1
Chemistry 1000 Practice Final Exam
Spring 2006
1)
2)
3)
4)
5)
6)
7)
INSTRUCTIONS
Read the exam carefully before beginning. There are 19 questions on pages 2 to 14 followed
by 2 pages of “Data Sheet” (including periodic table) and a blank page for any rough work.
Please ensure that you have a complete exam. If not, let Susan know immediately. All
pages must be submitted at the end of the exam.
If your work is not legible, it will be given a mark of zero.
You may use a calculator.
Show your work for all calculations. Answers without supporting calculations will not be
given full credit.
Marks will be deducted for improper use of significant figures and for answers with
incorrect/missing units.
Do not open the exam until you are told to begin. Beginning prematurely will result in
removal of your exam paper and a mark of 0.
You have 3 hours to complete this exam. Nobody may leave the exam room during the first
hour or the last 15 minutes of the exam.
Q
Mark
Q
Mark
1
/ 14
11
/7
2
/5
12
/4
3
/5
13
/4
4
/9
14
/5
5
/4
15
/8
6
/ 18
16
/8
7
/8
17
/7
8
/8
18
/8
9
/4
19
/6
10
/8
Q
Total
Mark
/ 140
2
1.
Answer the following questions in the space provided.
(a)
In one atom, what is the maximum number of orbitals with l = 2?
5
(b)
In one atom, what is the maximum number of electrons with n = 4?
32
(c)
How many nodes does a 5d orbital have?
# nodal surfaces =
2
# radial nodes =
[14 marks]
2
(d)
Which element has a higher ionization energy, oxygen or sulphur?
oxygen
(e)
Which of the following molecules is least stable: H2, He2 or Li2?
He2
(f)
Give the symbol for a halogen that is a liquid at room temperature.
Br
(g)
Give the symbol for a chalcogen that is also a metalloid.
Te
(h)
Give the symbol for the most stable ion that can be formed from beryllium.
Be2+
(i)
Would you expect the chemistry of In to be most similar to that of I, Sn, Al, or Ir?
Al
(j)
What hybrid orbital set is used by the central atom of a trigonal pyramidal molecule? sp3
(k)
Which has a larger radius, Ca or Ca2+?
Ca
(l)
Which has a higher electron affinity, phosphorus or silicon?
phosphorus
(m)
Name the neutral element with the electron configuration 1s22s22p63s1.
sodium
2.
Is the molecule arsenic trichloride polar or nonpolar? Provide evidence by showing the
structure, dipoles and the approximate direction of the net dipole.
[5 marks]
..
trigonal pyramidal
As Cl
Cl
Cl
net dipole
polar molecule
3
3.
Complete the following table.
Name of Element
boron
[5 marks]
Number of
Protons
Number of
Neutrons
Number of
Electrons
B
5
5
5
27
32
27
34
43
36
Symbol
10
5
cobalt
59
27
Co
selenium
77
34
Se
2-
4.
Complete the following table using shorthand notation for the electron configurations.
(i.e. Do not use noble gas notation for core electrons.)
[9 marks]
(Valence electrons are underlined.)
Number of
Symbol
Electron Configuration
Valence Electrons
Al
1s22s22p63s23p1
3
Br -
1s22s22p63s23p64s23d104p6
8
Fe3+
1s22s22p63s23p63d5
13
5.
Complete the following table.
[4 marks]
Formula
Name
CrPO3 · 3H2O
chromium(III) phosphite trihydrate
N2O5
dinitrogen pentoxide
VC
vanadium(IV) carbide
Fe(ClO4)3
iron(III) perchlorate
4
6.
Complete the following table. For charged species, place the charge(s) on the
appropriate atom(s) in the Lewis dot structure. Use the symbol ~ to indicate any angles
that are approximate (as opposed to exact).
[18 marks]
Formula
Lewis Dot Structure
Name of
Molecular Shape
Predicted
Bond Angle(s)
trigonal planar
120˚
square pyramidal
~90˚
linear
180˚
tetrahedral
~109.5˚
..O..
NO3-
ICl5
-
N3
..
.. O
..
-1
.. -1
..O ..
N
+1
.. .
.. ..
