Quantitative Analysis (CHM 235)
Dr. S.A. Skrabal
Exam 1—14 February 2006
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. Docetaxel (C43H53NO14; FW = 807.9 g mol-1) is a chemotherapy drug derived from the needles of the yew
plant. Suppose a chemotherapy infusion solution contains 0.500 g L-1 of docetaxel. What is the
concentration of docetaxel in the solution, in units of (a) %(w:v) and (b) mol L-1? (12)
(a) % w:v) = (mass solute (g) / volume solution (mL) ) 100
(0.500 g/L)(1 L/1000 mL) 100 = 0.0500% (w:v)
(b) (0.500 g docetaxel/L)(1 mol docetaxel/807.9 g docetaxel) = 6.19 x 10-4 mol L-1
2. During an autopsy of a victim of a suspected poisoning, a medical examiner removes 102.8 mL of fluid
from the victim’s stomach for the analysis of arsenic (FW = 74.92). A volume of 10.00 mL of the stomach
fluid was diluted to exactly 150.0 mL before analysis. The arsenic concentration in the diluted solution was
25.8 ppb. What is the arsenic concentration in the entire sample of stomach fluid in (a) ppb and (b) mol L-1?
(7 each)
(a) Cconc Vconc = Cdil Vdil
Cconc = (Cdil Vdil) / Vconc = (25.8 ppb)(150.0 mL) / (10.00 mL) = 387 ppb
(b) 387 ppb = 387 µg L-1
(387 µg As/L)(g/106 µg)(1 mol As/74.92 g As) = 5.17 x 10-6 mol L-1
3. Two different brands of cat food were analyzed for concentrations of methylmercury (CH3Hg+), a species
of mercury that is potentially very toxic to animals. Nine measurements of the “Happy Kitty” brand yielded a
mean concentration (± one standard deviation) of 1.322 ± 0.098 µg g-1. Eight measurements of the
“Fabulous Feline” brand yielded a mean concentration (± one standard deviation) of 1.100 ± 0.08 5 µg g-1.
(a) Use the F-test to determine whether or not the standard deviations of the two sets of measurements are
significantly different at the 95% confidence level. (7)
Fcalc 
s12
s 22

(0.09 8 ) 2
(0.085 ) 2
 1.329
df = (9 -1) = 8 for s1 and (8 – 1) = 7 for s2.
F(df = 8,7; 95% CL) = 3.73
Since Fcalc < Ftable, the standard deviations are not significantly different at the 95% CL.
(b) Determine whether or not the mean concentration of methylmercury is significantly different in the two
brands of cat food at the 95% confidence level. (12)
s pooled 
t calc 
s12 (n1 1)  s 22 (n2 1)
n1  n2  2
x1  x 2
n1 n2
s pooled
n1  n2


(0.09 8 ) 2 (9 1)  (0.08 5 ) 2 (8 1)
982
1.32 2 1.10 0
0.0922
(9)(8)
98
 0.0922
 4.955
df = 9 + 8 – 2 = 15
t(df = 15,95% CL) = 2.131
Since tcalc > ttable, the means are significantly different at the 95% CL.
4. A certified reference material (CRM) for the determination of total polychlorinated biphenyls (PCBs) in
harbor sediment is available from the National Research Council of Canada. The concentration of total
PCBs in the CRM is 21.8 µg kg-1 dry sediment. A new analyst in an environmental laboratory analyzes this
CRM seven times for total PCBs, obtaining the following results (in g kg-1 dry sediment): 20.9, 19.8, 22.8,
19.5, 19.9, 20.7, 21.0.
(a) Test these data to see if any outliers can be rejected at the 90% confidence level. (8)
22.8, 21.0, 20.9, 20.7, 19.9, 19.8, 19.5
22.8 is the possible outlier
Qcalc = gap/range = (22.8 – 21.0) / (22.8 – 19.5) = 0.545
Q(n = 7,90%CL) = 0.507
Since Qcalc > Qtable, reject the outlier at the 90% CL
2
4(b) Using only the acceptable data, determine whether or not the analyst is obtaining the “correct” value
for total PCBs in the CRM at the 95% confidence level (12)
Exclude the outlier.
Using calculator:
x  20.30 mg g 1
s  0.6 4 mg g 1
t calc 
known value  x
s
n

21.8  20.30
0.6 4
6
 5.741
df = 6 – 1 = 5
t(df = 5, 95% CL) = 2.571
Since tcalc > ttable, the analyst is not obtaining the certified or “correct” value at the 95% CL.
5. Six analyses of the phosphate content of a dishwasher detergent yielded the following results (in mg g -1):
14.2, 15.0, 15.2, 14.9, 14.5, 14.9. Calculate (a) the mean, (b) standard deviation, (c) relative standard
deviation, (d) average deviation, and (e) relative average deviation for these data. (15)
By calculator:
(a) x  14.7 8 or 14.8 mg g 1
(b) s
 0.36 or 0.4 mg g 1
(c) rsd = (0.36/14.78) 100 = 2.4 or 2%
(d)
d

14.2 14.7 8  15.0 14.7 8  15.2 14.7 8  14.9 14.7 8  14.5 14.7 8  14.9 14.7 8
6

1
0.2 9 or 0.3 mg g
(e) rad = (0.29 mg g-1) / 14.78 mg g-1) 100 = 1.9 or 2%
3
6. Concentrations of dissolved copper in rainwater samples were analyzed by graphite furnace atomic
absorption spectrometry. Standard concentrations ranged from 0 to 50.0 nM. The data were analyzed
using the LINEST function in Excel®, and the following unedited parameters of the regression were
calculated:
slope: 0.0005447 ± 0.0000398
y-intercept: 0.0012655 ± 0.0003313
std. dev. of y values: 0.0013555
An unknown sample was analyzed, yielding an absorbance of 0.023. The blank was found to be 0.001.
(A) Write the linear regression equation in the form y (± sy) = m (± sm) x + b (± sb) using the correct number
of significant figures. (7)
Applying real rule on sig. figs.:
y (± 0.0013) = 0.000544 (± 0.000039) x + 0.00126 (± 0.00033)
or
y (± 0.001) = 0.00054 (± 0.00004) x + 0.0013 (± 0.0003)
(B) Determine the copper concentration (in nM) of the unknown and the absolute and relative uncertainties
of the concentration (13)
Rearrange equation to solve for x = conc.
Blank- corrected absorbance = 0.023 – 0.001 = 0.022.
x ( s x ) 
y ( s y )  b ( sb )
m ( s m )
 x ( s x ) 
0.022 ( 0.0013 )  0.0012 6 ( 0.00033 )
0.00054 4 ( 0.000039 )
 38.1  s x nM
enum  (0.0013 ) 2  (0.00033 ) 2
%edenom 
 0.0013
%enum 
0.0013
100  6.2 %
(0.022  0.0012 6 )
0.000039
100  7.1%
0.00054 4
%eoverall  (6.2 %) 2  (7.1%) 2
 9.4 %
Convert to absolute uncertainty and use real rule on sig. figs. to round correctly:
(9.4%)(38.1 nM) = (0.094) (38.1 nM) = 3.5 nM
Final answer: 38 ± 3.5 nM and 38 nM ± 9.4% or 38 ± 4 nM and 38 nM ± 9%
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
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
5
Q = gap/range
6