Chemistry 222
Fall 2015
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 problems
you do not want graded. 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. In
our discussion of statistics, I stated several times 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?
2. In producing a calibration curve, raw data is typically subjected to a “linear least squares”
analysis. Dissect the phrase “linear least squares” and describe qualitatively what is done in
a linear least squares analysis. Why “linear”? “Least squares” of what? No calculations are
necessary.
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3. We tend to 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.
Do four of problems 4-8. Clearly mark the problem you do not want graded. (15 pts each)
4. The composition of a sample containing an unknown amount of sodium carbonate in
combination with an inert material was determined by dissolving the sample in 20.0 mL of
water and titrating the resulting solution with standardized nitric acid solution. Using the
information below, determine the percent by mass of sodium carbonate in the original
sample, with its absolute uncertainty. You may assume that the contribution of molar masses
to the overall uncertainty is negligible.
Concentration of nitric acid standard
0.2026  0.0006 M
Mass of carbonate-containing sample
0.9113  0.0005 g
Initial buret reading
1.28  0.05 mL
Final buret reading
29.74  0.05 mL
2
5. You need to prepare a 500.0 mL of solution that is 100.0 ppm calcium. 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 calcium nitrate
b. starting with a 0.100 M calcium nitrate solution
6. You have run a series of titrations to determine the unknown concentration of KHP in a solid
sample. The results of titrations indicate KHP concentrations of 36.14%, 35.69%, 30.15%,
35.55%, 36.07%, 35.98%. The "true" value for KHP in this sample is 36.29%. Evaluate the
data and determine if your results differ from the true value at the 95% confidence level.
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.
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 sample contain significantly more nitrite than rainwater
sample (at the 95% confidence level)?
Replicate
Rainwater (ppb)
Drinking Water (ppb)
1
55.1
74.6
2
59.6
81.0
4
3
63.1
87.3
4
66.4
91.8
5
71.5
93.2
Possibly Useful Information

d 
m' 1  a 
d
w
m 
 da 
1 

d 

Density of balance weights = 8.0 g/ml
ts
x 
y
n
eC  e  e
2
A
t calculated 
Density of air = 0.012 g/ml
s
x1  x 2
spooled
n
n1n 2
n1  n 2
spooled 
sy
m
s 2y  n
s12 n1  1  s 22 n 2  1
n1  n2  2
 di  d
i

D
2
n 1
 di  d
n2
sb2 
2

2
 di
n2
s 2y  x i2
D

s1 2
Fcalculated 
s2 2
yLOD = yblank + 3s
Q calculated 
n 1
sd 
sy 
2
i

1 1
( y  y )2
 
k n m 2  x i  x 2
2

sm
2
  eB 
   
 B
2
 x i  x 
s
d
t calculated 
n
sd
sx 
 2
2
2
e  ( x   ) 2
e
eC  C  A
 A
2
B
known value  x
t calculated 
1
gap
range
G calculated 
5
suspect value  x
s
Values of Q for rejection of data
Values of Student’s t
# of
Observations
4
5
6
Confidence Level (%)
Degrees of
Freedom
1
2
3
4
5
6
7
8
9
10

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
1.645
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
1.960
127.32
14.089
7.453
5.598
4.773
4.317
4.029
3.832
3.690
3.581
2.807
636.61
31.598
12.924
8.610
6.869
5.959
5.408
5.041
4.781
4.587
3.291
Q
(90% Confidence)
0.76
0.64
0.56
Grubbs Test for Outliers
# of
Gcritical
Observations At 95% confidence
4
1.463
5
1.672
6
1.822
Critical Values of F at the 95% Confidence Level
Degrees of freedom for s1
Degrees of
freedom for s2
2
3
4
5
2
3
4
5
6
7
8
9
10
19.0
9.55
6.94
5.79
19.2
9.28
6.59
5.41
19.2
9.12
6.39
5.19
19.3
9.01
6.26
5.05
19.3
8.94
6.16
4.95
19.4
8.89
6.09
4.88
19.4
8.84
6.04
4.82
19.4
8.81
6.00
4.77
19.4
8.79
5.96
4.74
6