Cl
Cl
.. .. .. .
.. I .. .
..Cl
.. .
.. .. .. Cl
Cl
..
..
..N
-1
+1
. .-1
N..
N
H
CHCl3
..
.. Cl
..
..
C
.. Cl ..
..
Cl ..
..
5
7.
(a)
Balance the following chemical reaction equations, and indicate whether or not each is a
redox reaction.
[8 marks]
Sb4O6 +
6C
→
4 Sb + 6 CO
redox
(b)
→
3 Cu(NO3)2 + 2 K3PO4
Cu3(PO4)2 + 6 KNO3
not redox
(c)
2 HNO3 +
→
Mg(OH)2
Mg(NO3)2 +
2 H2O
not redox
(d)
4 NH3 + 5 O2
→
4 NO + 6 H2O
redox
8.
[8 marks]
+2 +6 -2
+1 -1
+1 -1
+1 +6 -2
0
2 CuSO4 + 4 KI → 2 CuI + 2 K2SO4 + I2
(a)
On the reaction equation above, assign oxidation states to each element.
(b)
Identify which atom(s) are oxidized and which are reduced.
oxidized:
(c)
I
reduced:
Cu
reducing agent:
KI
Identify the oxidizing and reducing agent(s).
oxidizing agent:
CuSO4
6
9.
Lithium metal reacts with molecular nitrogen to give lithium nitride. Write the balanced
chemical equation for this reaction. Assign oxidation states to each element, and identify
which atom(s) are oxidized and which are reduced.
[4 marks]
0
0
+1 -3
6 Li + N2 → 2 Li3N
oxidized:
10.
Li
reduced:
N
The structural formula for hydrogen cyanide (HCN) is shown below.
H
C
[8 marks]
N ..
(a)
How many σ bonds are there in one molecule of HCN?
2
(b)
How many π bonds are there in one molecule of HCN?
2
(c)
What set of hybrid atomic orbitals is used by C in HCN?
sp
(d)
What set of hybrid atomic orbitals is used by N in HCN?
sp
(e)
Draw diagram(s) that clearly show(s) the π bonding in HCN.
Diagram(s)
should
clearly show two π
bonds at 90˚ to each
other. It’s up to you
whether you can show
this most clearly on
one diagram or using
one diagram per π
bond.
7
11.
The structural formula of the allyl cation is shown below.
Note:
(c)
C
C
C
Calculate the average C–C bond order.
H
Bond Order =
(b)
H
+
H
(a)
H
[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.
8
12.
How many atoms of oxygen are there in a 1.135 kg box of sodium bicarbonate?
a.
sodium bicarbonate = NaHCO3
b.
nNaHCO3 = mNaHCO3
MNaHCO3
[4 marks]
nNaHCO3
c.
= 1.135 kg NaHCO3 × 1000 g × 1 mol NaHCO3 .
1 kg
84.007 g NaHCO3
= 13.51 mol NaHCO3
nO = nNaHCO3 × mole ratio
= 13.51 mol NaHCO3 ×
nO = 40.53 mol O
d.
(84.007 g/mol)
3 mol O
1 mol NaHCO3
.
atoms O = nO × NA
= 40.53 mol O × 6.022 × 1023 atoms O
1 mol O
atoms O = 2.441 × 1025 atoms O
13.
***4 sig. fig.***
There are three stable isotopes of argon. Their atomic masses and abundances are listed
below. Calculate the average atomic mass of argon.
[4 marks]
Isotope
Natural Abundance (%)
Isotopic Mass (µ)
36
0.337
35.967546
38
0.063
37.962732
40
99.60
39.962383
Ar
Ar
Ar
MAr
= 0.337 × MAr36
100
+
0.063 × MAr38
100
+
99.60 × MAr40
100
0.337 × 35.967546 µ + 0.063 × 37.962732 µ + 99.60 × 39.962383 µ
100
100
100
=
MAr
0.121 µ
+
0.024 µ
+
= 39.95 µ
***2 decimal places therefore 4 sig. fig.**
39.80 µ
9
14.
