KEY
Directions: There are 4 questions, each worth equal points. Some are more difficult than others. Read over all questions before you start to answer. Answering the harder questions last will insure that you obtain the maximum score. You must show your work and pertinent formula's to get full credit. The final number answer is only ten percent of the total score.
Statistics formulas:
Useful data:
Values of Q @ 90% confidence
Q |
0.94 |
0.74 |
0.64 |
0.56 |
0.51 |
0.47 |
0.44 |
0.41 |
N |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Values of Student's t @ 95% confidence
t |
12.71 |
4.303 |
3.182 |
2.776 |
2.571 |
2.447 |
2.365 |
2.306 |
2.262 |
2.228 |
d.f. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Values of F @ 90% confidence
Degrees of freedom s2 |
Degrees of freedom s1 |
||||
2 |
3 |
4 |
5 |
6 |
|
2 |
9.00 |
9.16 |
9.24 |
9.29 |
9.33 |
3 |
5.46 |
5.39 |
5.34 |
5.31 |
5.28 |
4 |
4.32 |
4.19 |
4.11 |
4.05 |
4.01 |
5 |
3.78 |
3.62 |
3.52 |
3.45 |
3.90 |
6 |
3.46 |
3.29 |
3.18 |
3.11 |
3.05 |
1. Analysis, Concentrations, and Calibration
A) What are the major steps in a chemical analysis?
Understand problem
Sampling
Separation into components
Conversion to measurable form
Measurement
Data Analysis
Report
B) A 48% (w/w) solution of HBr (FW 80.917) in water has a density
of 1.50 g/mL.
What is the formal concentration of HBr?
What volume of 48% (w/w) solution is needed to prepare 500 mL of a 0.16 F HBr solution?
F1V1= F2V2
(8.9 F) V1 = (0.16 F)(500 mL)
V1 = 9.0 mL
What mass of the 48% (w/w) HBr is required to prepare this solution?
9.0 mL soln. (1.5 g/mL) = 13.5 g soln.
C) Write definitions or give formulas to describe the following terms
Accuracy: indication of error or difference between true and measured values
Analyte: substance being analyzed
Molar: concentration unit in moles of substance per liter of solution
Parts-per-thousand (ppth): weight fraction unit in grams of substance per kilogram or
Weight percent: weight fraction in grams of substance per 100 grams or
2. Statistics
A. Use the Q test to determine if the 3.483 value can be dropped from the following data set
3.274, 3.258, 3.265, 3.258, 3.350, 3.483
Qcalc = gap/range
Qcalc = (3.483-3.350)/(3.483-3.258) = 0.591
Qtable = 0.56
Since Qtable< Qcalc datum can be rejected from the data set
B. Report the mean, standard deviation, and relative standard deviation based on the results from part A. (values not rejected). Be sure to show your work.
The mean is
The standard deviation is
The relative standard deviation is
C. Calculate the confidence interval and report the statistically probable mean value at the 95% confidence level
The table t value for 5-1=4 degrees of freedom is 2.776 Using the mean and standard deviation
D. Calculate the variances of the data in part A with and without the 3.483 value. Apply the F-test to determine if there is a significant difference in the two variances (@ 90% confidence). Does your F-test result agree with the Q test in part A?
Without the rejected datum, there are N=6 points, with a mean and standard deviation of 3.315 and s1=0.08958 The calculated F is
The whole data set, used to calculate s1, has 6-1=5 degrees of freedom. The data set with one point rejected, used to calculate s2, has 5-1=4 degrees of freedom. Ftable = 4.05
Since Ftable < Fcalc the two standard deviations are statistically different.
This result agrees with that of the Q test. The later indicated that one datum could have come from a different measurement population. Including this datum results in an overall standard deviation that is statistically different than that obtained with the rejected datum.
3. Chemical Equilibrium
Calomel, Hg2Cl2, (FW 472.09) dissolves to form Hg22+ and chloride with a Ksp=1.2´ 10-18. Calculate the mass of calomel that will dissolve in...
The equilibrium equation is
Hg2Cl2 (s) « Hg22+ (aq) + 2Cl- (aq)
Using the table method,
Hg2Cl2 (s) |
Hg22+ (aq) |
Cl- (aq) |
|
initial |
solid |
0 |
0 |
change |
-x |
+x |
+2x |
final |
solid |
x |
2x |
Solve the Ksp equation for the molar solubility, x
Now find the weight in dissolved in 1.00 L pure water
The equilibrium equations are
Hg2Cl2 (s) « Hg22+ (aq) + 2Cl- (aq)
NaCl (s) ® Na+ (aq) + Cl- (aq)
Using the table method,
Hg2Cl2 (s) |
Hg22+ (aq) |
Cl- (aq) |
|
initial |
solid |
0 |
0.030 |
change |
-x |
+x |
+2x |
final |
solid |
x |
0.030 + 2x |
Since the chloride concentration produced from Hg2Cl2 dissolution in pure water is only 2x = 1.34x10-6, 2x can be neglected relative to the 0.030 M chloride from NaCl. We solve the Ksp equation for the molar solubility, x
Now find the weight in the 0.030 M chloride solution
4. Gravimetric Analysis
A) A 0.5962 g iron ore sample was dissolved in hot perchloric acid and then filtered. The iron was oxidized to the ferric state, Fe3+ (AW 55.85). The solution was made basic with ammonium hydroxide, NH4OH, and the iron subsequently precipitated as ferric hydroxide. This gel was filtered in a cistern-glass crucible, rinsed with dilute ammonium hydroxide, ignited, cooled in a desiccator, and weighed. The resulting ferric oxide, Fe2O3 (FW 159.69) weighed 0.3210 g.
1. Why is the solution filtered after perchloric acid dissolution?
To remove insoluble matrix material (like silica, alumina, etc.)
2. Why rinse with ammonium hydroxide solution?
Ammonium hydroxide is a volatile electrolyte solution. The volatile electrolyte replaces counter ions that may have caused colloid coagulation, thereby minimizing peptization (colloid dissolution). The original counter ions should be eliminated to avoid gravimetric errors.
3. Why use a desiccator during cooling?
The desiccator is used to keep the sample dry. Its use during cooling avoids weight gain due to water adsorption.
4. Calculate the weight percent iron in the ore.
B) Gravimetric analysis of nickel by precipitation with dimethylglyoxime gave the linear (y=mx+b) calibration shown below. Regression values for the slope and intercept, and their standard deviations, are given in the inset box. Calculate the weight % nickel, and estimate the error using propagation formulas, for a sample with precipitate weight 5.120± 0.001 g.
Using the line equation
y (weight precipitate) = m x (weight % Ni) + b
We solve for weight % Ni as a function of weight of precipitate
x (weight % Ni) = [y (weight precipitate) - b ]/m
x (weight % Ni) = [5.120 - 0.35]/0.297 = 16.56 % Ni
To estimate the error, we use the standard deviations given for the regression and the uncertainty in the precipitate weight. For addition/subtraction the variance is the sum of the variances. For multiplication and division, the square RSD (relative standard deviations) of the result is the sum of squared RSD. Using these rules in the above equation
The error is 1.65 wt % Ni
This page was last edited on Friday, August 25, 2006