AP Review Experimental Error
Hint: Write the equation for property in question; ask, is the “experimentally determined [quantity]” too large
or too small, and how does that affect property in question.
Multiple Choice
1994
70. To determine the molar mass of a solid monoprotic acid, a student titrated a weighed sample of the acid with
standardized aqueous NaOH. Which of the following could explain why the student obtained a molar mass
that was too large?
I. Failure to rinse all acid from the weighing paper into the titration vessel
II. Addition of more water than was needed to dissolve the acid
III. Addition of some base beyond the equivalence point
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Ans: (A) MM =
mass acid
.
L NaOH × M NaOH
Not all the massed acid was used, so the experimentally determined mass of
acid was too large, so MM will be too large.
€
1999
72. After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student
reported a value of 38 percent. The correct value for the percentage of water in the hydrate is 51 percent.
Which of the following is the most likely explanation for this difference?
(A) Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
(B) The dehydrated sample absorbed moisture after heating.
(C) The amount of the hydrate sample used was too small.
(D) The crucible was not heated to constant mass before use.
(E) Excess heating caused the dehydrated sample to decompose.
Ans: (B) %H2O = mass hydrate − mass dehydrated sample × 100% . If the residue absorbs water before weighing, the
mass hydrate
experimentally determined mass of dehydrated sample is too high, so %H2O is too low.
€
1983
The molar mass of a monoprotic acid HX was to be determined by titration with a standardized solution of
NaOH. The NaOH solution was standardized against oxalic acid dihydrate, H2C2O4·2H2O. Discuss the effect of
the calculated molecular weight of HX if the sample of oxalic acid dihydrate contained a nonacidic impurity.
Ans: Experimentally determined mass of H2C2O4·2H2O is too high, so moles H2C2O4·2H2O is too high, so
, the calculated MM is too small.
moles NaOH is too large => moles HX is too high. Since MM =
1
1993
I. 2 Mn2+ + 4 OH- + O2(g) → 2 MnO2(s) + 2 H2O
II. MnO2(s) + 2 I- + 4 H+ → Mn2+ + I2(aq) + 2 H2O
III. 2 S2O32- + I2(aq) → S4O62- + 2 I-
3. The amount of oxygen, O2, dissolved in water can be determined by titration. First, MnSO4 and NaOH are
added to a sample of water to convert all of the dissolved O2 to MnO2, as shown in equation I above. Then
H2SO4 and KI are added and the reaction represented by equation II proceeds. Finally, the I2 that is formed is
titrated with standard sodium thiosulfate, Na2S2O3, according to equation III.
(b) A student found that a 50.0-milliliter sample of water required 4.86 milliliters of 0.0112-molar Na2S2O3 to
reach the equivalence point. Calculate the number of moles of O2 dissolved in this sample.
(c) How would the results in (b) be affected if some I2 were lost before the S2O32- was added? Explain.
= 4 mol S2O32-
ANS:
2-
(0.00486 L)(0.0112 M S2O3 )(1 mol O2)/(4 mol S2O32-) = 1.36×10-5 mol O2
2Less I2 means less thiosulfate ion (S2O3 ) required, thus indicating a lower amount of dissolved oxygen
1999
5. A student performs an experiment to determine the molar mass of an
unknown gas. A small amount of the pure gas is released from a pressurized
container and collected in a graduated tube over water at room temperature, as
shown in the diagram above. The collection tube containing the gas is allowed
to stand for several minutes, and its depth is adjusted until the water levels inside
and outside the tube are the same.
(e) if the student fails to use information from the table of the equilibrium vapor
pressures of water in the calculation, the calculated value for the molar mass of
the unknown gas will be smaller than the actual value. Explain.
ANS:
PV = nRT and n =
OR molar mass =
Pressure will be larger, therefore number of moles will be larger. Molar mass = mass/moles, therefore
calculated molar mass will be smaller.
OR, The pressure is larger, or the number of moles is larger, or since Ptotal=Punknow –Pwater) we know that Ptotal >
Punknown.
2000
5. The molar mass of an unknown solid, which is nonvolatile and a nonelectrolyte, is to be determined by the
freezing-point depression method. The pure solvent used in the experiment freezes at 10°C and has a known
molal freezing-point depression constant, Kƒ.
(c) Suppose that during the experiment a significant but unknown amount of solvent evaporates from the test
tube. What effect would this have on the calculated value of the molar mass of the solid (i.e., too large, too
small, or no effect)? Justify your answer.
