Final Exam Review
Semester 2
Chapters: 8,9,10,11,13,14
Chapter 8 Test Review
• Covalent bond
• Endothermic
reaction
• Lewis Exothermic
reaction
• structure
• Molecule
• Pi bond
• Sigma bond
• Resonance
• Structural formula
• VESPR model
• Polar covalent bond
• Non-polar bond
• Chemical equation
• Chemical reaction
Vocabulary
•Coefficient
•Product
•Reactant
•Combustion reaction
•Decomposition reaction
•Single replacement reaction
•Double replacement reaction
•Synthesis reaction
•Precipitate
•Aqueous
•Complete ionic equation
•Net ionic equation
•Solute
•Solvent
•Spectator Ion
Covalent Bonds
• How many covalent bonds can elements in the
following groups form:
–
–
–
–
–
–
–
–
Group 1 (alkali metals)
Group 2 (alkali earth metals)
Group 3
Group 4
Group 5
Group 6
Group 7 (halogens)
Group 8 ( noble gases)
The Periodic Table of the Elements
IA
VIIIA
1
2
H
He
1.008
4.003
IIA
IIIA
5
IVA
3
4
9
10
Be
B
C
N
O
F
Ne
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
IIIB
IVB
VIB
VIIB
VIIIB
IB
16
IIB
19
20
21
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10
40.08
44.96
47.87
50.94
52.00
54.94
55.85
58.93
58.69
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
39
22
VB
15
8
VIIA
Li
14
7
VIA
6.941
13
6
VA
37
38
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.47
87.62
88.91
91.22
92.91
95.94
(98)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
55
56
57
72
73
74
75
76
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
137.3
138.9
178.5
180.9
183.8
186.2
190.2
192.2
77
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
87
88
89
104
105
106
107
108
109
110
111
112
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
Uun
Uuu
Uub
(223)
(226)
(227)
(261)
(262)
(266)
(264)
(269)
(268)
(271)
(272)
(277)
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.1
140.9
144.2
(145)
150.4
152.0
157.3
158.9
162.5
164.9
167.3
168.9
173.0
175.0
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.0
(231)
238.0
(237)
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(259)
(262)
Ionic Compounds
Molecular
Compounds
Crystal Lattice
Molecule
Metal with non-metal Non-metal with
or polyatomic ions
non-metal
Solid
Solid, liquid or gas
Types of
Elements
Physical
State
Melting Point High
> 300 C
Solubility in Generally high
water
Electrical
Good conductor
conductivity
of solution
Low
<300 C
Generally low
Poor to none
Properties of Covalent bonds
• Bond length decreases as number of covalent
bonds increases.
• Bond strength increases as number of covalent
bonds increases
– Ex.
• Bond length increases as number of covalent
bonds decreases
• Bond strength decreases as number of covalent
bonds decreases.
– Ex.
Sigma and Pi bonds
• Sigma– Single covalent bond
• Single bond- 1 sigma
• Pi
– Multiple covalent bonds
• Double bond- 1 sigma, 1 pi bond
• Triple bond- 1 sigma, 2 pi bonds
Diatomic Molecules
• List the 7 diatomic molecules:
Diatomic Molecules
• Molecules made up of two atoms.
• There are 7 diatomic molecules.
• H2,
N2, O2,
F2, Cl2, Br2, I2
How many atoms in each formula?
1.
2.
3.
4.
5.
6.
7.
8.
9.
CH3OH
CH4
PF3
OF2
NO2BH3
SO4 2CNN2H2
Common Prefixes
Number of
atoms
1
2
3
4
5
Prefix
MonoDiTriTetraPenta-
Number of
atoms
6
7
8
9
10
Prefix
HexaHeptaOctaNonaDeca-
Naming Molecules
• SiS4
• PCl5
• CCl4
• NO
Writing Formulas
• Sulfur difluoride
• Silicon tetrachloride
• Chlorine trifluoride
• Tetrasulfur heptanitride
Steps to doing lewis structures with covalent
bonds
1.Count the valence electrons for all atoms
2.Put the least electronegative atom in the center. Hydrogen is always
on outside
3.Assign 2 electrons to each atom
4.Complete octets on outside atoms
5.Put remaining electrons in pairs on central atom
6.If central atom doesn’t have an octet, move electrons from outer
atoms to form double or triple bonds
Lewis Structures and Octet
•
•
•
•
•
•
Practice by drawing
H2
O2
+
N2
H2O
CO2
+
-
Lewis structures
• CH3OH
• BH3
• N2H2
Lewis Structures with polyatomic ions
• SO4 2-
• CN-
Tetrahedral Based Shapes
Tetrahedral
Trigonal Pyramidal
Bent
Non-Tetrahedral Based Shapes
Linear
Trigonal Planar
FEWER ELECTRONS THAN OCTET!
