Quantitative Analysis (CHM 235)
Dr. S.A. Skrabal
Exam 1—8 February 2005
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. A solution containing vanadium was prepared by carefully weighing out 0.1005 g of high-purity solid
V2O5 (vanadium pentoxide), transferring it quantitatively to a 200.0 mL volumetric flask, dissolving the solid
V2O5 completely in HCl, then bringing the flask to volume with deionized H2O and mixing.
(A) What is the concentration of vanadium (V) in the solution in units of ppm? (8)
Remember 1 ppm = 1 mg/L
(0.1005 g V2O5/200.0 x 10-3 L)(1 mol V2O5/181.88 g V2O5)(2 mol V/1 mol V2O5)(50.94 g V/1 mol V)(1000
mg/g) = 281.47 or 281.5 mg/L or ppm
(B) What is the concentration of V in the solution on a %(w:v) basis? (7)
%(w:v) = [mass solute (g)/vol. soln. (mL)] 100
(281.47 mg V/L)(g/1000 mg)(1 L/1000 mL)100 = 0.028147% or 2.815 x 10-2 % (w:v)
2. List three methods that can be used to detect and correct for systematic error. (9)
Analyze blanks
Analyze certified (or standard) reference materials (CRM, SRM)
Perform interlaboratory comparison
Calibrate instruments
Analyze for same analyte using different methods
3. In preparation for a clinical determination of lead (Pb) in human blood, a standard reference material
(SRM) was analyzed using a common laboratory method. The SRM is certified to contain 13.9 μg dL-1
(micrograms per deciliter) of Pb. A clinical analyst obtained the following values for Pb in the SRM: 15.7,
16.2, 13.2, 14.8, and 15.1 μg dL-1.
(A) Determine whether or not any values should be rejected as questionable values using the Q test at the
90% confidence level. (7)
13.2, 14.8, 15.1, 15.7, 16.2
13.2 appears to be the outlier.
Qcalc = gap/range = (14.8 – 13.2)/(16.2 – 13.2) = 0.533
Qtable(n = 5, CL = 90%) = 0.642.
Since Qcalc < Qtable, do not reject the data
(B) What are the mean, standard deviation, and relative standard deviation of the acceptable results? (10)
Using calculator and including all data
Mean = 15.0 or 15 μg/dL
Std. dev. = 1.1 or 1 μg/dL
RSD = 1.1/15.0)100 = 7.3 or 7%
Note rounding using “real rule” on sig. figs.---first uncertain digit is last sig. fig.
(C) Using the appropriate t-test, determine whether or not the analyst obtained a Pb concentration in the
SRM that was significantly different from the certified value at the 95% confidence level. (10)
t calc 
t calc 
known value  x
s
n
13.9 g / dL  15.0 g / dL
1.1 g / dL
5
 2.236
df = n -1 = 5 -1 = 4.
ttable(df = 4, 95% CL) = 2.776
Since tcalc < ttable, value is not significantly different from the certified value at the 95% CL.
4. A 10-year-old child began treatment for lead poisoning on 15 February 2000. On that day, the child’s
blood was analyzed (4 measurements) and contained 32.7 ± 0.9 μg dL-1, where the uncertainty is the
standard deviation. After 30 days of treatment, the child’s blood was reanalyzed (5 measurements) and
found to contain 27.5 ± 0.8 μg dL-1.
(A) Use the F-test to verify that the standard deviations of the two sets of data are not significantly different
at the 95% confidence level. (6)
Fcalc = s12/s22 where s1 > s2
Fcalc = (0.9 μg/dL)2/(0.8 μg/dL)2 = 1.226
df = 4 -1 = 3 and 5 -1 = 4. Ftable(df = 3,4; 95% CL) = 6.59
Since Fcalc < Ftable, std. devs. are not significantly different at the 95% CL.
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4. (B) Use the appropriate t-test to determine whether or not the lead level in the child’s blood was
significantly different at the 95% confidence level after the 30 days of treatment. (10)
t-test for comparison of means
s pooled 
t calc 

