07/19/16
ANSWER KEY
CHEM 1A Midterm Exam
Name: __________________________________
Open-End Questions and Problems
Please read the following. To receive full credit for a question or a problem, in addition to the
correct answer, you must show a neat, complete, and logical method of solution where each
number is labeled with the appropriate unit and the final answer is rounded to the correct number
of significant digits. The correct answer without any work shown will generally get zero credit!
When an explanation is required, it should be brief, but accurate and complete. There are 10
questions for a total of 120 points. (The part of the exam with multiple choice questions is worth
80 points. Both parts combined are 200 points total.)
1. (12 points) Fill the blanks in the following table.
0.500 mol CO2 + 0.300 mol O2 + 0.200 mol CO
The total number of molecules
1.000 mol of molec. = 6.022×1023 molec.
The total number of atoms
(0.500×3 + 0.300×2 + 0.200×2) mol of atoms ×
6.022×1023 atoms/mol = 1.505×1024 atoms
The number of elements
2 (carbon and oxygen)
The number of substances
3 (carbon dioxide, oxygen, and carbon
monoxide)
The number of compounds
2 (carbon dioxide and carbon monoxide)
The number of elementary substances
1 (oxygen)
The total mass in grams
(Show work!)
0.500 mol CO2 ×
44.01 g
1 mol
= 22.0 g
0.300 mol O2 ×
32.00 g
1 mol
=9.60 g
0.200 mol CO ×
28.01 g
1 mol
= 5.60 g
22.0 g + 9.60 g + 5.60 g = 37.20 g
PV = nRT
The total volume in liters (at room
temperature and normal pressure)
(Show work!)
V=
V = nRT/P
1.000 mol×0.08206 atm∙L/(mol∙K) ×295K
= 24.2 L
1 atm
=
2. (12 points) Perform the following conversions. Record the results either in decimal notation
or in scientific notation whichever of the two is more appropriate. (You don’t have to show
your work.)
2.20×10–4 km =
2.20
dm
786 nm =
7.0×1022 mm2 =
7.0×1040
pm2
0.003600 cm3 =
846 mL =
1 yd3 (exactly) =
8.46
dL
46,656
1.00 µL =
in3
1 L (exactly) =
7.86×10−5
cm
3.600×10−9
1.00
0.001
m3
mm3
m
3. (12 points) Give two examples of each. Note: (i) do not use the same substance as an
example more than once; (ii) paper, plastic, wood, a stick, gasoline, milk, food are not pure
chemical substances and should not be used as examples.
Physical Change (a change that involves
a specific pure substance)
Chemical Change (a change that involves
a specific pure substance)
Example 1:
ice melts
Example 1:
potassium reacts with water to
produce potassium hydroxide and
hydrogen gas
Example 2:
sulfur burns in air to produce sulfur
dioxide
Example 2:
sugar (sucrose) dissolves in water
Physical Property (specific property of
a specific pure substance)
Chemical Property (specific property of
a specific pure substance)
Example 1:
density of aluminum is 2.70 g/cm3
Example 1:
chlorine reacts with iron to produce
iron(III) chloride
Example 2:
ammonia is gas at room temperature
and normal pressure
Example 2:
neon does not react with other
chemical elements to form any stable
compounds
4. (12 points) A 0.1005-g sample of an organic compound composed of C, H, and O is burned in
pure oxygen gas, producing 0.2829 g of CO2 and 0.1159 g of H2O.
(a) What is the mass percent of oxygen in the original compound?
0.2829 g CO2
×
0.1159 g H2O ×
12.01 g C
44.01 g CO2
2×1.008 g H
18.01 g H2O
mass of O = 0.1005 g
% O =
0.0103 g
0.1005 g
= 0.07720 g C
= 0.01297 g H
– 0.07720 g – 0.01297 g H g = 0.0103 g
× 100%
= 10.3%
(b) How many grams of oxygen gas have been consumed in the reaction?
0.1005 g CxHyOz + X g O2 = 0.2829 g CO2 + 0.1159 g H2O
X = 0.2829 + 0.1159 − 0.1005
0.2983 g O2
5. (12 points) 50.0 mL of 0.600 M chloric acid is poured over 0.795 grams of copper(II) oxide.
(a) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and NIE) for the
reaction occurred. In each equation, indicate physical state of each reactant and product:
(s), (l), (g), or (aq).
FFE: CuO(s) + 2 HClO3(aq) → Cu(ClO3)2(aq) + H2O(l)
CIE: CuO(s) + 2 H+(aq) + 2 ClO3−(aq) → Cu2+ + 2 ClO3−(aq) + H2O(l)
NIE: CuO(s) + 2 H+(aq) → Cu2+ + H2O(l)
(b) How many grams of a solid substance will remain at the completion of the reaction?
Show work.
0.795 g CuO / (79.55 g/mol) = 0.0100 mol CuO = 10.0 mmol CuO
50.0 mL × 0.600 M = 30.0 mmol HClO3
CuO is the L.R.; it will dissolve completely; no solid remains after the reaction
(c) Which ions and at what molar concentration will remain in the solution at the completion
of the reaction? Show work.