Club soda is an aqueous solution of carbon dioxide. A sample of club soda is titrated
with 0.04202 M NaOH(aq) according to the reaction equation below:
[5 marks]
CO2(aq) + 2 NaOH(aq)
→
Na2CO3(aq)
44.010 g/mol
If it takes 32.14 mL of 0.04202 M NaOH(aq) to react with a 25.00 mL sample of club
soda, what is the concentration of CO2 in club soda (in g/L)?
a.
nNaOH = MNaOH × VNaOH
= 0.04202 mol/L × 32.14 mL ×
nNaOH = 1.351 × 10-3 mol
b.
nCO2
nCO2
c.
1L .
1000 mL
= nNaOH × mole ratio
= 1.351 × 10-3 mol NaOH × 1 mol CO2
2 mol NaOH
-4
= 6.753 × 10 mol CO2
mCO2 = nCO2 × MCO2
= 6.753 × 10-4 mol CO2 × 44.010 g/mol
mCO2 = 0.02972 g
d.
cCO2
= m
V
= 0.02972 g × 1000 mL
25.00 mL
1L
cCO2
= 1.189 g/L
***4 sig. fig.***
.
10
15.
Two aqueous solutions are mixed (25.00 mL of 0.5461 M Hg2(NO3)2 and 10.00 mL of
2.165 M KI), and 5.92 g of Hg2I2 precipitates. Calculate the percent yield for this
reaction.
[8 marks]
Hg2(NO3)2(aq) + 2 KI(aq)
→
Hg2I2(s) + 2 KNO3(aq)
654.99 g/mol
nHg2(NO3)2
.
nHg2(NO3)2
n*Hg2(NO3)2
n*Hg2(NO3)2
= MHg2(NO3)2 × VHg2(NO3)2
= (0.5461 mol/L)(0.02500 L)
nKI
= MKI × VKI
= (2.165 mol/L)(0.01000 L)
= 0.01365 mol
nKI
= 0.02165 mol
= nHg2(NO3)2
1
= 0.01365 mol
1
= 0.01365 mol
n*KI
= nKI
2
= 0.02165 mol
2
= 0.01082 mol
n*KI
KI is the LIMITING REAGENT
nHg2I2 = nKI × mole ratio
= (0.02165 mol KI) × (1 mol Hg2I2).
(2 mol KI)
nHg2I2 = 0.01082 mol Hg2I2
mHg2I2 = nHg2I2 × MHg2I2
= (0.01082 mol)(654.99 g/mol)
mHg2I2 = 7.090 g
This is the THEORETICAL YIELD.
% Yield =
actual yield
× 100%
theoretical yield
= 5.92 g × 100%
7.090 g
% Yield = 83.5 %
***3 significant figures***
11
16.
(a)
A balloon is filled with 1.26 g of helium (4.0026 g/mol) at STP (1 atm, 273.15 K) and
tied tightly.
[8 marks]
Calculate the density of the helium in the balloon (in g/L).
d = m
therefore
V = m
V
d
and
PV = nRT
therefore:
m = nRT
d
P
therefore:
d
=
mP
nRT
therefore:
d
=
MP
RT
=
therefore
= m × P .
n
RT
V = nRT
P
but M = m
n
(4.0026 g·mol-1)(1 atm)
.
-1
-1
(0.082057 L·atm·mol ·K )(273.15 K)
d = 0.17858 g/L
***5 sig. fig. (1 atm is considered an exact number)***
(b)
Assuming that the volume of the balloon is fixed, calculate the pressure in the heated
balloon if the helium is heated to 85 ˚C.
R = P1V1 = P2V2
but
V1 = V2
and n1 = n2
n1T1
n2T2
P1 = P2
T1
T2
P2 = P1T2
T1
= (1 atm)(358 K)
(273.15 K)
T2 = 85 K + 273.15 K = 358 K
P2 = 1.31 atm
***3 sig. fig. (1 atm is considered an exact number)***
(c)
Calculate the root-mean-square speed of a helium atom in the heated balloon.
vrms
vrms
=
3 RT
M
=
3 (8.3145 J·mol-1·K-1)(358 K) × 1 kg·m2/s2 × 1000 g
(4.0026 g·mol-1)
1J
1 kg
=
1.49 × 103 m/s
***3 sig. fig.***
12
17.
