. After looking at them, you should realized that zinc provide a signal a one wavelength while copper provides a signal at both wavelengths. We assume and a casual inspection of the spectra confirms it is true that the spectra all have the same shape meaning they have the same linewidth and are Gaussian in form. This is to be expected when the lineshape is governed by the instrument as is typically the case in spectrometry. It is also reasonable to assume that the same pathlength is used in each case. We can therefore relate the signal intensity to the peak height. It should be peak area but since the peaks all have the same linewidth, their heights are all proportional to their areas. While it would be possible to solve this problem when both calibration solutions have a measurable signal at both wavelengths it is just a problem of two equations in two unknowns this case is made particularly simple by having only one entity give rise

6 to a signal and on of the wavelengths. Therefore, Cu 2 can be identified by comparing the unknown to the calibration signal at nm. Then, knowing the Cu 2 concentration, we can find the Zn 2 from the signal at nm. I (Cu(10.0 ppm)) = 240 I (unknown) = Cu = 10.0 ppm # = 6.29 ppm I (Zn(1.00 ppm)) = 4.2 I (Cu(10.0 ppm)) = 1.3 I (unknown) = Zn 6.7 = 4.2 f unknown p 6.29 ppm 1.3 d n 1.00 ppm 10.0 ppm 6.7 = 4.26 Zn unknown Zn unknown = = 1.4 ppm 5. (a) A 12-bit device means that it has 2 12 = 4096 resolution elements. Since these 4096 resolution steps are spread over 10 V in this case (device range is from 0-10 V), then each digital step spans a voltage step of = V= 2.44 mv which is the voltage resolution of the device. (b) The decimal number (meaning a base-10 number) which is an integer between 0 and 4095 can be found in two ways. Either we can use the step size of 2.44 mv and see how many steps would be needed to reach the desired voltage of 2.50 V, or we can realize that the range of integers is spread smoothly of over the voltage range. Both procedures give us the result of I will accept 1024 ± 1. R V int S 2.50 V W= 1024 steps V S step W T X or 2.50 V # 4096 steps = 1024 steps 10.0 V 6. (a) Limit of Quantitation. This is the lower concentration limit at which an experimental technique can provide a reliable quantitative result. It is chosen to be a concentration equal to 10 times the standard deviation of the blank signal. LOQ = 10 s blank. (b) Dynamic range. The concentration range between the LOQ and the upper limit of linearity LOL, or the point after which the instrumental method s response is no longer linear. A technique is able to readily measure concentration throughout its dynamic range. (c) Impedance. The vector sum of a circuit s resistance and reactance. The overall tendency for the circuit to impede the flow of current. I = (R 2 X 2 ) 1/2.

7 (d) 1/f noise. Also known as drift or pink noise. It is more pronounced at low frequencies (hence the reciprocal dependence on frequency). Its origins are uncertain and difficult to specify. Try to shift the frequency range of an experiment away from the DC condition. (e) Alias. A higher frequency signal masquerades as a lower frequency signal due to undersampling when digitizing. Anything above the Nyquist frequency (given by the reciprocal of twice the sampling rate) will appear as an alias and will be folded over into the spectrum. (f) Auxochrome. A molecular feature which is not itself photoactive, but whose presence does alter the frequency and intensity of nearby chromophores. 7. PROBLEM!! The raw data given in this problem did not give rise to the parameters as given. Sorry about that. You will be given full marks if you refit the data and used the correct slop and intercept values, etc. or if you just took my word for it and used the data as given into the equations. (a) The least squares parameters for the fitting line readily give the concentration for the unknown once we find the average signal. The average of the five measurements of the unknown is We solve the equation for the line given the slope and intercept in the problem. We have S = mc b C x = S - b m = = 4.25 ppm (b) To calculate the confidence limits, we need the standard deviation of the result, given as s calc on the equation sheet. While the problem gives s r as part of the data, we also need to calculate S xx First we calculate S xx ; the x i are the concentration values for each point on the calibration curve, of which there are 7, it also being the variable N in the calculation. We need to find the square of each concentration value, and then sum all the concentration values (and square them) and then sum all the squares of the concentration values. `!x i j 2 Sxx =! x i2 - N = ^33.47h 2 7 = = We can now find s calc. M = 5, the number of measurements of the unknown; N = 7, the number of points in the calibration plot; m = slope of calibration curve. We also need to find the average of the y values for the calibration curve. It is Scalc = m sr M 1 N 1 `y unknown - y calib j 2 m 2 Sxx = ^ h 2 = ^3.407h 2 ^81.76h We now use the equation with Student s t to specify the confidence limits. We measured the average with 5 experiments. The degrees of freedom is 4. At 95%, we find that t = Note that the equation uses N and not N-1. We obtain Confidence limits =! t s ^ =! 2.776h^0.074h = N 5 The final answer is 4.25 ± ppm with 95% confidence.