Cu2+
mmol before the reaction
change: + or – mmol
mmol after the reaction
molar concentration (M)
0
H+
30.0
ClO3−
30.0
+ 10.0
− 20.0
0
10.0
10.0
30.0
0.200
0.200
0.600
6. (12 points) In the Kjeldahl method, the nitrogen contained within organic material is
converted into ammonia. A 0.7535g sample of wheat flour was analyzed. The ammonia was
distilled into 25.00 mL of 0.06211 M hydrochloric acid and the excess of the acid was then
back-titrated with 9.95 mL of 0.03512 M sodium hydroxide. Calculate the mass of nitrogen
in the sample and the percentage protein in the flour given that proteins are 5.70% nitrogen
by mass.
NH3(g) + HCl(aq) → NH4Cl(aq)
HCl(aq; excess) + NaOH(aq) → NaCl(aq) + H2O(l)
nHCl = 25.00 mL × 0.06211 M = 1.5528 mmol
nNaOH = 9.95 mL × 0.03512 M = 0.3494 mmol
nNH3 = nHCl − nNaOH = 1.2033 mmol = nN
mN = 14.01 g/mol × 0.0012033 mol = 0.01685 g
mprotein = 0.01685 g ×
100 g
5.70 g
= 0.2956 g
% protein =
0.2956 g
0.7535 g
× 100% = 39.2%
7. (12 points) Oxalic acid dihydrate, H2C2O4∙2H2O, is used as a primary standard in acid-base as
well as in oxidation-reduction titrations. In a standardization procedure, 0.1305 g of the
dihydrate were dissolved is about 50 mL of water, the resulting solution was acidified with
sulfuric acid and titrated with a solution of potassium permanganate, KMnO4.
KMnO4 + H2C2O4 + H2SO4 → MnSO4 + CO2 + K2SO4 + H2O (not balanced)
(a) Using the half-equation method, derive the balanced full-formula equation for the reaction
used in the titration experiment.
MnO4− + 8 H+ + 5e− → Mn2+ + 4 H2O × 2
H2C2O4 → 2 CO2 + 2 H+ + 2e−
×5
2 MnO4− + 5 H2C2O4 + 6 H+ → 2 Mn2+ + 10 CO2 + 8 H2O
<== NIE
+ 2 K+ + 3 SO42−
+ 2 K+ + 3 SO42−
<== spectator ions
2KMnO4(aq) + 5H2C2O4(aq) + 3H2SO4(aq) → 2MnSO4(aq) + 10CO2(g) + K2SO4(aq)
+ 8H2O(l)
(b) Using the balanced chemical equation from part (a), calculate the molarity of the KMnO4
solution if 19.05 mL of it was used in the experiment?
nH2C2O4 = 0.1305 g / (126.07 g/mol) = 0.001035 mol
nKMnO4 = 0.001035 mol / 2.5 = 0.004141 mol
MKMnO4 = 0.004141 mol / (0.01905 mL) = 0.02174 mol/L = 0.02174 M
8. (12 points) A titration experiment is being conducted. The buret is filled with a solution of
sodium hydroxide of unknown concentration. Oxalic acid dihydrate is used as a standard.
(a) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate, but failed to completely dissolve it prior to titration. How will it affect the
calculated molarity of sodium hydroxide? Explain.
Smaller volume NaOH(aq) will be used. Moles of NaOH are calculated from
the weighed amount of H2C2O4∙2H2O. Therefore, the calculated molarity of
NaOH will be HIGHER: molarity = moles / liters of solution.
(b) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate, but used 50% more water to dissolved it as compared to the recommended
amount. How will it affect the calculated molarity of sodium hydroxide? Explain.
Extra water used to dissolve H2C2O4∙2H2O WILL NOT AFFECT the results of
the experiment. The calculations are based on the moles of H2C2O4∙2H2O, not
on its concentration.
(c) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate and completely dissolved it in the recommended amount of water, however the
student failed to notice that prior to the first run of titration the tip of the buret was not
completely filled with sodium hydroxide solution, but had an air bubble there. How will it
affect the calculated molarity of sodium hydroxide? Explain.
It would appear AS IF A LARGER VOLUME of NaOH(aq) was added to the
flask than the volume required to neutralize all of the H2C2O4∙2H2O.
Therefore, the calculated molarity of NaOH will be LOWER:
molarity = moles / liters of solution.
9. (12 points) A gas has a density of 1.86 g/L at STP. What will be its density at 655 mm Hg and
120°C? Show work.
m = 186 g
V1 = 1 L
T1 = 273 K
P1 = 760 torr
T2 = 120 + 273 = 393 K
P2 = 655 torr
(1) V2 = ?; (2) d2 = ?
P1V1
T1
V2 =
=
d2 =
P2V2
T2
=
P1V1T2
T1P2
=
760 torr × 1 L × 393 K
273 K × 655 torr
m
V2
=
186 g
1.67 L
= 1.67 L
= 1.11 g/L
10. (12 points) The dew point is the temperature at which the water vapor in a sample of air at
constant barometric pressure condenses into liquid water at the same rate at which it
evaporates. The following weather conditions were reported: temperature is 71°F and relative
humidity is 43%. What was the dew point in °F?
°C =
71 − 32
1.8
= 22°
PH2O = 19.8 mmHg × 0.43 = 8.5 mmHg
At dew point: 8.5 mmHg = PH2O(saturated)
From the vapor pressure table: ∆P/∆T = (9.2 - 6.5) mmHg / 5°C = 0.54 mmHg/°C
(8.5 – 6.5) mmHg / 0.54 mmHg/°C = 3.7°C
6.5°C + 3.7°C = 10.2°C = 50°F