1. Final Exam Review Semester 2 Chapters: 8,9,10,11,13,14

2. Chapter 8 Test Review

3. Vocabulary Covalent bond Endothermic reaction Coefficient Product
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

4. 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)

5. The Periodic Table of the ElementsIA
The Periodic Table of the Elements
VIIIA H 1
1.008
IIA
IIIA
IVA VA
VIA
VIIA He 2
4.003 Li 3
6.941 Be 4
9.012 B 5
10.81 C 6
12.01 N 7
14.01 O 8
16.00 F 9
19.00 Ne 10
20.18 Na 11
22.99 Mg 12
24.31
IIIB
IVB VB
VIB
VIIB
VIIIB IB
IIB Al 13
26.98 Si 14
28.09 15 P
30.97 16 S
32.07 Cl 17
35.45 Ar 18
39.95 19 K
39.10 Ca 20
40.08 Sc 21
44.96 Ti 22
47.87 23 V
50.94 Cr 24
52.00 Mn 25
54.94 Fe 26
55.85 Co 27
58.93 Ni 28
58.69 Cu 29
63.55 Zn 30
65.39 Ga 31
69.72 Ge 32
72.61 As 33
74.92 Se 34
78.96 Br 35
79.90 Kr 36
83.80 Rb 37
85.47 Sr 38
87.62 39 Y
88.91 Zr 40
91.22 Nb 41
92.91 Mo 42
95.94 Tc 43
(98) Ru 44
101.1 Rh 45
102.9 Pd 46
106.4 Ag 47
107.9 Cd 48
112.4 In 49
114.8 Sn 50
118.7 Sb 51
121.8 Te 52
127.6 53 I
126.9 Xe 54
131.3 Cs 55
132.9 Ba 56
137.3 La 57
138.9 Hf 72
178.5 Ta 73
180.9 W 74
183.8 Re 75
186.2 Os 76
190.2 77 Ir
192.2 Pt 78
195.1 Au 79
197.0 Hg 80
200.6 Tl 81
204.4 Pb 82
207.2 Bi 83
209.0 Po 84
(209) At 85
(210) Rn 86
(222) Fr 87
(223) Ra 88
(226) Ac 89
(227) Rf
104
(261) Db
105
(262) Sg
106
(266) Bh
107
(264) Hs
108
(269) Mt
109
(268)
Uun
110
(271)
Uuu
111
(272)
Uub
112
(277) Ce 58
140.1 Pr 59
140.9 Nd 60
144.2 Pm 61
(145) Sm 62
150.4 Eu 63
152.0 Gd 64
157.3 Tb 65
158.9 Dy 66
162.5 Ho 67
164.9 Er 68
167.3 Tm 69
168.9 Yb 70
173.0 Lu 71
175.0 Th 90
232.0 Pa 91
(231) 92 U
238.0 Np 93
(237) Pu 94
(244) Am 95
(243) Cm 96
(247) Bk 97 Cf 98
(251) Es 99
(252) Fm
100
(257) Md
101
(258) No
102
(259) Lr
103

6. 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

7. 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.

8. Sigma and Pi bonds Sigma- Pi 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

9. Diatomic Molecules
List the 7 diatomic molecules:

10. Diatomic Molecules Molecules made up of two atoms.
There are 7 diatomic molecules.
H2, N2, O2, F2, Cl2, Br2, I2

11. How many atoms in each formula?
CH3OH
CH4
PF3
OF2
NO2-
BH3
SO4 2-
CN-
N2H2

12. Common Prefixes 1 Mono- 6 Hexa- 2 Di- 7 Hepta- 3 Tri- 8 Octa- 4 Tetra-
Number of atoms
Prefix 1
Mono- 6
Hexa- 2
Di- 7
Hepta- 3
Tri- 8
Octa- 4
Tetra- 9
Nona- 5
Penta- 10
Deca-

13. Naming Molecules
SiS4
PCl5
CCl4 NO

14. Writing Formulas Sulfur difluoride Silicon tetrachloride
Chlorine trifluoride
Tetrasulfur heptanitride

15. 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

16. Lewis Structures and Octet
Practice by drawing H2 O2 N2
H2O
CO2 + - - +

17. Lewis structures
CH3OH
BH3
N2H2

18. Lewis Structures with polyatomic ions
SO4 2-
CN-

19. Tetrahedral Based Shapes
Trigonal Pyramidal
Bent

20. Non-Tetrahedral Based Shapes
Linear
Trigonal Planar
Fewer electrons than octet!

21. Summary of Common Molecular Shapes
Molecule
Total pairs
Shared pairs
Lone pairs
Hybrid orbital
Molecular shape
BeCl2 2 sp
Linear
AlCl3 3
sp2
Trigonal planar
CCl4 4
sp3
Tetrahedral
NH3 1
Trigonal pyramidal
H2O
Bent
NbBr5 5
sp3d
Trigonal bipyramidal
SF6 6
sp3d2
Octahedral