s12 (n1 1)  s 22 (n2 1)
n1  n2  2
x1  x 2
n1 n2
s pooled
n1  n2


32.7  27.5
0.8443
(0.9) 2 (4  1)  (0.8) 2 (5  1)
452
(4)(5)
45
 0.8443
 9.181
df = n1 + n2 – 2 = 4 + 5 - 2 = 7
ttable(df = 7, 95% CL) = 2.365
Since tcalc > ttable, the mean values on the two days are significantly different at the 95% CL.
5. Twenty-six measurements of the oxygen partial pressure in an experimental chamber were made over a
period of 60 days. The mean and standard deviation of these measurements was 0.2095 ± 0.0008 atm
(atm = atmosphere). Calculate the 99% confidence interval of these measurements. (10)
x
ts
n
df = n -1 – 26 – 1 = 25
ttable(df = 25, 99% CL) = 2.787
  0.2095 
(2.787)(0.0008)
26
 0.2095  0.0004 3 atm
3
6. A student prepared a set of vanadium standards (0-50 μM) which she analyzed using atomic absorption
spectroscopy. Using EXCEL, the student plotted her results, then used the LINEST function to obtain the
parameters of the regression equation, including the slope (m), the uncertainty or 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 in the spreadsheet depicted below.
[V] (μM)
Abs.
Corr. Abs.
0
5.00
10.0
20.0
50.0
0.001
0.023
0.040
0.079
0.192
0
0.022
0.039
0.078
0.191
m=
sd of m=
R^2 =
0.003797468
3.30894E-05
0.999772274
13170.73171
0.02278481
0.001443038
0.000813892
0.001315279
3
5.18987E-06
=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. (6)
Apply “real rule” on sig. figs.
y (± 0.0013) = 0.003797 (± 0.000033) x + 0.00144 (± 0.00081)
or
y (± 0.001) = 0.00380 (± 0.00003) x + 0.0014 (± 0.0008)
(B) Suppose an unknown sample was analyzed for vanadium in the same manner as the standards, giving
an absorbance of 0.112. Determine the vanadium concentration (in μM) of the unknown sample. (8)
Blank-corrected absorbance = 0.112 – 0.001 = 0.111
x = (y – b) / m = (0.111 – 0.00144) / 0.003797 = 28.85 μM
Keeping 3 sig. figs. so far until uncertainty analysis is performed
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6. (C) Using the LINEST results, determine the absolute and relative uncertainties of the concentration of
the unknown sample. Be sure to express your final answers with the correct number of significant figures.
(9)
Rewrite equation in (B) including uncertainties:
x ( s x ) 
y ( s y )  b ( sb )
m (  s m )

0.111( 0.0013 )  0.0014 4 ( 0.00081 )
0.00379 7 ( 0.000033 )
 28.85  s x
Propogation of uncertainty for mixed operations:
enum  (0.0013 ) 2  (0.00081 ) 2
%enum 
 0.0015
0.0015
100  1.3 %
(0.111  0.0014 4 )
%edenom 
0.000033
100  0.8 6 %
0.00379 7
%eoverall

(1.3 %) 2  (0.8 6 %) 2
 1. 5 %
Convert to absolute uncertainty and express the answer with the correct number of sig. figs.
(1.5%) (28.85 μM) = (0.015) (28.85 μM) = 0.43 μM
Answers: 28.85 ± 0.43 μM or 28.8 ± 0.4 μM
and 28.85 μM ± 1.5% or 28.8 μM ± 2%
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Useful information
Standard deviation: s 
 ( xi  x ) 2
s
100
x
%RSD = 
i
Confidence interval:   x 
n 1
ts
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


( s 2 / n  s 22 / n2 ) 2

  2 1 21
( 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
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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