September 13, 2010 Introduction to Analytical ChemistryExam 1 (20 pts. Total) (CHM 3120 B51, Fall 2010)

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Answer the following questions, showing all appropriate work and writing FINAL answer in the spaces provided. Point values for each question are given. Useful equations and tables of statistics are given on the last three pages. Make sure to read the questions carefully before attempting to answer them. You have 50 minutes. Good luck!!

In order to monitor possible oil leaks from pipelines in the Gulf of Mexico, you use a method based on fluorescence to measure hydrocarbons (as indicators of the presence of oil) in seawater. The following series of questions are based on this analytical technique.

1. You decide to use the molecule, fluorene, as a representative fluorescent hydrocarbon found in crude oil, to create a calibration curve of fluorescence response versus concentration.

a. You first prepare a stock solution of fluorene by accurately weighing 2.655 x 10-2 g of fluorene, and dissolving it in 250.0 mL of “artificial seawater” (a prepared solution of salts that chemically resembles seawater) in a 250-mL volumetric flask. If the molecular mass of fluorene is 166.2185 g/mol. What is the molar concentration of the solution? Show correct significant figures. (2 pts.)

Answer: 6.389 x 10 -4 M

2.655 x 10-2 / 166.2185 g/mol = 1.5973 x 10-4 mol1.5973 x 10-4 mol/2.500 x 10-1 L = 6.389 x 10-4 M

a. What is the concentration of the solution prepared (in the previous question) in ppm? NOTE: The density of seawater is 1.025 g/mL, as opposed to pure water that is 1.000 g/mL. (2 pts.)

Answer: 1.036 x 10 2

2.655 x 10-2 g/250.0 mL = [2.655 x 10-2 g/(250.0 mL x 1.025 g/mL)] x 106

=(1.036 x 10-4) x 106

= 1.036 x 102 ppm

b. If 10.0 mL of the solution prepared (in the previous questions) is transferred with a 10-mL volumetric pipet to a 250-mL volumetric flask, and diluted to 250.0 mL. What is the final concentration of this diluted solution in g/L. (2 pts.)

Answer: 4.248 x 10 -3 g/L

C1V1 = C2V2

(2.655 x 10-2 g/0.2500 L)(0.0100 L)=C2(0.2500 L)C2 (final concentration) = 4.248 x 10-3 g/L

2. You create a calibration curve based on the fluorescence response – specifically measured in fluorescence intensity units (FIUs) – for a series of standard solutions of fluorene, prepared in artificial seawater, covering a range of molar concentrations (measured in µM). You use the method of least squares to calculate the equation of the linear relationship (y=mx+b), including the slope (m) and y-intercept (b), as well as the uncertainty (i.e. ± 1 standard deviation, sm and sb) associated with both of these. The calibration curve, along with the calculated values, shown on the figure (below).

1. To test the accuracy of your proposed method, you use a standard reference material that is certified to contain 100.000 ± 0.001 µM fluorene in seawater. Based on the linear relationship from the calibration curve (above), what is the expected fluorescence response (in FIUs)? What is the associated uncertainty in this response (in other words, give your answer with ± one standard deviation)? Hint: Recall the linear relationship is defined as y=mx+b. (2 pts.)

Answer: 1.367 x 10 -3 ± 0.156 x 10 -3

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(1.225 x 10-5) ± (0.155 x 10-5) %e1 = 0.155 x 10-5/1.225 x 10-5=0.12653

x 1.00000 x 10 2 ± 0.00001 x 10 2 %e2 = 0.000001 x 102/1.00000 x 102

1.225 x 10-3 ± 0.155 x 10-3 = 0.00001%e3 = ((0.12653)2+(0.0001)2) [propogation of uncertainty mult./div.]

= 0.12653

e3 = (0.12653)(1.225 x 10-3) = 0.155 x 10-3

1.225 x 10-3 ± 0.155 x 10-3 1.225 x 10-3 ± 0.155 x 10-3

+1.420 x 10 -4 ± 0.182 x 10 -4 +0.1420 x 10 -3 ± 0.0182 x 10 -3 1.367 x 10-3 ± 0.156 x 10-3

ef = ((0.155 x 10-3)2 + (0.0182 x 10-3)2)

= 0.156 x 10-3

2. You measure the fluorescence response of the standard reference material seven (n=7) times, and based on your calibration curve calculate the following concentrations: 101.1 µM, 99.4 µM, 100.5 µM, 99.8 µM, 101.2 µM, 100.8 µM, 101.7 µM. Calculate the mean, and 95% confidence interval, for your measured fluorene concentrations? (3 pts.)

Answer: 100.6 ± 0.7 Mean

Mean = (101.1 + 99.4 + 100.5 + 99.8 + 101.2 + 100.8 + 101.7)/7=100.6

s = (101.1-100.6)2+(99.4-100.6)2+…(101.7-100.6))/(7-1)= 0.81

95% Confidence Interval = ts/n = (2.447)(0.81)/ (7) = 0.74

3. Based on the results (from the previous question) the measured concentration of fluorene is significantly different from the certified concentration (i.e. 100.000 µM) standard reference material? (1 pt.)

Answer (circle only one): Significantly Different Not Significantly Different

The expected concentration 100.000 µM falls within the 95% confidence interval (99.9 – 101.3) from the previous question.

4. To determine the limit of quantitation (LOQ) of your analytical technique, you measure the fluorescence response of a blank of artificial seawater seven (n=7) times, and calculate a mean response

(± 1 standard deviation) of 2.35 (± 0.12) x 10-6 FIUs. You then measure fluorescence response seven (n=7) times for a sample of 10.00 µM fluorene, and measure a mean response of 4.03 (± 0.13) x 10-6 FIUs. Based on this, and the calibration curve (above), what is the calculated LOQ? (2 pts.)

