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?