Summary of Common Molecular
Shapes
Molecule
Total
pairs
Shared
pairs
Lone
pairs
Hybrid
orbital
Molecular shape
BeCl2
2
2
0
sp
Linear
AlCl3
3
3
0
sp2
Trigonal planar
CCl4
4
4
0
sp3
Tetrahedral
NH3
4
3
1
sp3
Trigonal pyramidal
H2O
4
2
2
sp3
Bent
NbBr5
5
5
0
sp3d
Trigonal bipyramidal
SF6
6
6
0
sp3d2
Octahedral
Molecular Shapes
• CH4
• PF3
• OF2
• NO2-
Chapter 9 Review
List the following GENERAL equations
• Combustion
• Synthesis
• Decomposition
• Double replacement
• Single replacement
Identify the type of Chemical
Reaction
Name the type of chemical reaction
• 2SO2 + O2  2AL(OH)3 +
3CaSO4
• 2Be + O2  2BeO
• 2PbO2  2PbO + O2
• C2H6 + O2  CO2 + H2O
• Li + NaOH  LiOH + Na
What chemical reaction does this
picture show?
What chemical reaction does this
picture show?
Balance the following equations and
write the ratio of coefficients:
Activity Series
The activity series ranks the relative
reactivity of metals.
No, Ni is
It allows us to predict if certainYes,
chemicals
will
Li
is
below Na
undergo single displacement above
reactions.
Zn
Metals near the top are most reactive and
Yes, Al is
will displace metals near the bottom.
Q: Which of these will react? above Cu
Fe
Ni
Li
Al
+
+
+
+
CuSO4
NaCl
ZnCO3
CuCl2




Yes, Fe is
Cu + Fe2above
(SO4)3Cu
NR (no reaction)
Zn + Li2CO3
Cu + AlCl3
K
Na
Li
Ca
Mg
Al
Zn
Fe
Ni
Sn
Pb
H
Cu
Hg
Ag
Au
Chapter 10
The Mole
Define the following:
1. Hydrate
2. Molecular formula
3. Empirical formula
4. Percent composition
5. Mole
Find the atomic mass for the following
atoms:
9. C
10. H
11. Cl
12. O
13. Fe
Converting Moles to Particles and
Particles to Moles
• You can use the mole as a conversion
factor in dimensional analysis problems.
• Since one mole is equal to 6.02 X 1023
particles or things the conversion
factors are:
1 mole__
6.02 X 1023
6.02 X 1023
1 mole
Using Molar Mass
• Molar mass can be used in dimensional analysis
to convert the # of moles of a substance into
grams or vice versa.
Molar mass conversion factors:
1 mol of substance= Molar Mass of
Substance (g)
Molar Mass of Substance (g)
1 mol of substance
1 mol of substance___
Molar mass of substance (g)
MASS
(g)
MOLES
(mol)
Particles
(atoms or
molecules
or F.U.’s)
Find the molar mass for the following
compounds:
14. CH2O
15. CaSO4
16. Na3PO4
Solve the following molar conversions:
23. How many atoms are in 0.750 moles of zinc?
24. How many moles of magnesium is 3.01 x 1022
atoms of magnesium?
25. Find the mass of 1.00 x 1023 molecules of N2
Percent Composition
Determine the percentage composition of sodium carbonate
(Na2CO3)?
Molar Mass
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106 g
Percent Composition
46.0 g
% Na =106 g
12.0 g
% C = 106 g
48.0 g
% O = 106 g
x 100% = 43.4 %
x 100% = 11.3 %
x 100% = 45.3 %
Solve the following percent
composition problems:
26. Mg(NO3)2
27. (NH4)2S
Formulas
Percent composition allow you to calculate the simplest
ratio among the atoms found in compound.
Empirical Formula – formula of a compound that expresses
lowest whole number ratio of atoms.
