NAME:____________________________
Spring 2006
Student Number:______________________
Chemistry 1000 Practice Test #3
INSTRUCTIONS:
1) Please read over the test carefully before beginning. You should have
6 pages of questions, and a formula/periodic table sheet (7 pages total).
2) If your work is not legible, it will be given a mark of zero.
3) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
4) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
5) You may use a calculator.
6) You have 60 minutes to complete this test.
1.
Balance the following reaction equations:
(a)
____UO2 + __4__HF
[2 marks]
→ ____UF
(b) ____Na2SiO3 + __8__HF
2.
____/ 50 marks
→
4
+ __2__H2O
____H2SiF6 + __2__NaF + __3__H2O
Complete the following table.
[5 marks]
Molecular Formula
Name
Co(NO3)2·6H2O
cobalt(II) nitrate hexahydrate
(NH4)3PO3
ammonium phosphite
SnF4
tin(IV) fluoride
PbSO4
lead(II) sulphate
NaClO4
3.
or
lead(II) sulfate
sodium perchlorate
The structural formula of glycine is shown below. Name the hybrid orbital set used by
each ‘central’ atom.
[4 marks]
sp3
sp3
H
H
..O..
N
C
C
H
H
..
..
O
..
sp3
H
sp2
Suggested Approach:
1. Add lone pairs to convert structural formula to Lewis dot structure.
2. Determine shape family for each 'central' atom.
3. Choose hybrid orbital set based on shape family.
NAME:____________________________
4.
Student Number:______________________
The structural formula of formaldehyde (CH2O) is shown below.
..
.O.
[8 marks]
H
C
H
(a)
Name the hybrid orbital set used by C.
(3 atoms bonded + 0 lone pairs ∴ trigonal planar ∴ sp2)
sp2
(b)
Name the hybrid orbital set used by O.
(1 atom bonded + 2 lone pairs ∴ trigonal planar ∴ sp2)
sp2
(c)
Consider the C–O bond according to valence bond theory. Use the pictures below as
skeletons on which to draw the atomic orbitals involved in bonding and the bonding
molecular orbitals formed for the C–O bond only.
(Choose whichever view will give the clearest picture of the orbitals. Please use a
different picture for each molecular orbital – you may draw another copy of a view if
necessary.)
atomic orbital(s)
molecular orbital(s)
H
O
C
H
C
O
H
H
σ
sp2 + sp2
O
C
H
H
C
O
H
H
π
p + p
(d)
top view
Label each molecular orbital drawn in part (c) to indicate the type of bond.
see σ and π labels in part (c)
side view
NAME:____________________________
5.
The structural formula of the allyl cation is shown below.
Note:
(c)
H
C
C
C
Calculate the average C–C bond order.
H
Bond Order =
(b)
H
+
H
(a)
Student Number:______________________
[7 marks]
H
# bonds = 3 = 1.5
# C-C links
2
Draw a molecular orbital diagram for the π bonds only of the allyl cation.
Be sure to: (i) show the relative energies of the molecular orbitals,
(ii) draw a picture of each molecular orbital,
(iii) label each molecular orbital as bonding, nonbonding or antibonding,
(iv) include electrons on your diagram.
#eπ = #etotal – #eσ = # electrons in Lewis dot structure – 2 (# σ bonds)
=
[3(4) + 5(1) + (-1)]
–
2(7)
=
16
–
14
#eπ
= 2 π electrons to put on molecular orbital diagram
Based on your molecular orbital diagram, would adding two more π electrons change the
average C–C bond order? Why or why not?
No. The next two electrons go in πnonbonding, and electrons in a nonbonding
orbital don’t affect bond order.
NAME:____________________________
Student Number:______________________
6.
When burned without proper ventilation, propylene gas (C3H8) reacts with molecular
oxygen to produce carbon monoxide and water.
(a)
Write a balanced chemical equation for this reaction including states of matter. [3 marks]
(b)
If burning 4.67 g propylene produces 4.67 g of carbon monoxide, calculate the percent
yield of this reaction.
[5 marks]
(a)
2 C3H8(g) + 7 O2(g)
(b)
nC3H8 = mC3H8
MC3H8
=
(4.67 g)
.
(44.096 g/mol)
nC3H8 = 0.106 mol
→
6 CO(g) + 8 H2O(g)
nCO = nC3H8 × mole ratio
= (0.106 mol C3H8) × (6 mol CO) .
