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September 16, 2011 Introduction to Analytical ChemistryExam 1 (20 pts. Total) (CHM 3120 B51, Fall 2011)

Name: Signature:

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

1. You add 50.0 microliters (L) of a 0.100 M solution of NaCl to a 1-liter volumetricflask, and bring the final volume to 1.000 L of water. What is the final concentrationof NaCl in parts per million (ppm)? Give correct significant figures. The

molecular/formula mass for NaCl is 58.44 g/mol. Show all work for partial credit. (2pt.)

Answer:

2. What volume of a 100 mM solution would you need to make (by dilution) 100 mL ofa 5 M solution? Show all work for partial credit. (2 pts.)

Answer:

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3. A Class A 50-mL buret is certified by the manufacturer to deliver volumes within atolerance (i.e. uncertainty) of 0.05 mL. The smallest graduations on the buret are0.1 mL. You use the buret to titrate a solution, adding four (4) successive volumesto the solution. The following volumes were added:

Addition Volume (mL)1 6.732 8.923 7.524 2.48

What is the total volume added, and what is the uncertainty associated with thisfinal volume? Give appropriate significant figures. (2 pts.)

Answer:

4. You are validatinga method for measuring mercury in fish tissue by the method ofatomic fluorescence. Which of the following pieces of information would support theACCURACY of your method? Give all choices that apply. (1 pt.)

Answer:

a. Atomic fluorescence responses increases linearly with increasingconcentration of Hg; the correlation coefficient is high (R2 = 0.99)

b. The concentration of Hg in a 1.000 nM Hg standard reference solutionmeasured by the atomic fluorescence method is not significantly different(with 95% confidence) from the expected value.

c. A spike of 0.5 nM Hg is recovered (99.99% recovery) by the atomicfluorescence method.

d. Measure of Hg using the atomic fluorescence method conducted by three,independent laboratories provided double-blind samples of a 1.000 nMstandard reference solution agreed within 0.021 nM.

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5. In order to measure the amount of lead ions (Pb+) in water, you add potassiumiodide (KI) which forms a precipitate of lead iodide (PbI) according to the followingreaction:

KI + Pb+

K+

+ PbI

You collect the PbI, as a precipitate from the reaction, by filtering into a pre-weighedGooch crucible, and dry the crucible (with the collected precipitate) in a 110C oven.You determine the amount of PbI by re-weighing the dried crucible with precipitate,and subtracting the pre-determined weight of the crucible (i.e. weigh by difference).Which of the following would be expected to reduce the accuracy in your finalmeasurement of the amount of precipitate? Give all answers that apply. (1 pt.)

Answer(s):

a. Not cooling the crucible properly prior to weighing itb. Handling your crucible with your hands and leaving fingerprintsc. Drying the crucibles (both with and without precipitate) in an oven that consistently

heats at a temperature that is 1C too highd. Using a different analytical balance to weigh the crucible without precipitate than the

one used to weigh the crucible with precipitate.e. Using the same analytical balance for both weights (with and without collected

precipitate), but one that consistently measures weights that are 0.010 g too high

6. You repeat the measurement described in Question 4 for a total of six times. Youmeasure the following amounts of precipitate: 38.24 mg, 39.65 mg, 38.12 mg, 39.08

mg, 37.95 mg, 38.92 mg.

a. What is the average weight measured? (2 pts.)

Answer:

b. Give the 95% confidence interval for the average weight of precipitate.(2 pts.)

Answer:

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7. You are measuring the effects of a toxic compound on the enzyme, glutathione-S-transferase (GST), in human liver cells. You expose five (5) test tubes of cells to100 L of a 5 ppm aqueous solution of the toxic compound (=Treated), and exposefive (5) test tubes of cells to an equal volume (100 L) of water alone (Untreated).You then measure the GST activity of cells in each test tube; enzyme activity is in

units of mol/minute. The following are the measurements of GST activity:

Treated Untreated

GST Activity GST ActivityTube (mol/min) Tube (mol/min)1 3.25 1 5.842 3.98 2 6.593 3.79 3 5.974 4.15 4 6.255 4.04 5 6.10

Average: 3.84 Average: 6.15Standard StandardDeviation: 0.36 Deviation: 0.29

a. Is the variance of the measured GST activity of cells exposed to the toxic compoundequal to that of cells exposed to water alone? Show any relevant work. (2 pts.)

