HOUR EXAM I RK
CHEM 103 OCTOBER 2003
Page 1 of 13
NAME:______________________________
EXAM IRK
OCTOBER 16 2003
CH 103
DIRECTIONS
• There is only one correct answer to each question unless otherwise noted. Any questions for which more than one response
has been selected will not be counted
• Your score is based solely on the number of questions you answer correctly. It is to your advantage to answer every
question.
• The best strategy is to arrive at your own answer to a question before looking at the choices. Otherwise, you may be misled
by plausible, but incorrect, responses.

IF YOU SELECT “NONE OF THE ABOVE” as an answer, write your numerical answer, including units,
next to it.
 YOUR SCORE FOR THE 1st HOUR EXAM WILL BE EITHER THE AVERAGE OF YOUR SCORE ON EXAM I &
EXAM IR, OR YOUR SCORE ON EXAM I, WHICHEVER IS THE HIGHER.
IF YOU FIND YOURSELF BAULKED BECAUSE THERE IS A PIECE OF INFORMATION YOU DON’T HAVE,
RAISE YOUR HAND. YOU MAY BE ABLE TO BUY THE INFORMATION FOR A POINT.
TAKE A DEEP BREATH AND LET YOUR EYES FOCUS.
RELAX YOUR SHOULDERS
WHEN IN DOUBT CONVERT TO MOLES!
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CHEM 103 OCTOBER 2003
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INORGANIC NOMENCLATURE:
1. Give formulae for the following elements:
a. N2O ______________Dinitrogen monoxide or nitrous oxide____________________
b. CO
____carbon monoxide ___________________________
c. Co
____Cobalt _________________________
d. RbF ____ Rubidium fluoride_____________________________
e.
CaCO3 _____Calcium carbonate ___________________________
5 pts: ____
2. Name, by IUPAC rules, the following compounds:
a. Triphosphorus hexaoxide ______________ P3O6___________________
b. Aluminum sulfide: ___ Al2S3_____________________
c. Calcium phosphide: ______ Ca3P2_________________
d. Lithium hydride: _____ LiH_______________
e. Barium phosphate:
______ Ba3(PO4)2______________
5 pts: ____
BONUS: Give the formula of the Mercury(I) ion Hg22+
10
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CHEM 103 OCTOBER 2003
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3. Which of the following will have the LOWEST % by mass sodium?
A Na2Te
2 mole Na (22.99 g/mol Na) x 100 = 74.2 % Na in Na2O
B. Na2Se
((22.9 x 2) +16) g/mol
C. Na2S
D. Na2O
2 mole Na (22.99 g/mol Na) x 100 = 26.5 % Na in Na2Te
E. All have equal % by mass Na.
((22.9 x 2) +127.6) g/mol
4. Which of the following compounds are held together ENTIRELY by COVALENT bonds?
A.
B
C.
D.
E.
Cobalt(III) iodide, CoI3
Potassium iodate, KIO3
Sodium iodide, NaI
Thyroxine, C15H11I4NO4
None of the above
5. 1 erg = 1 g cm2/sec2. How many joules is this?
A.
B.
C.
D.
E.
1 x 107 joules/erg
1 x 105 joules/erg
1 x 10-7 joules/erg
1 x 10-5 joules/erg
None of the above
1 J = 1 kg m2/sec2
1 g cm2/sec2 [1 kg/1000 g][1 m/100 cm]2 = C
6. A single 11H + ion speeding about in high vacuum has a kinetic energy of 6.022 x 10-30 J.
How fast is it going? (there is information that you will need to calculate mass on page 11)
A
B
C
D
E
0.00719 m/sec
0.0848 m/sec
11.8 m/sec
139 m/sec
None of the Above
A single H+ in vacuo is a bare proton. Mass = 1.673 x 10-27 kg
KE = ½ mv2
6.022 x 10-30 kg m2/sec2 = ½ (1.673 x 10-27 kg) v2
v2 = 2 x 6.022 x 10-30 kg m2/sec2/1.673 x 10-27 kg = 7.199 x 10-3 m2/sec2
v v=B
Bonus: Is an erg a measure of force or energy? Energy!
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20/30
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HOUR EXAM I RK
CHEM 103 OCTOBER 2003
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NAME:______________________________
7. The correct Lewis structure for the Francium ion is:
..
A. . Fr -
:+
B. :Fr
..
C.
Fr+
8. The number of electrons in the
(A) 99
9.
(B) 102
(C) 249
..
D. :Fr : +
E. None of the above
..
