Chemistry 222
Fall 2013
Exam 1: Chapters 1-4
Name__________________________________________
80 Points
Complete two (2) of problems 1-3 and four (4) of problems 4-8. CLEARLY mark the problem you
do not want graded. You must show your work to receive credit for problems requiring math.
Report your answers with the appropriate number of significant figures and with the appropriate
units.
Do two of problems 1-3. Clearly mark the problem you do not want graded. (10 pts each)
1. A statistical analysis is an essential component in the evaluation of experimental results.
Early in our discussion of statistics, I mentioned that statistics only tell us about the precision
of a measurement, not the accuracy. Why is this so? If this is true, how can we use the
confidence interval to predict how close our results are to a “true” or accepted value?
When we refer to quality of results, we are typically considering the accuracy and precision of
a value. In terms of precision, statistics are a useful tool to evaluate how reproducible our
data are, with a standard deviation serving as an estimate of the scatter of the data. The
challenge comes in the fact that we typically have a very small data set and are forced to rely
on that small set to approximate the standard deviation. The confidence interval also allows
us to make some inferences about the accuracy of a method by taking the fact that we have
a small data set into account; assuming only random errors are impacting our
measurement. Therefore, we must rely on good experimental design to remove systematic
errors.
2. Choose ONE of the following pairs of terms and briefly (but clearly) compare and contrast the
two concepts.
a. TC vs. TD
b. Determinate Error vs. Indeterminate Error
a. TC = to contain. TC glassware is designed to hold a fixed volume of material (within its
tolerance) when filled to the mark. The glassware is calibrated for a given temperature
(usually 20C). Complete transfer of all the material from TC glassware requires rinsing.
TD = to deliver. TD glassware is designed to dispense a fixed volume of solution (within
its tolerance) after being filled to the mark. It is also calibrated at a fixed temperature.
Care must be taken to use TD glassware properly and not blow out all the liquid unless
the glassware was calibrated as "blow out" (etched stripe)
b. Determinate or systematic error affects accuracy. This error is usually constant and can
be identified and corrected.
Indeterminate or random error cannot be removed but can be evaluated and minimized
with appropriate experiment design and running multiple samples. Random errors impact
the precision of a measurement.
1
3. We ignore the contribution of buoyancy in virtually all of the mass measurements we make in
the laboratory. How can we get away with this? Identify one situation where we would be
unable to ignore buoyancy-introduced error.
The buoyancy correction accounts for the varying volume of air displaced when a sample is
weighed compared to the volume displaced when the balance was calibrated with calibration
weights. When the density of the sample being weighed is similar to the density of the
balance weights (8 g/mol), the error due to buoyancy is minimal (remember the plot we
discussed in class). In general buoyancy errors are minimal because we have been
weighing solid samples and because we do our critical weighing by difference. If we were to
weigh samples of very low density (like water or organic solvents or especially gases), we
should account for buoyancy errors.
Do four of problems 4-8. Clearly mark the problem you do not want graded. (15 pts each)
4. In an experiment, you use a solution of standardized potassium hydroxide ion to determine
the concentration of a solution of sulfuric acid. The data below was obtained for such a
titration. Based on this information, calculate the concentration of H2SO4 (with its associated
absolute uncertainty) in the solution.
Concentration of KOH standard
0.1117  0.0005 M
Volume of H2SO4 solution used
20.00  0.03 mL
Initial buret reading
1.46  0.05 mL
Final buret reading
33.54  0.05 mL
Uncertainty in the volume delivered by the buret:
(33.54  0.05 mL) - (1.46  0.05 mL) = 32.08  e1 mL
e1 = [(0.05)2 + (0.05)2]1/2 = 0.0707 mL
Concentration calculation: (DON’T FORGET THE STOICHIOMETRY!)
1
= 0.08958e2 M
0.11170.0005 mol KOH+ x 32.080.07mL x 1 mol H2SO4 x
1L
2 mol KOH
20.000.03mL
2
2
2
 0.0005   0.03   0.07 
e2  0.08958 M 
 