Yes No (circle one)

b. Is the average GST activity measured for cells exposed to the toxic compoundsignificantly different (at 95% confidence level) than that measured for cells exposedto water alone? Show all relevant calculations, including any t-statistic. (3 pts.)

Yes No (circle one)

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8. The amount of chlorophyll in lake water is used to estimate the density of algae andother photosynthetic organisms living in lakes. Specifically, the fluorescence ofchlorophyll can be used to measure its concentration. The amount of fluorescence ismeasured for the following concentrations of chlorophyll (in tap water): 1, 2, 5, 10,and 20 ppm. The results are shown in the graph below:

You also measure the fluorescence of seven (7) samples of tap water (blank), andseven (7) samples of chlorophyll in tap water at the concentration of 3 ppm. Theaverage ( one standard deviation) calculated fluorescence for the blank was 0.0019( 0.0071), and the average ( one standard deviation) calculated fluorescence for 3ppm chlorophyll was 0.9002 ( 0.0067).

For concentrations of1, 2, 5 and 10 ppm only (NOT including 20 ppm), you

calculate a best fit line for the relationship between fluorescence and chlorophyllconcentration with a slope (m) of 0.298 and y-intercept (b) of 0.005. The calculatedR2-value for this line is 0.99. This line is shown on the graph above. This linearrelationship (y=mx+b) between measured fluorescence and chlorophyll concentrationis used to measure chlorophyll concentration in water.

The data presented above support which of the following statements regarding theproposed method of measuring chlorophyll? Give all answers that apply. (2 pts.)

Answer(s):

a. The method can be used to determine unknown concentrations of chlorophyll ashigh as 20 ppm.

b. Chlorophyll concentrations of 0.15 ppm are within the limit of quantitation (LOQ)of the method.

c. Chlorophyll concentrations of 0.15 ppm are within the limit of detection (LOD) ofthe method.

d. The use of the calibration curve, based chlorophyll in tap water, is sufficient tomeasure chlorophyll in lake water.

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9. Which of the following directly addresses inaccuracies due to matrix effect? Give allchoices that apply. (1 pt.)

Answer(s):

A. Use of an internal standardB. Spike recoveryC. Use of appropriate blanksD. Following Standard Operating Procedures (SOPs)E. Calibrating instrumentation

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Propagation of Uncertainty for Random Error:

Addition/Subtraction: ef = (ei)2

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

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

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

Variance = s2

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

Confidence Interval: = x ts/ n

F-Test: Fcalculated = s12/s2

2 (s1 is the larger of the two variances)

Q-Test: Qcalculated = gap/range

t-calculated (forequalvariance)

t-calculated = |x1 x2| n1n2spooled n1 + n2

spooled = s12(n1 -1) + s2

2(n2-1)n1 + n2 -2

Degrees of Freedom = n1 + n2 -2

t-calculated (forunequalvariance)

t-calculated = |x1 x2|

s12/n1 + s2

2/n2

Degrees of Freedom = (s12/n1 + s2

2/n2)2 - 2

(s12/n1)

2 + (s22/n2)

2

n1 + 1 n2 + 1

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t-calculated (Paired data)

sd = (di d)2

n - 1

t-calculated = |d| nsd

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 = (di)/n-2

sm2 = sy

2n/D

sm2 = sy

2(xi2)/D

*Note: Equation for D would be given to you.

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

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 finalconcentrations of the standard (S) and (X), and I x and Is+x are the response of theunknown 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 standardadded and V is the final volume (of unknown plus standard).

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