Quantitative Analysis (CHM 235)
Dr. S.A. Skrabal
Exam 1—26 Sept. 2006
Name:_____________________________________
Instructions: Read each question carefully. Show calculations when required to receive full credit. It is not
necessary to show work for the mean and standard deviation when your calculator functions are used. Partial
credit is given when warranted. Please box in your final answer. Points will be taken off for incorrect significant
figures. Value of each question is given in parentheses after the question. Total points = 100. Note that important
information is given on the back pages of the exam. Good luck!
1. Sally in the quant lab prepares a solution of aluminum trichloride (AlCl3) for use in an experiment. She pipets
2.50 (± 0.01) mL of 10.0 (± 0.2) µM AlCl3 into a volumetric flask and dilutes the solution with deionized water to a
final volume of 100.0 (± 0.1) mL.
(a) What is the concentration of AlCl3 in the final diluted solution in units of µM? (7)
(a) Cconc Vconc = Cdil Vdil
Cconc = (Cdil Vdil) / Vconc = (10.0 μM) (2.50 mL) / (100.0 mL) = 0.250 μM
(b) What are the absolute and relative uncertainties of the concentration in the final diluted solution? (10)
[10.0 (± 0.2) μM] [2.50 (± 0.01) mL] / [100.0 (± 0.1) mL] = 0.250 μM ± ?
Convert all the absolute uncertainties to relative uncertainties:
(0.2/10.0)100 = 2.0%
(0.01/2.50)100 = 0.40% (0.1/100.0)100 = 0.10%
%e  (2.0 %) 2  (0.4 0 %) 2  (0.10 %) 2
 2.0 %
Convert back into absolute uncertainty and use “real rule” of sig. figs.
(2.0%) (0.250 μM) = (0.020) (0.250 μM) = 0.005 μM
Final answers: 0.250 ± 0.005 μM and 0.250 μM ± 2.0%
(c) What is the concentration of Cl- in the final diluted solution in units of parts per billion? (5)
Remember 1 ppb = 1 μg L-1
(0.250 μmol/L)(1 mol/106 μmol)(3 mol Cl-/1 mol Al)(35.45 g Cl-/mol Cl-)(106 μg/g) = 26.58 or 26.6 μg L-1 or ppb
2. What is the molality (m) of a 10.0 %(w:w) solution of sodium bromide (NaBr)? (10)
Consider 100 g of this solution.
10.0 %(w:w) means 10.0 g NaBr/100 g soln.
In 100 g of solution, therefore, there are 10.0 g NaBr and 90.0 g H2O, which is the solvent.
Since molality = mol solute/kg solvent, we have
(10.0 g NaBr/90.0 g H2O) (1000 g/kg) (1 mol NaBr/102.89 g NaBr) = 1.08 mol NaBr/kg H2O or 1.08 m
3. The recent doping controversy regarding Floyd Landis, the 2006 Tour de France winner, is centered on the
unnaturally high ratio of to testosterone to epitestosterone found in his urine. High ratios, above 4:1, suggest that
doping with the male hormone testosterone has occurred. (Landis’s sample showed a ratio of 11:1.) Suppose that
during an interlaboratory comparison exercise, two testing laboratories, one in France and one in Belgium, have
been given portions of the exact same urine sample to analyze for the ratio of the concentrations of testosterone to
epitestosterone. In 5 analyses, the French lab obtains a mean ratio of 7.41 with a standard deviation of 0.75. In 7
analyses, the Belgian lab obtains a mean ratio of 8.52 with a standard deviation of 0.56.
(a) Use the F-test to determine whether or not the standard deviations (i.e., the precisions) of the two sets of
measurements are significantly different at the 95% confidence level. (5)
Fcalc 
s12
s 22

(0.07 5 ) 2
(0.05 6 ) 2
 1.79
df = (5 -1) = 4 for s1 and (7- 1) = 6 for s2.
F(df = 4,6; 95% CL) = 4.53
Since Fcalc < Ftable, the standard deviations are not significantly different at the 95% CL.
(b) Determine whether or not the testosterone-to-epitestosterone ratios obtained by the two laboratories are
significantly different at the 95% confidence level. (10)
s pooled 
t calc 
s12 (n1 1)  s 22 (n2 1)
n1  n2  2
x1  x 2
n1 n2
s pooled
n1  n2


