NAME:____________________________
Spring 2006
INSTRUCTIONS:
1.
Student Number:______________________
Chemistry 1000 Test #3
____/ 50 marks
1) Please read over the test carefully before beginning. You should have
7 pages of questions, and a formula/periodic table sheet (8 pages total).
The question on page 7 is a bonus question.
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 50 minutes to complete this test.
Complete the following table.
[5 marks]
Molecular Formula
Name
(NH4)2SO4
ammonium sulfate
CoF2·4H2O
cobalt(II) fluoride tetrahydrate
Ca(OCl)2
calcium hypochlorite
P2O5
diphosphorus pentoxide (or diphosphorus pentaoxide)
NiS
nickel(II) sulfide
2.
Balance each of the following reaction equations:
(a)
____La2O3 + __3__H2O
→
__2__La(OH)3
(b)
____PCl5
→
____H3PO4 + __5__HCl
(c)
__4__CH3NH2 + __9__O2
+
__4__H2O
→ __4__CO
1
[3 marks]
2
+ __10__H2O + __2__N2
NAME:____________________________
3.
Student Number:______________________
DNA consists of three types of molecules connected together. The “coding” molecules
are called nitrogenous bases (because they are bases that contain nitrogen). The molecule
shown below is adenine, one of the four nitrogenous bases in DNA.
[5 marks]
sp3
H
..
H
N
sp2
..
C
N
C
C
C
N
C
H
N
H
N
H
sp2
(a)
Name the hybrid orbital set used by each of the three atoms in bold.
(b)
How many σ bonds are there in one molecule of adenine? 16 (16 bonds of any type)
(c)
How many π bonds are there in one molecule of adenine? 4 (4 double bonds; 0 triple
bonds)
4.
Use diagram(s) to explain how sp orbitals are formed. Clearly indicate the number, type
and geometry of all orbitals involved.
[4 marks]
An s orbital and a p orbital in the same shell of the same atom combine to make two sp
orbitals:
2
NAME:____________________________
5.
Student Number:______________________
Consider the cyanide anion (CN-) according to valence bond theory. Draw both the
atomic orbitals involved in bonding and the bonding molecular orbitals formed. Label
each atomic orbital and each molecular orbital.
[10 marks]
(Please use a different picture for each molecular orbital. Clearly indicate if any of the
pictures are looking at the molecule from a different angle than the others.)
atomic orbital(s)
molecular orbital(s)
3
NAME:____________________________
6.
(a)
Student Number:______________________
An unknown solid is found to contain 52.26% Fe, 44.91% O and 2.83% H by mass.
[6 marks]
Calculate the empirical formula for this unknown solid.
(b)
Assuming that the empirical and molecular formulae are the same, name the unknown
solid.
(a)
100 g of unknown molecule contains 52.26 g Fe, 44.91 g O and 2.83 g H.
nFe = mFe
MFe
=
nO = mO
MO
(52.26 g) .
(55.845 g/mol)
nFe = 0.9358 mol
(b)
=
nH = mH
MH
(44.91 g)
.
(15.9994 g/mol)
nO = 2.807 mol
(2.83 g) .
1.00079 g/mol
nH = 2.81 mol
Therefore,
nFe : nO : nH = 0.9358 mol Fe : 2.807 mol O : 2.81 mol H
0.9358
0.9358
0.9358
nFe : nO : nH = 1.000 mol Fe : 3.000 mol O : 3.00 mol H
Therefore,
the empirical formula is FeO3H3
FeO3H3 = Fe(OH)3 = iron(III) hydroxide
4
NAME:____________________________
7.
Student Number:______________________
An impure sample (1.25 g) of sodium chloride was dissolved in 25.00 mL of water.
Silver(I) nitrate solution was then added until no more precipitate formed. The
precipitate was found to be 1.25 g of pure silver(I) chloride. Calculate the purity (in
percent-by-mass) of the initial sodium chloride sample.
[6 marks]
NaCl(aq) + AgNO3(aq)
→
AgCl(s) + NaNO3(aq)
Step 1: Calculate the number of moles of AgCl produced.
nAgCl =
1.25 g AgCl = 0.00872 mol AgCl
143.3207 g/mol
Step 2: Calculate the number of moles of NaCl consumed.
nNaCl = 0.00872 mol AgCl × 1 mol NaCl = 0.00872 mol NaCl
1 mol AgCl
Step 3: Calculate the mass of pure NaCl (i.e. the mass of NaCl consumed) in the initial sample.
mNaCl = 0.00872 mol NaCl × 58.4425 g/mol = 0.510 g NaCl
Step 4: Calculate the purity of the initial (impure) sample of NaCl.
% mass = mNaCl × 100% = 0.510 g × 100% = 40.8 %
mimpure
1.25 g
5
NAME:____________________________
Student Number:______________________
8.
Copper metal (Cu) reacts with concentrated nitric acid (15 M HNO3) to produce
copper(II) nitrate, water and nitrogen dioxide gas.
[11 marks]
(a)
Write a balanced chemical equation for this reaction.
(b)
What mass of nitrogen dioxide will be produced when 6.3546 g of copper is reacted with
25.00 mL of concentrated nitric acid?
(a)
Cu + 4 HNO3
(b)
*Only round when you get to the final answer. I’m showing rounded values to let you
know where the sig. fig. come from, but I’m carrying through all digits in my calculator.*
→
Cu(NO3)2 + 2 H2O + 2 NO2
Step 1: Calculate the moles of each reactant and identify the limiting reagent.
nCu = 6.3546 g Cu = 0.10000 mol Cu
63.546 g/mol
nHNO3 = 15 mol/L × 0.02500 L = 0.38 mol HNO3
n*Cu = 0.10000 mol Cu = 0.10000 mol Cu
1
n*HNO3 = 0.38 mol HNO3 = 0.094 mol HNO3
4
∴ HNO3 is the limiting reagent
Step 2: Calculate the moles of NO2 produced.
nNO2 = 0.38 mol HNO3 × 2 mol NO2 = 0.19 mol NO2
4 mol HNO3
Step 3: Calculate the mass of NO2 produced.
mNO2 = 0.19 mol NO2 × 46.0055 g/mol = 8.6 g NO2
6
NAME:____________________________
Student Number:______________________
BONUS QUESTION
(+1 to your final “number grade” in the course; equivalent to 5 marks on this test)
The central C–C bond in butane (CH3CH2CH2CH3) is longer than the central C–C bond in
butadiene (CH2CHCHCH2). Use molecular orbital theory to explain why this is the case.
7
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
107
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)
Np
93
Pu
94
Am
Cm
95
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré
8