CHEM 31
Introductory Chemistry
EXAM #2
October 16, 2002
Name:
Keye, Onsur
SSN:
Lab T.A.:
INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the
questions. For questions involving calculations, show all of your work -- HOW you
arrived at a particular answer is MORE important than the answer itself! Circle your
final answer to numerical questions.
The entire exam is worth a total of 150 points. Attached are a periodic table and a
formula sheet jam-packed with useful stuff!
Good Luck!
2
Possible
Points
20
Points
Earned
20
3
4
5
6
35
20
20
20
7
8
14
21
35
20
20
20
14
21
150
150
Page
TOTAL:
1. (10 pts) Caffeine (a stimulant found in coffee ☺) is 49.5% C, 5.15% H, 28.9 % N
and 16.5% O, by mass. Determine the empirical formula for caffeine.
In a 100-gram sample:
46.5 g C x 1 mol C
= 4.121 mol C ÷ 1.031 ≈ 4
12.011 g C
5.15 g H x 1 mol H
= 5.110 mol H ÷ 1.031 ≈ 5
1.0079 g H
28.9 g N x 1 mol N
= 2.063 mol N ÷ 1.031 ≈ 2
14.007 g N
16.5 g O x 1 mol O
= 1.031 mol O ÷ 1.031 ≈ 1
15.999 g O
Thus, the empirical formula is: C4H5N2O
2. (10 pts) The fizz produced when an Alka-Seltzer® tablet is dissolved in water is
due to the reaction between sodium bicarbonate (NaHCO3 – 84.0059 g/mol) and
citric acid (H3C6H5O7 – 192.1222 g/mol):
3 NaHCO3 (aq) + H3C6H5O7 (aq) → 3 CO2 (g) + 3 H2O (l) + Na3C6H5O7 (aq)
If 2.00 g of sodium bicarbonate and 3.00 g of citric acid are allowed to react,
which is the limiting reactant?
= 0.02381 mol NaHCO3
2.00 g NaHCO3 x
1 mol
84.0059 g NaHCO3
3.00 g H3C6H5O7 x
1 mol
= 0.015615 mol H3C6H5O7
192.1222 g H3C6H5O7
How many mol NaHCO3 are needed to react with all of the H3C6H5O7?
0.015615 mol H3C6H5O7 x
3 mol NaHCO3 = 0.04684 mol NaHCO3
1 mol H3C6H5O7
Since: mol NaHCO3 needed = 0.468 > 0.0238 mol NaHCO3 actual,
NaHCO3 is the limiting reagent
2
3. (5 pts each) Write balanced net ionic equations for the precipitation reactions
that may occur in each of the following cases. If no reaction occurs, then you
only need write NR.
a. Na2CO3 (aq) + MgSO4 (aq) → NaSO4(aq) + MgCO3(s)
Mg2+(aq) + CO32-(aq) → MgCO3(s)
b. NaOH (aq) + LiClO4 (aq) → no reaction
NR
c. AgNO3 (aq) + (NH4)2S (aq) → Ag2S(s) + 2NH3NO3(aq)
2Ag+(aq) + S2-(aq) → Ag2S(s)
4. (2 pts each) Determine the oxidation number for the indicated element in each
of the following compounds:
a. Sn in SnCl4:
+4
b. Cr in Cr2O72-:
+6
c. N in N2O:
+1
d. N in NO2:
+4
e. N in NO3-:
+5
5. (10 pts) Write a balanced molecular equation for the reaction that occurs when
solid calcium carbonate (the active ingredient in Tums™) reacts with an aqueous
solution of nitric acid (Hint: you saw this reaction as a demo in class!).
CaCO3(s) + 2HNO3(aq) → H2O(l) + CO2(g) + Ca(NO3)2(aq)
3
6. (3 pts each) For the following redox reactions, indicate which element is oxidized
and which element is reduced.
a. Ni(s) + Cl2(g) → NiCl2(s)
Oxidized:_______Ni______
Reduced:______Cl________
b. 3Fe(NO3)2 (aq) + 2 Al(s) → 3Fe(s) + 2 Al(NO3)3(aq)
Oxidized:_____Al______
Reduced:______Fe_______
c. 14KMnO4(s) + 4C3H5(OH)3(l) → 7K2CO3(s) + 7Mn2O3(s) + 5CO2(g) + 16 H2O(l)
Oxidized:_____C______
Reduced:_____Mn________
7. (5 pts) How many moles of HNO3 are present in 45.0 mL of a 1.68 M solution of
nitric acid?
45.0 mL x
1 L
1000 mL
x 1.68 mol HNO3 = 7.560 x 10-2 mol HNO3
L
= 7.56 x 10-2 mol HNO3
8. (6 pts) Calculate the molarity of a solution made by dissolving 0.2165 grams
Ca(OH)2 (MW = 74.0198 g/mol) in enough water to make exactly 250.00 mL of
solution.
0.2165 g Ca(OH)2 x
250.0 mL
mol
x 1000 mL = 1.169957 x 10-2 M
74.0198 g Ca(OH)2
L
= 1.170 x 10-2 M Ca(OH)2
4
9. (10 pts) The distinctive odor of vinegar is due to acetic acid, CH3COOH. As you
may recall from lab, acetic acid reacts with sodium hydroxide in the following
fashion:
CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)
If a 5.00-mL sample of vinegar requires 37.2 mL of 0.115 M NaOH to reach the
equivalence point in a titration, what is the molarity of acetic acid in the vinegar
sample?