A He+ cation is excited from the ground state to the n = 4 excited state.
Z=2
n1 = 1
n2 = 4
(a)
Calculate the energy of the photon absorbed by the ground state He+ cation.
Ephoton = En2 - En1
Ephoton = Ry × Z2 × [(n1)-2 - (n2)-2]
= (2.179 × 10-18 J) × (2)2 × [(1)-2 - (4)-2]
Ephoton = 8.171 × 10-18 J
***4 sig. fig. (Z, n1 and n2 are exact numbers)***
(b)
Calculate the wavelength of the absorbed photon.
c = λν
therefore
Ephoton = hν
Ephoton = hc
λ
λ
=
hc
.
Ephoton
=
(6.626 × 10-34 J·s)(2.9979 × 108 m/s)
(8.171 × 10-18 J)
λ
=
2.431 × 10-8 m × 109 nm
1m
λ
=
24.31 nm
***4 sig. fig.***
(c)
Calculate the momentum of the absorbed photon.
λ = h
ρ
ρ = h = (6.626 × 10-34 J·s) × 1 kg·m2/s2
λ
(2.431 × 10-8 m)
1J
ρ = 2.726 × 10-26 kg·m/s
***4 sig. fig.***
ν = c
λ
[7 marks]
13
18.
Corundum is a colourless crystal that, when contaminated with coloured transition metal
ions, makes precious gems including rubies and sapphires. A sample of corundum is
found to contain 53% Al and 47% O.
[8 marks]
(a)
Calculate the empirical formula for corundum.
100 g corundum contains 53 g Al and 47 g O.
nAl = mAl
MAl
=
(53 g)
(26.9815 g/mol)
nAl = 2.0 mol
nAl
:
nO = mO
MO
=
(47 g)
.
(15.9994 g/mol)
nO = 2.9 mol
nO
=
2.0 mol Al
: 2.9 mol O
Al2O3
(b)
Assume that the empirical formula for corundum is also its molecular formula. Give the
complete chemical name for the hydrate formed if 1.29 g of anhydrous corundum absorbs
water to give 2.44 g of hydrated corundum.
Al2O3
+
101.9612 g/mol
x H2O
→
Al2O3 · x H2O
8.0152 g/mol
nAl2O3 = mAl2O3
MAl2O3
=
(1.29 g)
.
(101.9612 g/mol)
nAl2O3 = 0.0127 mol
x = nH2O
nAl2O3
= (0.0638 mol H2O) .
(0.0127 mol Al2O3)
x = 5.05 mol H2O per mol Al2O3
mH2O = mAl2O3xH2O - mAl2O3
= (2.44 g) – (1.29 g)
mH2O = 1.15 g
nH2O
nH2O
= mH2O
MH2O
=
(1.15 g)
.
(18.0152 g/mol)
= 0.0638 mol
Al2O3 · 5H2O = aluminum oxide pentahydrate
14
19.
A 500 mL Dewar flask contains 50.00 g of methane at -78˚C (195 K).
(CH4 = 16.0426 g/mol)
[6 marks]
(a)
Calculate the pressure in the Dewar flask if the methane behaves as an ideal gas.
nCH4 = mCH4
MCH4
=
(50.00 g)
.
(16.0426 g/mol)
nCH4 = 3.117 mol
PV = nRT
P = nRT
V
= (3.117 mol)(0.082057 L·atm·mol-1·K-1)(195 K)
(0.500 L)
P = 99.8 atm
***3 sig. fig.***
(b)
Calculate the pressure in the Dewar flask if the methane behaves as a real gas.
n
P + a
V
2
(V - b n) = nRT
n
P + a
V
2
=
P =
nRT
(V - b n)
nRT
(V - b n)
- a
n
V
2
(3.117 mol)(0.082057 L.atm.mol-1.K1)(195 K)
P =
- (2.253 atm.L-2.mol-2)
[(0.500 L) - (0.04278 L.mol-1)(3.117 mol)]
P =
P =
136 atm
-
87.5 atm
48 atm
***0 decimal places therefore 2 sig. fig.***
(
2
)
(3.117 mol)
(0.500 L)