22. Molecular Shapes
CH4
PF3
OF2
NO2-

23. Chapter 9 Review

24. List the following GENERAL equations
Combustion
Synthesis
Decomposition
Double replacement
Single replacement

25. Identify the type of Chemical Reaction

26. 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

27. What chemical reaction does this picture show?

28. What chemical reaction does this picture show?

29. Balance the following equations and write the ratio of coefficients:

30. The 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 Hg Ag Ca Mg Al Zn Fe Ni Sn Pb H Au Li Na K
No, Ni is below Na
Yes, Li is above Zn
Yes, Al is above Cu
Yes, Fe is above Cu
Cu + Fe2(SO4)3
NR (no reaction)
Zn + Li2CO3
Cu + AlCl3

31. Chapter 10
The Mole

32. Define the following: Hydrate Molecular formula Empirical formula
Percent composition
Mole

33. Find the atomic mass for the following atoms:
11. Cl
12. O
13. Fe

34. 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

35. Using Molar Mass Molar mass conversion factors:
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)

36. MASS
(g)
MOLES
(mol)
Particles
(atoms or
molecules or F.U.’s)

37. Find the molar mass for the following compounds:
14. CH2O
15. CaSO4
16. Na3PO4

38. Solve the following molar conversions:
23. How many atoms are in 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

39. Percent Composition
Determine the percentage composition of sodium carbonate (Na2CO3)?
Molar Mass
Percent Composition
46.0 g
x 100% =
43.4 %
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= g
% Na =
106 g
12.0 g
x 100% =
11.3 %
% C =
106 g
48.0 g
x 100% =
45.3 %
% O =
106 g

40. Solve the following percent composition problems:
26. Mg(NO3)2
27. (NH4)2S

41. 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:
C6H12O6
- molecular
C4H10
- molecular
C2H5
- empirical
CH2O
- empirical

42. Calculating Empirical Formula
Example:
A g sample of cobalt reacts with g chlorine to form a binary compound. Determine the empirical formula for this compound.
4.550 g Co
1 mol Co
= mol Co
58.93 g Co
1 mol Cl
5.475 g Cl
= mol Cl
35.45 g Cl
mol Co
mol Cl
= 1
= 2
CoCl2

43. 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%

44. 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 g. What is the compound’s molecular formula?
Step 3: Multiply
Step 1: Molar Mass
P = 2 x g = 61.94g
O = 5 x 16.00g = g g
(P2O5)2 =
P4O10
Step 2: Divide MM by
Empirical Formula Mass g
= 2
141.94g

45. 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.

46. Which samples have the same empirical formula?
Which substances have the same empirical formula?
Which samples have the same empirical formula?
Sample
Formula 1
CH3OH 2
CH2O 3
C6H12O6 4
C2H4O2

47. 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

49. Chapter 11 Review

50. Define the following: Stoichiometry Mole ratio Excess reactant
Limiting reactant
Theoretical yield
Actual yield
Percent yield

51. Find the following molar masses:
AlPO4
NaCl
C6H5Cl
CuO

52. 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?

53. Mole to Mole: Given the following equation: 2 KClO3 –> 2 KCl + 3 O2
How many moles of O2 can be produced by letting moles of KClO3 react?

54. 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?

55. 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?

56. 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.

57. 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

58. Chapter 14 Review

59. Define the following: Suspension Colloid Brownian motion
Tyndall effect
Souble
Insoluble
Miscible
Immiccible
Condentration
Molartity
Dilution

60. Name the type of heterogeneous mixture:
Milk
Fog
Hairspray
Muddy Water
Blood
Orange juice
Gelatin

61. What is the percent by volume when 50 mL of ethanol is diluted to 140 mL with water?

62. What is the percent by mass of 21. 0 g of sodium acetate mixed with 40
What is the percent by mass of 21.0 g of sodium acetate mixed with 40.0 g of water

63. What mass of lithium chloride is found in 85 g of a 25% by mass solution

64. Calculate the molarity of a solution made by dissolving 20
Calculate the molarity of a solution made by dissolving 20. g of NaOH in enough water to make 5.0 Liters of solution?

65. Full strength hydrochloric acid is 11. 6 M
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?