Quantitative Analysis (CHM 235)
Dr. S.A. Skrabal
Exam 1—13 February 2007
Name: SOLUTIONS
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. Thiomersal (C9H9HgNaO2S; molar mass = 404.81 g mol-1) is a mercury-containing antibacterial and
antifungal compound that has been used as a preservative in vaccines, ophthalmic and nasal solutions,
tattoo inks, and other medical solutions. There have been recent concerns about the use of the compound
in children’s vaccines because of the toxicity of Hg. Suppose the concentration of thiomersal in a vaccine
solution is 0.010 %(w:v). (a) What is the molarity of Hg in the vaccine? (b) What is the concentration of Hg
in the vaccine in units of ppm? (12)
(a) (0.010 g thio./100 mL soln)(1000 mL/L)(1 mol thio./404.81 g thio.)(1 mol Hg/1 mol thio.) =
2.47 x 10-4 mol L-1 or 2.5 x 10-4 mol L-1
Remember 1 ppm = 1 mg/L
(b) (2.47 x 10-4 mol/L)(200.6 g Hg/mol Hg)(1000 mg/g) = 49.5 mg L-1 or ppm or 50 ppm
2. Suppose an infant with a blood volume of 275 mL is given 1.00 mL of a vaccine that contains
25.0 mg L-1 of Hg. After complete mixing in the blood, what would be the Hg concentration in the infant’s
blood in units of (a) mg L-1 and (b) mol L-1? (12)
(a) CconcVconc = CdilVdil
Cdil = (CconcVconc) / Vdil = (25.0 mg L-1)(1.00 mL) / 276 mL = 9.057 x 10-2 mg L-1 or 9.06 x 10-2 mg L-1
(b) (9.057 x 10-2 mg/L)(g/1000 mg)(1 mol Hg/200.6 g Hg)(106 µmol/mol) = 0.4514 µmol L-1 or 0.451 µmol L-1
3. A new forensic chemist is learning a technique for determining cocaine in hair for use in criminal
investigations. Prior to analyzing real forensic samples, she is given a certified reference material that
contains 0.250 mg L-1 cocaine. She analyzes the solution eight times, obtaining the following results
(in mg L-1): 0.239, 0.245, 0.231, 0.219, 0.252, 0.244, 0.253, 0.247.
(a) Using the Q-test at the 90% confidence level, determine whether or not any outliers can be rejected from
this data set. (5)
0.219, 0.231, 0.239, 0.244, 0.245, 0.247, 0.252, 0.253
0.219 is the possible outlier
Qcalc = gap/range = (0.231 – 0.219)/(0.253-0.219) = 0.353
Qtable(90%CL, n = 8) = 0.468.
Since Qcalc < Qtable, data point can not be rejected at the 90% CL.
(b) Calculate the mean, standard deviation, and % relative standard deviation of the acceptable data. (9)
Use all data. Using calculator:
Mean = 0.2412 mg L-1 - Looking at std. dev. and considering “real rule” of sig. figs., the mean is rounded
to 0.241 mg L-1
Std. dev. = 0.011 mg L-1
%rsd = (std. dev./mean) 100 = (0.011/0.241)100 = 4.5% or 5%
(c) Determine whether or not the new forensic chemist is obtaining results which are significantly different
from the certified value at the 95% confidence level. (10)
t calc 
known value  x
s
n

0.250  0.241
0.011
8
 2.314
df = n – 1 = 8 – 1 = 7.
t(95% CL, df = 7) = 2.365
Since tcalc < ttable, there is no significant difference between the certified value and the chemist’s result.
(d) Calculate the 90% confidence interval for the acceptable data. (6)
df = n – 1 = 8- 1 = 7
t(90% CL, df = 7) = 1.895
  0.241 mg L1 
(1.895)(0.011 )
8
 0.241  0.007 3 mg L1
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4. A known cocaine user was arrested a few hours after a felony crime had been discovered. Hair found at
the crime scene was suspected to have come from the head of the arrested individual. After receiving the
suspect’s consent, samples of hair from the crime scene and from the suspect’s head were analyzed for
cocaine content at the crime lab. Results of the analyses are tabulated below:
Source of hair
Crime scene
Suspect’s head
Mean (ng mg-1)
30.23
28.5
n
4
6
Std. dev. (ng mg-1)
0.98
1.2
(a) Use the F test to determine whether or not the standard deviations of the two data sets are different at
the 95% confidence level. (5)
Fcalc 
s12
s 22