ANS: The molar mass is too small. If some of the solvent evaporates, then the (kg solvent) term is larger than
the actual value. If the (kg solvent) term used is too large, then the value calculated for the molar mass
will be too small.
molar mass =
or
2
If some of the solvent evaporates, then the concentration (molality) of the solute will be greater than we
think it is. More moles of solute results in a smaller molar mass (or since ΔT = iKf m, then the ΔTobs would
be greater than it should be). Since the molar mass of the unknown solute is inversely proportional to ΔT,
an erroneously high value for ΔT implies an erroneously low value for the molar mass (calculated molar
mass would be too small).
2002
H+(aq) + OH−(aq) → H2O(l)
5. A student is asked to determine the molar enthalpy of neutralization, ΔHneut , for the reaction represented
above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup
calorimeter. The heat released by the reaction is determined by using the equation q = mcΔT.
(e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this
have on the calculated value of the molar enthalpy of neutralization, ΔHneut? Justify your answer.
ANS: Heat lost to the air will produce a smaller ΔT. In the equation q = mcΔT a smaller ΔT will produce a
smaller value for q (heat released) than it should. In the following equation, the smaller magnitude of q and the
constant mol H2O means that ΔHneut will be less negative (more positive).
ΔHneut =
ΔH decreases because q decreases earns 1 point; ΔT decreases because ΔH decreases earns 1 point; no points
earned for ΔT decreases therefore q decreases.
2003
5. A student is instructed to determine the concentration of a solution of CoCl2 based on absorption of light
(spectrometric/colorimetric method). The student is provided with a 0.10 M solution of CoCl2 with which
to prepare standard solutions with concentrations of 0.020 M, 0.040 M, 0.060 M and 0.080 M.
(e) The student handles the sample container (e.g., test tube or cuvette) that holds the unknown solution and
leaves fingerprints in the path of the light beam. How will this affect the calculated concentration of the
unknown? Explain your answer.
ANS: Beer’s Law: A = a b c where a = molar absorptivity (not absorbance), b = path length of cuvette, c =
concentration. The presence of the fingerprints will scatter or absorb light. Since less light reaches the
detector, the solution will have a higher apparent absorbance, and therefore a higher reported concentration.
2004 Form B
5. An experiment is performed to determine the molar mass of an unknown solid monoprotic acid, HA, by
titration with a standardized NaOH solution.
(g) Explain what effect each of the following would have on the calculated molar mass of HA. Justify your
answers.
i. The original solid acid, HA, was not completely dry at the beginning of the experiment.
ii. The procedure called for 25 mL of H2O in the Erlenmeyer flask, but a student used 35 mL of H2O.
ANS:
i. molar mass of HA =
. Measured mass of HA is larger; so, calculated molar mass will be
higher than it should.
ii. No effect on calculated molar mass, because mathematical expression for molar mass does not
include amount of water used to dissolve solid HA. Both mass and number of moles of HA are
unaffected by the addition of water.
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2006 Form B
A student carries out an experiment to determine the equilibrium constant for a reaction by colorimetric
(spectrophotometric) analysis. The production of the red-colored species FeSCN2+(aq) is monitored.
Kc =
(d) If the student’s equilibrium FeSCN2+(aq) solution of unknown concentration fades to a lighter color before
the student measures its absorbance, will the calculated value of Kc be too high, too low, or unaffected? Justify
your answer.
ANS: The value of Kc will be too low; the lower absorbance reading indicates a lower [FeSCN2+] than actually
existed before the fading occurred, so substitution of a lower [FeSCN2+] into the equilibrium expression will
result in a lower value of Kc.
2007
5 Fe2+ + MnO4- + 8 H+ → 5 Fe3+ + Mn2+ + 4 H2O
5. The mass percent of iron in a soluble iron(II) compound is measured using a titration based on the balanced
equation above.
(e) What effect will adding too much titrant have on the experimentally determined value of the mass percent of
iron in the compound? Justify your answer.
mass % Fe =
where M = molarity of MnO4-; g = mass of compound
ANS: The experimentally determined mass percent of iron in the compound will be too large. V is too large ⇒
mass % Fe in expression above is too large.
2010
2. A student performs an experiment to determine the molar enthalpy of solution of urea, H2NCONH2. The
student places 91.95 g of water at 25°C into a coffee cup calorimeter and immerses a thermometer in the water.
After 50 s, the student adds 5.13 g of solid urea, also at 25°C, to the water and measures the temperature of the
solution as the urea dissolves. A plot of the temperature data is shown in the graph below. [plot shows an
endothermic heat of solution with ΔT = −3.2°C]
qsoln + mcΔT = 0 ⇒ qsoln = −mcΔT
ΔH
=
(f) The student performs a third trial of the experiment but this time adds urea that has been taken directly from
a refrigerator at 5°C. What effect, if any, would using the cold urea instead of urea at 25°C have on the
experimentally obtained value of ? Justify your answer.
ANS: There would be an increase in the obtained value for ΔH
larger negative temperature change.
because the colder urea would have caused a
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