Practice lab exam questions, CHML 235, Fall 2006
1. In the first part of a laboratory experiment, the 0-50.00 mL portion of a 50.00 mL buret was
calibrated. The results for three trials are shown below.
Buret
Tem
reading (ml) p.
(“apparent (C)
volume”)
Mass to
volume
correctio
n (mL/g)
Star End
t
0.00 49.99 23
1.0035
0.03 50.00 23
1.0035
0.01 49.98 23
1.0035
Mass of flask
(g)
Mass
of
water
(g)
Apparen
t volume
of water
(ml)
Actual
volume
of
water
(mL)
Calibratio
n
correction
(mL)
49.85
4
49.84
0
49.83
9
49.99
50.028
0.038
49.97
50.014
0.044
49.97
50.013
0.043
Empty Full
26.88
9
26.99
2
26.99
5
76.74
3
76.83
2
76.83
4
(a) Complete the eight missing entries in the table.
The actual volume is calculated by multiplying the mass of water by the mass-to-volume correction.
The calibration correction is (actual volume – apparent volume).
(b) Calculate the average calibration correction using the data in the first part of the experiment.
Remember to express this using the proper number of significant digits.
Answer: 0.041 or 0.04 mL
(c) In the second part of the experiment, a sample was titrated using the calibrated buret. The titration
required 41.15 mL, as read off of the buret. What should the volume on this titration be corrected to?
The calibration correction is added to the apparent titration volume to obtain the corrected volume.
Remember, the calibration correction can be a positive or negative number. Here the correction is
positive; i.e., the volume read off the buret is too low by 0.04 mL, so this volume needs to be added to
volumes read off the buret.
Answer: 41.19 mL
2. Polycyclic aromatic hydrocarbons (PAHs) are common contaminants in surface waters. A sample of
water draining an asphalt parking lot was obtained and analyzed for two PAHs, fluoranthene and
naphthalene, using the techniques of Expt. 3. A 112.0-mL sample was passed through a C18 solid
phase extraction cartridge. The PAHs were eluted off the column into 1.26 mL of acetonitrile and
quantified using high performance liquid chromatography (HPLC). A series of standard solutions
containing fluoranthene and naphthalene were analyzed and the following calibration equations were
obtained:
Fluoranthene:
Peak area = 46.081 (conc.) – 2.22
Naphthalene:
Peak area = 11.885 (conc.) + 1.03.
where the PAH concentrations are in mg L-1. The preconcentrated sample was analyzed, giving peak
areas of 114.943 and 93.212 for fluoranthene and naphthalene, respectively.
(a) What is the concentration (mg/L) of fluoranthene in the 1.26 mL of preconcentrated sample?
Rearrange regression equation to solve for conc.
Conc. = {114.943 – (-2.22)} / 46.081 = 2.54 mg L-1
Answer: 2.54 mg L-1
(b) What is the naphthalene concentration (mg/L) in the original 112.0 mL water sample?
Similar to (a) but must also take into account preconcentration factor.
Conc. = {93.212 – 1.03} / 11.885 = 7.76 mg L-1 in 1.26 mL preconcentrated sample.
Concentration factor = 112.0 mL / 1.26 mL = 88.9
So naphthalene concentration in original sample = 7.76 mg L-1 / 88.9 = 0.0873 mg L-1
Answer: 0.0873 mg L-1
3. In an analytical laboratory, two methods were used to determine theobromine, the biochemical
precursor of caffeine, which is found in such foods as chocolate, cocoa, and coffee. The concentration
of theobromine in samples of a cocoa powder was determined by the two techniques (volumetric
analysis and high performance liquid chromatography) and the data were statistically analyzed to
discern whether or not the two methods gave similar results. A summary of the analytical data is
shown in the table below.
Method
No. of measurements (n)
Volumetric analysis
HPLC
26
31
Mean concentration of
theobromine (mg/g)
0.267
0.342
Std. dev.(mg/g)
0.060
0.048
(a) Use the F-test to confirm that the standard deviations are not significantly different at the 95%
confidence level.
Fcalc = s12/s22 where s1 > s2
Fcalc = (0.060)2 / 0.048)2 = 1.56
Degrees of freedom = n1 – 1 and n2 -1 = (26 – 1) = 25 and (31 – 1) = 30
F(95% CL, df = 25,30) = 1.89
Answer: Since Fcalc < Ftable, std. devs. are not significantly different at the 95% CL.
(b) Then use the t-test for the comparison of means to determine whether or not the two methods give
values for the concentration of theobromine that are significantly different at the 95% confidence level.
s pooled 
t calc 
s12 (n1 1)  s 22 (n2 1)
n1  n2  2
x1  x 2
n1 n2
s pooled
n1  n2


(0.06 0 ) 2 (26 1)  (0.04 8 ) 2 (31 1)
26  31  2
0.26 7  0.34 2
0.002893
(26)(31
26  31
 0.002893
 97.486
df = 26 +31 – 2 = 55
t(df = 55, 95% CL) = 2.005 (Interpolated between df = 40 and df = 60).
Answer: Since tcalc > ttable, the means are significantly different at the 95% CL.
4. As a famous forensic anthropologist with expertise in analytical chemistry, you are given a small
sample of a grave marker and need to know its CaCO3 (calcium carbonate) content so you can
possibly determine the type of rock from which it was made. You finely grind the sample and dissolve
0.5730 g of it in acid. The Ca2+ in the sample was quantitatively precipitated as CaC2O4 (calcium
oxalate) by addition of ammonium oxalate and ammonia. The CaC2O4 precipitate was dissolved in a
solution of H2SO4, then titrated to the endpoint with 21.08 mL of 0.04175 M KMnO4 according to the
following reaction:
5C2O42- + 2MnO4- + 6H+  10CO2 + 2Mn2+ + 8H2O.
A blank titration required 0.03 mL of KMnO4 to reach the endpoint. The calibration correction for the 030.00 interval of the buret used was -0.02 mL.
(a) How many milligrams of Ca are in the sample?
Blank-corrected titration volume = 21.08 – 0.03 = 21.05 mL
Calibration-corrected volume = 21.05 + (-0.02) = 21.03 mL
(21.03 x 10-3 L)(0.04175 mol MnO4-/L)(5 mol C2O42-/2 mol MnO4-)(1 mol Ca/1 mol C2O42-)(40.08 g
Ca/mol Ca)(1000 mg /g) = 87.975 or 87.98 mg Ca
Answer: 87.975 or 87.98 mg Ca
(b) What is the weight percent CaCO3 in the sample?
(87.975 mg Ca)(g/1000 mg)(1mol Ca/40.08 g Ca)(1 mol CaCO3/1 mol Ca)(100.09 g CaCO3/mol
CaCO3)(100/0.5730 g) = 38.34%
Answer: 38.34%