252
102
No3+ ion is
(D) 252 (E) 351
(F) 354 (G) None of the Above
Magnesium Aspirin has the chemical composition:
Chemical composition of Magisal (Magnesium aspirin)
C
H
O
Mg
12.011
1.008
15.999
24.31
g/mol
g/mol
g/mol
g/mol
56.50
3.69
33.45
6.36
%
%
%
%
The empirical formula of Magnesium aspirin is
A
C128H1O16Mg24
B
C56H4O33Mg6
C
C15H1O9Mg2
D
E
C18H14O8Mg
None of the Above
56.5
3.69
33.45
6.36
g C (1 mol C/12.011g) = 4.704 mol C/.2616 = 18 C
g H (1 mol H/1.008 g) = 3.661 mol H/.2616 = 14 H
g O (1 mol O/15.999 g ) = 2.091 mol O/.26216 = 8 O
g Mg ( 1 mol Mg/24.31 g) = 0.2616 mol Mg/.2616 = 1 Mg
10. A large iron containing protein has a molecular weight of 64500 g/mol.
The protein is 0.346% iron. That is, here are 0.346 g Fe/100.0 g protein. Calculate how many
moles of Fe there are per mole of protein. CONVERT TO MOLES!
AW Fe = 55.85 g/
A.
B.
C.
D.
E.
.
0.346 g Fe (1 mole Fe/55.85 g)
400 moles Fe/mole protein 100.0 g protein(1 mole protein/64500 g)
223 moles Fe/mole protein
4 moles Fe/mole protein
1 mole Fe/mole protein
None of the above
= 0.006195 mol Fe
0.00155 mol protein = 4 Fe/1 protein
BONUS: How many valence electrons do the HALOGENS have? 7
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CHEM 103 OCTOBER 2003
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CH 3
H
O
O
O
CH
C
N
CH 2
C
H2
H2
C
C
CH
CH
H
C
NH
O
C
O
NH 2
H
H
O
H
11. Wildire toxin is produced by pseudomonas tobaci, the bacterium responsible for wildfire
disease in tobacco. It has the chemical formula C11H19N3O6.
What is the % by mass nitrogen in wildfire toxin?
C
H
N
O
MW Wildfire toxin
A
B
C
D
E
12.011
1.008
14.01
15.999
g/mol
g/mol
g/mol
g/mol
289.29
g/mol
4.843 %
7.692 %
14.53 %
35.92 %
None of the Above
3 mols N( 14.01 g N/mol N) X 100
289.29 g/mol toxin
= C
12. Which of the following are ISOELECTRONIC with neon (have the same number of
electrons)?
A.
B.
C.
D.
E.
Mg2+
O2Na+
All of the above
None of the above
BONUS: Is cesium iodide held together by covalent or ionic bonds?
IONIC
10/60
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CHEM 103 OCTOBER 2003
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13. Lorazepam is a mild tranquilizer. How many molecules of Lorazepam are there in
10.00 g of the tranquilizer?
MW Lorazepam = 321.16 g/mol
O
HN
H
C
C
H
C
10.00 g Lz (1 mol/321.16 g)(6.022 x 1023 molecules/mol) = C
C
HC
OH
C
C
C
H
Cl
N
C
Cl
HC
C
HC
CH
Lorazepam
C
H
5.171x 10-26
3.114 x 10-02
1.875 x 10+22
1.934 x 10+26
None of the above
A
B
C
D
E
Molecules
Molecules
Molecules
Molecules
O
14. Consider the reaction by which you made oil of wintergreen:
SalH + MeOH  SalOMe + H2O
C
C
C
C
C
C
H
O
C
+
CH3-OH
O
H
If you start with 4.0020g of salicyclic acid (SalH), how many grams
and mL of methanol (MeOH) will you have to use to react with it
completely?
O
C
C
C
C
CH 3
O
+
C
MW (g/mol)
C
C
density (g/mL)
H2O
O
H
salicylic acid (SalH)
138.12
methanol (MeOH)
32.04
methyl salicylate (SalOMe)
152.14
Water
18.0
A
B
C
D
0.9284
0.9284
17.2521
17.2521
E
None of the
above
g methanol, which is
g methanol, which is
g methanol, which is
g methanol, which is
1.180
0.7302
21.93
13.57
0.7866
1.0004.0020
mL
mL
mL
mL
g SalH(1 mol SalH/138.12 g) = 0.02897 mol SalH
0.02897 mol SalH( 1 MeOH/1 SalH) = 0.02897 mol
MeOH needed
0.02897 mol MeOH(32.04 g/mol MeOH) = 0.9284 g
MeOH
0.9284 g MeOH (1 mL/0.7866 g) = 1.18 mL MeOH
H3C
BONUS: What is the bond angle around the C-O-H bond in methanol?
The Lewis structure shown is not warranted to be geometrically correct. 109o
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..
O
..