 
  0.08958 M0.00520   0.00046 M
 0.1117   20.00   32.08 
e2 = 0.00046 = 0.0005 M so the H2SO4 concentration is 0.0896  0.0005 M
2
5. You need to prepare a 500.0 mL of solution that is 100.0 ppm magnesium. Clearly describe
how you would prepare this solution starting from the points below. Include the quantities of
each starting material that you would need
a. starting with solid magnesium nitrate
b. starting with a 0.100 M magnesium nitrate solution
a. Remember, magnesium nitrate is Mg(NO3)2 (FW = 148.315 g/mol)
100 mg Mg2+x1 mol Mg2+x1 mol Mg(NO3)2x148.315 g Mg(NO3)2x0.500 L=305.1 mg Mg(NO3)2
1L
24.305 g
1 mol Mg2+
1 mol Mg(NO3)2
So, dissolve 0.3051 g Mg(NO3)2
b.
Since each mole of Mg(NO3)2 that dissociates liberates 1 mole of Mg2+, a
0.100 M Mg(NO3)2 solution is also 0.100 M Mg2+
100 mg Mg2+ x 1 mol Mg2+ x 0.500 L x
1L
= 20.6 mL
24.305 g
0.100 mol Mg2+
1L
So, dilute 20.6 mL of 0.100 M MgCl2 solution in a small amount of water in a 500 mL
volumetric flask, mix well, dilute to the mark and mix well again. The 20.6 mL could be
delivered by pipet or buret.
6. You have run a series of titrations to determine the unknown concentration of KHP in a solid
sample. The results of six titrations indicate KHP concentrations of 46.14%, 45.69%,
40.15%, 45.55%, 46.07%, 45.98%. If the "true" value for KHP in this sample is 46.29%, do
your results differ at the 95% confidence level?
Looking at the data, it appears that the value 40.15% is an outlier so try a Q-test or a G-Test:
Qcalc = 45.55 - 40.15 = 0.90
46.14 - 40.15
Gcalc = 44.93-40.15 = 2.03
2.353
Qtable = 0.56 < Qcalc, and Gtable = 1.822 < Gcalc so the data point should be rejected.
Based on the remaining data, the mean for the data set is 45.886% with a standard deviation
of 0.25 %. Do a t-test:
46.29  45.88 6
t calculated 
5  3.553
0.254
ttable for 4 degrees of freedom is 2.776, since tcalc>ttable, the results do differ significantly.
(NOTE: if you do not do the Q-test, the standard deviation is large enough that is looks like
the results do not differ. Always look at the data!)
Alternatively, you could have calculated the range determined by the confidence limit and
shown that 46.29% lies outside this range. The 95% CI is 45.9 ± 0.3 %
3
7. Obtaining an accurate mass for solid samples can make or break an analysis. Given your
new job as a teaching assistant in Quantitative analysis lab, describe how you would teach a
new Quant. student the proper method to handle solid samples during an analysis in order to
obtain the best quantitative results.
Your discussion should include the following:
 Drying solid samples to constant mass (what does “constant mass” mean? How do you
know samples are dry?)
 Making mass measurements by “weighing by difference” (how?)
 Avoid sample loss during weighing
 Handle weighing bottles using lint-free and oil-free materials
 Cool solid samples before weighing
 Close balance doors before weighing
 Store samples in desiccator
8. Nitrite (NO2-) was measured in rainwater and unchlorinated drinking water using replicate
measurements of a single sample by an established spectrophotometric method. Based on
the results below, does drinking water contain significantly more nitrite than rainwater (at the
95% confidence level)?
Replicate
Rainwater (ppb)
Drinking Water (ppb)
1
55.1
74.6
2
59.6
81.0
3
63.1
87.3
4
66.4
91.8
5
71.5
93.2
mean
63.1
85.6
st. dev.
6.28
7.77
This is a comparison of two methods, using several runs of a single sample to establish the
uncertainty on each method. Since we have two means and standard deviations, use spooled
to perform a t-test. Check the standard deviations with an F-test first:
Fcalculated
2
s1 


s 2 2

7.762
6.522
 1.53
Since Fcalculated is less than Ftable (6.39), our “normal” equations will be fine.
spooled 
t calculated 
6.522 4  7.762 4  7.06
552
85.2  63.0
25
 5.02
7.06
55
ttable for (5+5-2) = 8 degrees of freedom is 2.306
Since tcalculated > ttable, the results are significantly different
4
Blank Space if You Need Extra Room
5
Possibly Useful Information

d 
m' 1  a 
dw 
m 
d 

1  a 
d 

y
n
known value  x
t calculated 
s
x1  x 2
spooled
n1n 2
n1  n 2
d
t calculated 
n
sd
Q calculated 
Density of balance weights = 8.0 g/ml
ts
x 
t calculated 
Density of air = 0.012 g/ml
gap
range
Fcalculated 
s12
s2 2
n
1
 2
2
 x i  x 
i
s
spooled 
2
2
e  ( x   ) 2
n 1
s12 n1  1  s 22 n 2  1
n1  n 2  2

sd 
G calculated 
 di  d
i
2
n 1
suspect value  x
s
2+2=4
6
Values of Q for rejection of data
Values of Student’s t
# of
Observations
4
5
6
7
8
9
10
Confidence Level (%)
Degrees
of
Freedom
1
2
3
4
5
6
7
8
9
10
20
30
40
60
120

90
95
99.5
99.9
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.752
1.697
1.684
1.671
1.658
1.645
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.086
2.042
2.021
2.000
1.980
1.960
127.32
14.089
7.453
5.598
4.773
4.317
4.029
3.832
3.690
3.581
3.153
3.030
2.971
2.915
2.860
2.807
636.619
31.598
12.924
8.610
6.869
5.959
5.408
5.041
4.781
4.587
3.850
3.646
3.551
3.460
3.373
3.291
Q
(90% Confidence)
0.76
0.64
0.56
0.51
0.47
0.44
0.41
Grubbs Test for Outliers
# of
Gcritical
Observations At 95% confidence
4
1.463
5
1.672
6
1.822
7
1.938
8
2.032
9
2.110
10
2.176
Critical Values of F at the 95% Confidence Level
Degrees of freedom for s1
Degrees of
freedom for s2
2
3
4
5
6
7
8
9
10
2
3
4
5
6
7
8
9
10
19.0
9.55
6.94
5.79
5.14
4.74
4.46
4.26
4.10
19.2
9.28
6.59
5.41
4.76
4.35
4.07
3.86
3.71
19.2
9.12
6.39
5.19
4.53
4.12
3.84
3.63
3.48
19.3
9.01
6.26
5.05
4.39
3.97
3.69
3.48
3.33
19.3
8.94
6.16
4.95
4.28
3.87
3.58
3.37
3.22
19.4
8.89
6.09
4.88
4.21
3.79
3.50
3.29
3.14
19.4
8.84
6.04
4.82
4.15
3.73
3.44
3.23
3.07
19.4
8.81
6.00
4.77
4.10
3.68
3.39
3.18
3.02
19.4
8.79
5.96
4.74
4.06
3.64
3.35
3.14
2.98
7