(1.2 ) 2
(0.9 8 ) 2
 1.499
df = (6 -1) = 5 for s1 and (4- 1) = 3 for s2.
F(df = 5,3; 95% CL) = 9.01
Since Fcalc < Ftable, the standard deviations are not significantly different at the 95% CL.
(b) Use a t test to determine whether or not the means of the two data sets 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


(1.2 ) 2 (6 1)  (0.9 8 ) 2 (4 1)
642
28.5  30.2 3
1.12256
(6)( 4)
64
 1.12256
 2.387
df = n1 + n2 -2 = 6 + 4 – 2 = 8
t(95% CL, df = 8) = 2.306
Since tcalc > ttable, means are significantly different at the 95% CL.
5. Suppose you are weighing an extract of a natural product on a balance whose absolute uncertainty is
0.00001 g. What is the minimum mass of the natural product that you must weigh if you are to have a
relative uncertainty in your mass of not more than 0.5%? (10)
relative uncertainty = (absolute uncertainty/measurement) 100
measurement = (absolute uncertainty/relative uncertainty) 100
measurement = (0.00001 g / 0.5) 100 = 0.002 g
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6. Concentrations of dissolved zinc were analyzed in samples from a lake using graphite furnace atomic
absorption spectrometry. Standard concentrations ranged from 0 to 60.0 nM. The data were analyzed
using the LINEST function in Excel®, and the following unedited parameters of the regression were
calculated:
slope: 0.002588 ± 0.0000222
y-intercept: 0.0016011 ± 0.0005773
std. dev. of y values: 0.0018101
An unknown sample was analyzed, yielding an absorbance of 0.057. The blank was found to be 0.002.
(A) Write the linear regression equation in the form y (± sy) = m (± sm) x + b (± sb) using the correct number
of significant figures. (5)
y (± 0.0018) = 0.002588 (± 0.000022) x + 0.00160 (± 0.00057)
(B) Determine the Zn concentration (in nM) of the unknown. (8)
Blank-corrected absorbance = 0.057 – 0.002 = 0.055
x = (y – b) / m = (0.055 – 0.00160) / 0.002588 = 20.6 or 21 nM
(We will use uncertainty analysis to figure out how many sig. figs we should keep.)
(C) Determine the absolute and relative uncertainties of the Zn concentration in the unknown. (8)
x ( s x ) 
0.055 ( 0.0018 )  0.0016 0 ( 0.00057 )
0.002588 ( 0.00002 2 )
enum  (0.0018 ) 2  (0.00057 ) 2
 0.0018
%enum 
0.0018
(100)  3.3 %
(0.055  0.0016 0 )
%edenom
 0.00002 2 
 (100)  0.85 %
 
 0.002588 
%eoveerall

(3.3%) 2  (0.85 %) 2
 3.4 %
(3.4%)(20.6 nM) = (0.034)(20.6 nM) = 0.70 nM
Use the uncertainty to round correctly: 20.6 ± 0.70 or 0.7 nM and 20.6 nM ± 3.4 or 3%
4
Useful information
 ( xi  x ) 2
Standard deviation: s 
Average deviation: d
s
100
x
%RSD = 
i
n 1

x
i
x
Relative avg. dev. (rad) =
i
Confidence interval:   x 
n
ts
d
100
x
where df = 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


( s12 / n1  s 22 / n2 ) 2

  2
2
( s 22 / n2 ) 2
  ( s1 / n1 )

  n1  1
n2  1

Comparing measured result to known result:
t calc 



2




known value  x
s
n
where degrees of freedom = n – 1.
Fcalc
s12
 2
s2
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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