Answer: 0.106 µM

LOQ = 10s/m = 10(0.13 x 10-6 FIU)/(1.225 x10-5 FIU/µM)* = 0.1061 µM*From calibration curve

5. You suspect the use of artificial seawater, rather than actual seawater, may affect your fluorescence response. You create a similar calibration curve for fluorene using seawater collected from the Gulf of Mexico. This calibration curve in actual collected seawater is shown below; for comparison, the calibration curve in artificial seawater is shown in the same figure. The calculated R2 for both is >0.99.

Which of the following statements is true with regards to the use of artificial seawater versus actual seawater? (1 pt.)

a. The use of actual seawater, rather than artificial seawater, has no effect on the accuracy of the measurement of concentrationb. The use of actual seawater, rather than artificial seawater, improves the accuracy of the measurement of concentrationc. The use of actual seawater, rather than artificial seawater, improves the precision of the measurement of concentrationd. The use of actual seawater, rather than artificial seawater, increases the sensitivity of the measuremente. All of the above.NOTE: The graph suggest an effect of using actual seawater, and therefore, it would be expected that using actual seawater (rather than artificial seawater) would improve accuracy

6. You decide to use the method of standard addition, specifically adding standards of fluorene to a sample of seawater containing an unknown

concentration of hydrocarbons. To do this, you measure the fluorescence response (Ix) of a 9.000 mL sample containing an unknown concentration, and find it to be 2.38 x 10-4 FIUs. You then add 1.000 mL of 2.156 mM standard of fluorene to the sample, and measure the fluorescence response of the final solution (Is+x) to be 4.35 x 10-4 FIUs. What is the concentration of hydrocarbons in the initial sample based on these measurements? (3 pts.)

Answer: 0.2324 mM

[X]i = Ix

[S]f + [X]f Is+x

[X]f = [X]i(Vo/V) = [X]i(9/10) = 0.9 [X]i

[S]f = [S]i(Vs/V) = (2.156 mM)(1/10) = 0.2156 mM

[X]I = (2.38 x 10 -4 FIU) (0.2156 mM) + 0.9 [X]i (4.35 x 10-4 FIU)

Solving for [X]i (the only unknown above), you get [X]I = 0.2324 mM

7. You want to determine if concentrations of hydrocarbons in seawater measured by fluorescence are significantly different than concentrations measured by a second method, specifically based on the use of gas chromatography/flame ionization detection (GC-FID). You measure the concentration of a certified standard reference material (100.0 µM) with both methods seven (n=7) times. Specifically, you first measure each sample by fluorescence, and then measure the same sample by GC-FID. The concentrations determined by the two methods are shown below.

[fluorene (µM)]Fluorescence GC-FID

1 100.2 101.12 100.9 100.53 99.9 100.24 100.1 100.25 100.1 99.86 101.1 100.77 100.0 99.9

Calculate the appropriate t-statistic to compare the two sets of measurements. (2 pts.)

Answer: 0.08

The comparison is a PAIRED t-test, so calculate as follows:

[fluorene (µM)]Fluorescence GC-FID d

1 100.2 101.1 -0.92 100.9 100.5 0.43 99.9 100.2 -0.34 100.1 100.2 -0.15 100.1 99.8 0.36 101.1 100.7 0.47 100.0 99.9 0.1

Mean d: -0.014

sd: 0.47

t-statistic = |dmean| n = (0.014)( 7) = 0.08sd (0.47)

Propagation of Uncertainty for Random Error:

Addition/Subtraction: ef = S (ei)2

Multiplication/Division: %ef = S (%ei)2

Mean/Average of the Sample (x) = S xi / n

Standard Deviation of the Sample (s) = S (xi – x)2 / n-1

Variance = s2

Coefficient of Variation (%CV) = s/x • 100

Confidence Interval: m = x ± ts/ n

F-Test: Fcalculated = s12/s2

2 (s1 is the larger of the two variances)

Q-Test: Qcalculated = gap/rangeGrubb’s Test: Gcalculated=|questionable value – mean|/s

Equation of a Line: y = mx + b, where “m” is the slope, and “b” is the y-intercept

Standard Deviation of a Calibration Curve:

sy = S(di)/n-2

sm2 = sy

2n/D

sb2 = sy

2S(xi2)/D

D = | S(xi2) Sxi |

| Sxi n |

Limit of Detection: LOD=3s/m S/N=3Limit of Quantitation: LOQ=10s/m S/N=10

t-calculated (for equal variance)

t-calculated = |x1 – x2| n1n2

spooled n1 + n2

spooled = s12(n1 -1) + s2

2(n2-1) n1 + n2 -2

Degrees of Freedom = n1 + n2 -2

t-calculated (for unequal variance)

t-calculated = |x1 – x2|

s12/n1 + s2

2/n2

Degrees of Freedom = (s12/n1 + s2

2/n2)2 - 2(s1

2/n1)2 + (s22/n2)2

n1 + 1 n2 + 1

t-calculated (Paired data)

sd = S(di – d)2

n - 1

t-calculated = |d| n sd

Standard Addition Equation(s)

[X]i = Ix

[S]f + [X]f Is+x

Where [X]I is the initial concentration of the unknown (X), [S]f and [X]f are the final concentrations of the standard (S) and (X), and Ix and Is+x are the response of the unknown and the unknown plus the standard, respectively.

[X]f = [X]i(Vo/V) [S]f = [S]i(Vs/V)

Where Vo is the initial volume of the unknown, Vs is the volume of standard added and V is the final volume (of unknown plus standard).