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Final Exam ReviewSemester 2
Chapters: 8,9,10,11,13,14
Chapter 8 Test Review
Vocabulary• 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
• 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)
IA
The Periodic Table of the ElementsVIIIA
1
H1.008
IIA
IIIA IVA VA VIA VIIA
2
He4.003
3
Li6.941
4
Be9.012
5
B10.81
6
C12.01
7
N14.01
8
O16.00
9
F19.00
10
Ne20.18
11
Na22.99
12
Mg24.31
IIIB IVB VB VIB VIIB VIIIB IB IIB
13
Al26.98
14
Si28.09
15
P30.97
16
S32.07
17
Cl35.45
18
Ar39.95
19
K39.10
20
Ca40.08
21
Sc44.96
22
Ti47.87
23
V50.94
24
Cr52.00
25
Mn54.94
26
Fe55.85
27
Co58.93
28
Ni58.69
29
Cu63.55
30
Zn65.39
31
Ga69.72
32
Ge72.61
33
As74.92
34
Se78.96
35
Br79.90
36
Kr83.80
37
Rb85.47
38
Sr87.62
39
Y88.91
40
Zr91.22
41
Nb92.91
42
Mo95.94
43
Tc(98)
44
Ru101.1
45
Rh102.9
46
Pd106.4
47
Ag107.9
48
Cd112.4
49
In114.8
50
Sn118.7
51
Sb121.8
52
Te127.6
53
I126.9
54
Xe131.3
55
Cs132.9
56
Ba137.3
57
La138.9
72
Hf178.5
73
Ta180.9
74
W183.8
75
Re186.2
76
Os190.2
77
Ir192.2
78
Pt195.1
79
Au197.0
80
Hg200.6
81
Tl204.4
82
Pb207.2
83
Bi209.0
84
Po(209)
85
At(210)
86
Rn(222)
87
Fr(223)
88
Ra(226)
89
Ac(227)
104
Rf(261)
105
Db(262)
106
Sg(266)
107
Bh(264)
108
Hs(269)
109
Mt(268)
110
Uun(271)
111
Uuu(272)
112
Uub(277)
58
Ce140.1
59
Pr140.9
60
Nd144.2
61
Pm(145)
62
Sm150.4
63
Eu152.0
64
Gd157.3
65
Tb158.9
66
Dy162.5
67
Ho164.9
68
Er167.3
69
Tm168.9
70
Yb173.0
71
Lu175.0
90
Th232.0
91
Pa(231)
92
U238.0
93
Np(237)
94
Pu(244)
95
Am(243)
96
Cm(247)
97
Bk(247)
98
Cf(251)
99
Es(252)
100
Fm(257)
101
Md(258)
102
No(259)
103
Lr(262)
Ionic Compounds Molecular Compounds
Crystal Lattice Molecule
Types of Elements
Metal with non-metal or polyatomic ions
Non-metal with non-metal
Physical State
Solid Solid, liquid or gas
Melting Point High> 300 C
Low <300 C
Solubility in water
Generally high Generally low
Electrical conductivity of solution
Good conductor 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. CH3OH2. CH4 3. PF3 4. OF2 5. NO2
-
6. BH3
7. SO4 2-
8. CN-
9. N2H2
Common PrefixesNumber of
atoms Prefix Number of atoms Prefix
1 Mono- 6 Hexa-2 Di- 7 Hepta-3 Tri- 8 Octa-4 Tetra- 9 Nona-5 Penta- 10 Deca-
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 SeriesThe activity series ranks the relative
reactivity of metals.It allows us to predict if certain chemicals will
undergo single displacement reactions. Metals near the top are most reactive and
will displace metals near the bottom.Q: Which of these will react?
Fe + CuSO4 Ni + NaCl Li + ZnCO3 Al + CuCl2
Cu + Fe2(SO4)3
Yes, Fe is above Cu
NR (no reaction)
No, Ni is below Na
Zn + Li2CO3Cu + AlCl3
Yes, Al is above Cu
CuHgAg
CaMgAlZnFeNiSnPbH
Au
LiNaK
Yes, Li is above Zn
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 CompositionDetermine the percentage composition of sodium carbonate (Na2CO3)?
Molar Mass Percent Composition
% Na =46.0 g106 g
x 100% =43.4 %
% C =12.0 g106 g
x 100% =11.3 %
% O =48.0 g106 g
x 100% =45.3 %
Na = 2(23.00) = 46.0C = 1(12.01) = 12.0O = 3(16.00) = 48.0 MM= 106 g
Solve the following percent composition problems:
26. Mg(NO3)2
27. (NH4)2S
Formulas
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
Percent composition allow you to calculate the simplest ratio among the atoms found in compound.
Examples:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
CH2O - empirical
Calculating Empirical FormulaExample: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 Co1 mol Co
58.93 g Co= 0.07721 mol Co
5.475 g Cl 1 mol Cl
35.45 g Cl= 0.1544 mol Cl
0.07721 mol Co 0.1544 mol Cl
0.07721 0.07721 = 2= 1
CoCl2
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 FormulaExample 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 1: Molar Mass
P = 2 x 30.97 g = 61.94gO = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by Empirical Formula Mass
238.88 g141.94g
= 2
Step 3: Multiply
(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
Which substances have the same empirical formula?
HydratesHydrated 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 ReactantGiven 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 YieldGiven 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?