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H
10/70
HOUR EXAM I RK
CHEM 103 OCTOBER 2003
Page 7 of 13
NAME:______________________________
15. Elements in Group IA of the periodic table:
A. Are metals
B. Are diatomic
C. Have filled valence shells;
D. Tend to gain a single valence electron;
E. Tend to lose two valence electrons;
For Q 16-17, consider the reaction:
3SrF2 + 2Rb3PO4  Sr3(PO4)2 + 6RbF
MW (g/mole)
SrF2
125.6
Rb3PO4
351.4
Sr3(PO4)2
452.9
RbF
106.5
Suppose you start with 500.0 mmoles (millimoles) of strontium fluoride and 400.0 mmoles
of rubidium phosphate.
16. Which is the limiting reagent?
A. SrF2
B. Rb3PO4
C. Sr3(PO4)2
500 mmol SrF2(10-3 mol/mmol) = 0.500 mol SrF2 (2 Rb3PO4/3SrF2) = 0.333 mole Rb3PO4
needed
400 mmol Rb3PO4(10-3 mol/mmol) = 0.400 mol Rb3PO4 Only 0.333 mol needed XS SrF2 is
LR
D. RbF
17. How many grams of strontium phosphate will you make?
A
B
C
D
75.48
90.58
113.2
407610
g
0.500 mol SrF2 (1 Sr3(PO4)2/3 SrF2)(452.9 g/mol Sr3(PO4)2 = A
g
g
g
E None of the above
BONUS: What is the charge on the phosphate ion? -3 PO43Page 7 of 13
15/85
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CHEM 103 OCTOBER 2003
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18. Consider the reaction:
Pb(NO3)2(aq) + 2K2CrO4(aq)  2KNO3(aq) + PbCrO4(s) (Chrome yellow, King’s yellow)
If you start with 1.0000 g of Pb(NO3)2 and 1.0000 g of K2CrO4, what is the theoretical yield of
Chrome Yellow (PbCrO4)?
MW (g/mole)
Pb(NO3)2
351.23
K2CrO4
194.2
PbCrO4
323.22
KNO3
V
A
0.9203
A
g
B
1.664
g
C
3.329
g
D
E
0.8322
g
None of the above
101.1 to moles
Convert
1.0000 g Pb(NO3)2 ( 1 mol/351.23 g) = 0.002847 mol Pb(NO3)2
1.0000 g of K2CrO4 (1 mol/194.2 g) = 0.005149 mol K2CrO4
Determine LR
0.002847 mol Pb(NO3)2(1 K2CrO4/ 2 mol Pb(NO3)2) = 0.001424 mol K2CrO4
needed
0.005149 mol K2CrO4 is more than enough, Pb(NO3)2 is LR
work with LR
0.002847 mol Pb(NO3)2(1 PbCrO4/1 Pb(NO3)2(323.22 g/mol PbCrO4) = A
19. Consider the reaction:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)
If you make 10.00 kg of CO2, how many grams of ethane (C2H6) did you start with? Assume
oxygen is in excess.
MW (g/mole)
C2H6
30.0
O2
32.0
CO2
44.0
H2O
18.0
A
B
C
D
E
10.00 kg CO2(1000 g/kg)(1 mole CO2/44.0 g) = 2.27 x 102 mol CO2
2.27 x 102 mol CO2(2 C2H6 /4 CO2)(30.0 g/mol C2H6) = B
3.41 x 10-03
g
3410
g
6820
g
13640
g
None of the above
BONUS: How much does 1 molecule of CO2 weigh? 44 g/molCO2(1 mol/6.022 x 1023
molecules) = 7.31 x 10-23 g
10/95
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CHEM 103 OCTOBER 2003
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20. The Lewis structure of dicysteine may be represented as:
+
:O:
H3N
CH
O
..
HN
H2C
C
.. :
O
..
H2C
SH
C
CH
SH
A
+
O-
H3N
CH
HS
C
CH 2
+
H2C
HN
CH
SH
..
O
..
C
B
..
:O
..
3
Which of the following is FALSE?
A. Structure A and Structure B are 2 resonance structures of the SAME molecule.
B. A better picture of reality is that the C-O bond and C-N bond are intermediate in length
and bond energy between single & double bonds.
C. A and B flip back and forth.
D. The circled O has 3 lone pairs of electrons.
21. Draw the Lewis structure of PO43- and indicate shape and bond angles.
PO43- has 5 + 4(6) + 3 =32 valence electrons
3O
P
O
O
O
H
C
C
BONUS: Draw another resonamce structure for
C
H
H
H
H
H
H
C
C
C
C
C
C
C
H
Page 9 of 13
H
H
C
C
H
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CHEM 103 OCTOBER 2003
Page 10 of 13
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CHEM 103 OCTOBER 2003
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22. If a nmole (nanomole) of K+ ions and a μmole (micromole) of F- ions are 1.0000
km away from each other, the force of attraction will be:
1 K+ ion has a charge of + 1.602 x 10-19 C.
1 F- ion has a charge of – 1.602 x 10-19 C
A
B
C
D
E
8.37 x 10-02
N
-34
2.31 x 10
N
8.37 x 10+10
N
8990
N
None of the above
F Fes = -8.99 x 109 Nm2/C2 q1 q2 /r2
r = 1.0000 km (1000 m/km) = 1000 m;
q1 = +1.602 x10-19 C/K+(1 nmole K+)(10-9 mol/nmol)(6.022 x 1023 K+/mol) = 9.647 x 10-5 C
q2 = -1.602 x10-19 C/F-(1 umole F-)(10-6 mol/nmol)(6.022 x 1023 F-/mol) = -9.647 x 10-2 C
Fes = A
Klaque’s Bonus: The smallest diamond has 8 C atoms (AW C = 12.011 g/mol) in a volume of
45.3 x 10-24 cm3. Calculate the density of diamond.
8 C Atoms(1 mol C/6.022 x 1023 atoms)(12.011 g/mol C) = 1.596 x 10-22 g/45.3 x 10-24 cm3
= 3.52 g/cm3
END OF EXAM!
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CHEM 103 OCTOBER 2003
Page 12 of 13
SPECIAL UNITS & CONVERSION FACTORS
Acceleration due to gravity (g)
Avogadro’s number
Bohr radius
Boltzmann’s constant
Electronic charge-to-mass ratio
Elementary charge
Faraday constant
Gravitational constant (Universal)
Mass of an electron
Mass of a neutron
Mass of a proton
Molar volume of an ideal gas @ STP
Permittivity of vacuum
Planck’s constant
Rydberg constant
Speed of light in vacuo
Universal gas constant
g
No
ao
kB
-e/me
e
F
G
me
Mn
Mp
Vm
εo
h
R∞
c
R
9.80 m/sec2
6.022 x 1023 particles/mol
0.529 x 10-10 m
1.381 x 10-23 J/K
1.759 x 1011 C/kg
1.602 x 10-19 C
9.649 x 104 C/mol
6.673 x 10-11 m3/kg sec2
9.109 x 10-31 kg
1.675 x 10-27 kg
1.673 x 10-27 kg
22.41 L/mol
8.85 x 10-12 C2/N m2
6.626 x 10-34 J sec
2.18 x 10-18 J
2.998 x 108 m/sec
8.314J/mol K
0.08205 L atm/K mol
Fgrav = -Gm1m2/r2
Felectrostatic = -[1/4πεo] q1 q2/r2
Arrhenius Equation
=
-(8.99 x 109 m2/C2 N) q1 q2/r2
k = Ae-Eact/RT
1 N = 1 kgm/sec2
1 J = 1 Nm = 1 kg m2/sec2
Formal Charge = # valence e- -lone pair e- -1/2 bonding e-
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CHEM 103 OCTOBER 2003
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PERIODIC TABLE OF THE ELEMENTS
1A
8A
1
H
1.008
2A
3A
4A
5A
6A
7A
2
He
4.003
3
Li
6.941
4
Be
9.012
5
B
10.81
6
C
12.01
7
N
14.01
8
O
16.00
9
F
19.00
10
Ne
20.18
11
Na
22.99
12
Mg
24.31
3B
4B
5B
6B
7B
14
Si
28.09
15
P
30.97
16
S
32.07
17
Cl
35.45
18
Ar
39.95
19
K
39.10
20
Ca
40.08
21
Sc
44.96
22
Ti
47.88
23
V
50.94
24
Cr
52.00
25
Mn
54.94
26
Fe
55.85
27
Co
58.93
37
Rb
85.47
38
Sr
87.62
39
Y
88.91
40
Zr
91.22
41
Nb
92.91
42
Mo
95.94
43
Tc
(98)
44
Ru
101.1
55
Cs
132.9
56
Ba
137.3
57
La
138.9
72
Hf
178.5
73
Ta
181.0
74
W
183.8
75
Re
186.2
87
Fr
(223)
88
Ra
226.0
89
Ac
227.0
104
Unq
(261)
105
Unp
(262)
106
Unh
(263)
107
Uns
(262)
58
Ce
140.
1
90
Th
232.
0
59
Pr
140.
9
91
Pa
231.
0
60
61
62
63
64
65
66
67
68
69
70
71
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
144. (145) 150. 152. 157. 158. 162. 164. 167. 168. 173. 175.
2
4
0
3
9
5
9
3
9
0
0
92
93
94
95
96
97
98
99
100
101
102
103
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
238. 237. (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
0
0
Page 13 of 13
1B
2B
13
Al
26.98
28
Ni
58.69
29
Cu
63.55
30
Zn
65.39
31
Ga
69.72
32
Ge
72.61
33
As
74.92
34
Se
78.96
35
Br
79.90
36
Kr
83.80
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222)
108
Uno
(265)
109
Une
(266)
8B
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