37.2 mL NaOH sol’n x
L
x 0.115 mol NaOH x
1000 mL
L
1 mol HAc= 4.2780 x 10-3
1 mol NaOH
mol HAc
4.2780 x 10-3 mol HAc x 1000 mL = 0.8556 M HAc
5.00 mL Sol’n
L
= 0.856 M HAc
10. (10 pts) The Hindenburg was a famous hydrogen-filled dirigible that burned in
1937. If the Hindenburg held 2.0 x 105 m3 of hydrogen gas (H2) at 25. oC and
1.0 atm, what mass (kg) of hydrogen was present?
V = 2.0 x 105 m3 x 1000 L = 2.0 x 108 L
1 m3
n = PV =
(1.0 atm)(2.0 x 108 L)
= 8.174547 x 106 mol H2
RT (0.08206 L-atm/mol-K)(298.15 K)
8.1745 x 106 mol H2 x 2.01594 g H2 x
1 kg = 1.6479 x 104 kg
mol H2
1000 g
= 1.6 x 104 kg H2
5
11. (10 pts) Calculate the molar mass (g/mol) of a gas if 3.75 g occupies 0.935 L at
430. torr at 35.0 oC.
430 torr x
1 atm
760 torr
= 0.56579 atm
(0.56579 atm)(0.935 L)
= 2.09205 x 10-2 mol
n = PV =
RT (0.08206 L-atm/mol-K)(308.15 K)
Molar mass =
3.75 grams
= 179.25 g/mol = 179. g/mol
-2
2.09205 x 10 mol
12. (10 pts) Analysis of an air bubble trapped in amber formed 80 million years ago
revealed that the atmosphere before the dinosaurs became extinct was 30.0%
O2 and 70.0% N2 (by mass). Calculate the partial pressure (torr) of oxygen in
this prehistoric air sample if the total pressure of the air was 1.00 atm.
30.0 g O2 x
mol O2
= 0.937535 mol O2 = nO2
31.9988 g O2
70.0 g N2 x
mol N2
= 2.49880 mol N2 = nN2
28.0134 g N2
nT = nO2 + nN2 = 0.9375 + 2.49880 = 3.436335 mol
XO2 = nO2/nT = 0.937535/3.46335 = 0.27283
PO2 = XO2PT = (0.27283)(1.00 atm) = 0.27283 atm
In torr:
0.27283 atm x 760 torr
1 atm
= 207.35 torr
= 2.07 x 102 torr
6
13. (5 pts) A balloon made of rubber that is permeable to small molecules is filled
with helium to a pressure of 1 atm. This balloon is then placed in a box that
contains pure hydrogen (H2), that is maintained at a pressure of 1 atm. Will the
balloon expand or contract? Explain.
The balloon will EXPAND. The rate of effusion of the H2 is greater than
that of He, due its lower molar mass (recall: r1/r2 = [(M2/M1)]½), so H2 will
enter the balloon faster than He will leave the balloon and it will expand.
14. Vessel A contains CO gas at 20 oC and 1 atm. Vessel B contains CO2 gas at 20 oC
and 0.5 atm. The two vessels have the same volume.
a. (3 pts) Which vessel contains more molecules? Briefly explain.
Vessel A
Vessel B
Same
(circle one)
At constant Temp and Volume, mol ∝ Pressure; vessel A has a
higher pressure than vessel B, so it has more molecules of gas.
b. (3 pts) In which vessel is the average kinetic energy of the molecules
greater? Briefly explain.
Vessel A
Vessel B
Same
(circle one)
Temperature is a measure of the average K.E. of a gas – since
the two vessels are at the same temperature, the average K.E. of
the molecules in the two vessels must also be the same.
c. (3 pts) In which vessel is the rms speed of the molecules greater? Briefly
explain.
Vessel A
Vessel B
Same
(circle one)
If the K.E. of the molecules in the two vessels are the same,
then the rms speed of the molecules will be inversely proportional
to the square root of the MASS of the molecules. Since CO is
lighter than CO2, CO (in vessel A) will have a greater rms
velocity.
7
15. A balloon is heated by adding 400 J of heat. It expands, doing 422 J of work on
the atmosphere. For this process:
a. (4 pts) Calculate the change in internal energy (∆E) of the system.
∆E = q + w = +400 J + (-422 J)
= -22 J
b. (2 pts) This process is:
Endothermic
or
Exothermic
(circle one)
16. (5 pts) In the space below, write the name (first and last!) of the T.A. for your
lab section – you must spell their name correctly in order to receive credit for
your response!
Rachel Burdge
Mike Calichman
Gerald (Jerry) Love
Sean McCarthy
17. (10 pts) The specific heat of copper metal is 0.385 J/g-K. How many J of heat
are necessary to raise the temperature of a 2.26 kg block of copper from 25.0 oC
to 100.0 oC?
m = 2.26 kg x 1000 g = 2.26 x 103 g
kg
∆T = 100.0 – 25.0 oC = 75.0 oC = 75.0 K
cs = 0.385 J/g-K
q = mcs∆T = (2.26 x 103 kg)(0.385 J/g-K)(75.0 K)
= 6.525750 x 104 J
= 6.53 x 104 J
8