(0.7 5 ) 2 (5 1)  (0.5 6 ) 2 (7 1)
572
7.41  8.5 2
0.6428
(7)(5)
75
 0.6428
 2.949
df = 5 + 7 – 2 = 10
t(df = 10, 95% CL) = 2.228
Since tcalc > ttable, the means are significantly different at the 95% CL.
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4. The antidepressant Lexapro™ (escitalopram HBr) is manufactured in 10 mg tablets. Suppose that as part of a
quality control program, 121 tablets of Lexapro 10-mg tablets were analyzed, yielding a mean drug content of 9.95
mg with a standard deviation of 0.094 mg. Calculate the 90% and 98% confidence interval for these analyses. (10)
x
ts
n
df = n -1 – 121 – 1 = 120
90% CI:
ttable(df = 120, 90% CL) = 1.658
  9.95 
(1.658)(0.09 4 )
121
 9.95  0.014 or 9.95  0.014 mg
98% CI:
ttable(df = 120, 98% CL) = 2.358
  9.95 
(2.358)(0.09 4 )
121
 9.95  0.020 or 9.95  0.02 0 mg
5. The National Research Council of Canada sells a standard reference material (SRM) for metals in dried
sediment called MESS-3. Suppose you analyze the MESS-3 SRM for cadmium (Cd) content. For eight analyses,
you obtain the following results (in mg kg-1): 0.21, 0.12, 0.22, 0.19, 0.24, 0.23, 0.18, 0.21. The certified value for
Cd in the SRM is 0.24 mg kg-1.
(a) Test any possible outliers for elimination using the Q-test at the 90% confidence level. (7)
Arranged data: 0.12, 0.18, 0.19, 0.21, 0.21, 0.22, 0.23, 0.24 mg kg-1
The questionable data point is 0.12.
Qcalc = gap/range = (0.18 – 0.12) / (0.24 – 0.12) = 0.50
Q(90% CL, n = 8) = 0.468
Since Qcalc > Qtable, data point can be rejected at the 90% CL.
(b) Calculate the mean, standard deviation, and relative standard deviation of the acceptable data. (10)
The rejected point is excluded from the following calculations.
By calculator:
Mean = 0.211 mg kg-1
Std. dev. = 0.021 mg kg-1
RSD = (0.021 / 0.211) 100 = 9.9 % or 1 x 101 %
(c) Determine whether or not you are obtaining a Cd concentration in the SRM that is significantly different from
the certified value at the 90% confidence level. (10)
t calc 
known value  x
s
n

0.24  0.211
0.021
7
 3.654
df = n – 1 = 7 – 1 = 6
t(df = 6, 90% CL) = 1.943
Since tcalc > ttable, the obtained value is significantly different that the certified value at the 90% CL.
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6. A student prepared a set of standards for dissolved aluminum ranging in concentrations of 0 to 16 µM. These
were analyzed using a spectrophotometric technique. Using EXCEL, the student plotted his results, then used the
LINEST function to obtain the parameters of the regression equation, including the slope (m), the standard
deviation of the slope (sm), the y-intercept (b), the standard deviation of the y-intercept (sb), and the standard
deviation of the y values (sy). These results are shown below.
[Al] (µM)
Abs.
Corr. Abs.
0
1
5
10
16
0.001
0.048
0.244
0.484
0.791
0.000
0.047
0.243
0.483
0.790
m=
sd of m =
R^2 =
0.0492596
0.0003783
0.9998231
16959.6848341
0.4299771
-0.0026614
0.0033062
0.0050352
3.0000000
0.0000761
=b
= sd of b
= sd of y
(a) Write the linear regression equation in the form y (± sy) = m (± sm) x + b (± sb) using the correct number of
significant figures. (8)
Applying real rule on sig. figs.:
y (± 0.0050) = 0.04925 (± 0.00037) x - 0.0026 (± 0.0033)
or
y (± 0.005) = 0.0493 (± 0.0004) x + 0.003 (± 0.003)
(b) Suppose an unknown sample was analyzed for Al in the same manner as the standards, giving an absorbance of
0.538. Determine the Al concentration (in μM) of the unknown sample. (8)
Rearrange equation to solve for x = conc. and y = absorbance
Blank- corrected absorbance = 0.538 – 0.001 = 0.537
x 
y b
m
 x 
0.537  (0.002 6 )
0.0492 5
 10.9 5 or 11.0 M
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Useful information
Standard deviation: s 
 ( xi  x ) 2
s
100
x
%RSD = 
i
Confidence interval:   x 
n 1
ts
where degrees of freedom = n - 1.
n
Comparison of means (when standard deviations are not significantly different):
t calc 
x1  x 2
n1 n2
s pooled
n1  n2
s pooled 
s12 (n1 1)  s 22 (n2 1)
n1  n2  2
where degrees of freedom = n1 + n2 - 2
Comparison of means (when standard deviations are significantly different):
t calc

x1  x 2
s12 / n1  s 22 / n2
DF
Comparing measured result to known result:


( s 2 / n  s 22 / n2 ) 2

  2 1 21
( s 22 / n2 ) 2
  ( s1 / n1 )

  n1  1
n2  1

t calc 
known value  x
s



2




n
where degrees of freedom = n – 1.
Fcalc 
s12
s 22
where s1 > s2 and degrees of freedom is n1 - 1 and n2 - 1
en  e12  e22  e32  ...  en21
%en  %e12  %e22  %e32  ...  %en21
y = mx + b
Q = gap/range
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