Molecular Formula – actual formula of a compound showing
the number of atoms present
Examples:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
CH2O
- empirical
Calculating Empirical Formula
Example:
A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a
binary compound. Determine the empirical formula for this
compound.
4.550 g Co 1 mol Co
= 0.07721 mol Co
58.93 g Co
5.475 g Cl 1 mol Cl
35.45 g Cl
0.07721 mol Co
=1
0.07721
CoCl2
= 0.1544 mol Cl
0.1544 mol Cl
0.07721
=2
Solve the following empirical formula
problem:
What’s the empirical formula of a molecule
containing 65.5% carbon, 5.5% hydrogen, and
29.0% oxygen.
Element
Percent Composition
C
65.5%
H
5.5%
O
29.0%
Calculating Molecular Formula
Example 1:
A white powder is analyzed and found to have an
empirical formula of P2O5. The compound has a molar
mass of 283.88g. What is the compound’s molecular
formula?
Step 3: Multiply
Step 1: Molar Mass
P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by
Empirical Formula Mass
238.88 g
=2
141.94g
(P2O5)2 =
P4O10
Solve the following molecular formula
problem:
A compound with an empirical formula of
C2H8N and a molar mass of 46 grams per
mole. Find the molecular formula.
Which samples have the same
empirical formula?
Sample
Formula
1
CH3OH
2
CH2O
3
C6H12O6
4
C2H4O2
Hydrates
Hydrated salt – salt that has water molecules trapped
within the crystal lattice
Examples:
CuSO4•5H2O
CuCl2•2H2O
Anhydrous salt – salt without water molecules
Examples: CuCl2
Chapter 11 Review
Define the following:
•
•
•
•
•
•
•
Stoichiometry
Mole ratio
Excess reactant
Limiting reactant
Theoretical yield
Actual yield
Percent yield
Find the following molar masses:
•
•
•
•
•
•
O2
O
AlPO4
NaCl
C6H5Cl
CuO
Create the following mole ratios:
• __Ag(s) + __H2S(g) + __O2(g) __Ag2S(s) + __H2O(l) (Equation
must first be balanced.)
• Ag : H2S
• O2 : Ag2S
•
•
•
•
Ag2S : H2O
O2 : H2S
Ag : O2
H2O : H2S
How many ratios can this
equation form?
Mole to Mole:
Given the following equation:
2 KClO3 –> 2 KCl + 3 O2
• How many moles of O2 can be produced by
letting 12.00 moles of KClO3 react?
Mole to Mass:
Given the following equation:
2 KClO3 –> 2 KCl + 3 O2
• How many grams of O2 can be produced by letting 3
moles of KClO3 react?
Mass to Mass:
Given the following equation:
2 KClO3 –> 2 KCl + 3 O2
• How many grams of O2 can be produced by letting
34.7g of KClO3 react?
Limiting Reactant
Given the following equation:
Al2(SO3)3 + 6 NaOH  3 Na2SO3 + 2 Al(OH)3
• If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH, determine
the limiting reactant for how many grams of Al(OH)3 is formed.
Percent Yield
Given the following equation:
2 FePO4 + 3 Na2SO4  1 Fe2(SO4)3 + 2 Na3PO4
• What is the percent yield of this reaction if takes
place with 25g of FePO4 and an excess of Na2SO4, and
produces 18.5g of Fe2(SO4)3
Chapter 14 Review
Define the following:
•
•
•
•
•
•
•
•
•
•
•
Suspension
Colloid
Brownian motion
Tyndall effect
Souble
Insoluble
Miscible
Immiccible
Condentration
Molartity
Dilution
Name the type of heterogeneous
mixture:
•
•
•
•
•
•
•
Milk
Fog
Hairspray
Muddy Water
Blood
Orange juice
Gelatin
What is the percent by volume when 50
mL of ethanol is diluted to 140 mL with
water?
What is the percent by mass of 21.0 g of
sodium acetate mixed with 40.0 g of water
What mass of lithium chloride is found in
85 g of a 25% by mass solution
Calculate the molarity of a solution made by dissolving
20. g of NaOH in enough water to make 5.0 Liters of
solution?
Full strength hydrochloric acid is 11.6 M. How
many liters of this concentrated solution is
required to make 1.0 L of a 1.0 M solution?