(2 mol C3H8)
nCO = 0.318 mol CO
mCO = nCO × MCO
= (0.318 mol)(28.010 g/mol)
mCO = 8.90 g
***This is the theoretical yield***
% yield =
actual yield × 100%
theoretical yield
= (4.67 g)(100%)
(8.90 g)
% yield = 52.5 %
***3 significant figures***
NAME:____________________________
Student Number:______________________
7.
In2S3(s) can be converted to elemental indium through a two-step process. In the first
step, the In2S3(s) is reacted with oxygen gas to give In2O3(s) and sulfur dioxide gas. In the
second step, the In2O3(s) is reacted with solid carbon to give In(s) and carbon dioxide gas.
(a)
Write a balanced reaction equation for each step.
(b)
Write a balanced reaction equation for the overall process.
(c)
Calculate the mass of indium produced if 75.00 kg In2S3 is reacted with 4.75 kg carbon
and 30.00 kg oxygen.
[7 marks]
(a)
step 1:
2 In2S3(s) + 9 O2(g)
step 2:
2 In2O3(s) +
(b)
overall:
2 In2S3(s) + 9 O2(g) + 3 C(s)
(c)
nIn2S3 = mIn2S3 = (75.00 kg)(1000 g/kg)
MIn2S3
(325.84 g/mol)
nIn2S3 = 230.2 mol
n*In2S3 = nIn2S3 = (230.2 mol)
2
2
n*In2S3 = 115.1 mol
nO2 = mO2 = (30.00 kg)(1000 g/kg)
MO2
(31.9988 g/mol)
nO2 = 937.5 mol
n*O2 = nO2 = (937.5 mol)
9
9
n*O2 = 104.2 mol
nC = mC = (4.75 kg)(1000 g/kg)
MC
(12.011 g/mol)
nC = 395 mol
n*C = nC = (395 mol)
3
3
*
n C = 132 mol
3 C(s)
→
→
[4 marks]
[1 mark]
2 In2O3(s) + 6 SO2(g)
4 In(s) + 3 CO2(g)
→ 4 In
(s)
+ 6 SO2(g) + 3 CO2(g)
lowest n*
therefore
limiting
reactant
nIn = nO2 × mole ratio
= (937.5 mol O2) × (4 mol In)
(9 mol O2)
nIn = 416.7 mol In
mIn = nIn × MIn
= (416.7 mol)(114.82 g/mol)
= 4.784 × 104 g × 1 kg .
1000 g
mIn = 47.84 kg
***4 significant figures***
NAME:____________________________
Student Number:______________________
8.
Analysis of an unknown molecule shows that it contains 28.79% Ti and 71.21% Se.
(a)
Find the empirical formula of the unknown molecule.
(b)
If the molecular formula is the same as the empirical formula, name the unknown
molecule.
[1 mark]
(a)
100 g of unknown molecule contains 28.79 g Ti and 71.21 g Se.
nTi = mTi
MTi
nSe = mSe
MSe
=
=
nTi
(b)
(28.79 g) .
(47.88 g/mol)
= 0.6013 mol
nSe
(71.21 g) .
(78.96 g/mol)
= 0.9018 mol
Therefore,
nTi : nSe = 0.6013 mol Ti : 0.9018 mol Se
0.6013
0.6013
= 1.000 mol Ti : 1.500 mol Se
nTi : nSe = 2.000 mol Ti : 3.000 mol Se
Therefore,
the empirical formula is Ti2Se3
Ti2Se3 = titanium(III) selenide
[3 marks]
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit
1.6605 × 10-24 g
Avogadro's number
6.02214 × 1023 mol–1
Bohr radius
5.29177 × 10-11 m
Coulomb constant
8.998 × 109 N·m2·C-2
Electron charge (e)
1.6022 × 10-19 C
Electron mass
5.4688 × 10-4 µ
6.626 × 10-34 J·s
1.0072765 µ
1.0086649 µ
1.097 x 107 m-1
2.179 x 10-18 J
2.9979 x 108 m·s-1
Planck's constant
Proton mass
Neutron mass
Rydberg Constant (R)
Rydberg unit (Ry)
Speed of light in vacuum
Formulae
v = υλ
(often, c = υ λ )
d =
m
V
E = hυ
U =
m
n
2
rn = a 0 n
Z
F = k
λ = h
ρ
∆x . ∆ρ >
En = -1 Ry
1
(n+e)(n-e)
d2
ρ = mv
Z2
n2
E = k
(n+e)(n-e)
d
= R
1 - 1
n12
n22
1
h
4π
λ
1 - 1
n 22
n12
∆E = En2 - En1 = Ry . Z 2
Chem 1000 Standard Periodic Table
18
4.0026
1.0079
H
He
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
2
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
107
Np
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré