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

Chapter 8 Test Review

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

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 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

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.

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

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? CH3OH CH4 PF3 OF2 NO2- BH3 SO4 2- CN- N2H2

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-

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 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 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

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:

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

Chapter 10 The Mole

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

Find the atomic mass for the following atoms: 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 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)

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 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= 106 g % Na = 106 g 12.0 g x 100% = 11.3 % % C = 106 g 48.0 g x 100% = 45.3 % % O = 106 g

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: C6H12O6 - molecular C4H10 - molecular C2H5 - empirical 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 1 mol Cl 5.475 g Cl = 0.1544 mol Cl 35.45 g Cl 0.07721 mol Co 0.1544 mol Cl = 1 = 2 0.07721 0.07721 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 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 (P2O5)2 = P4O10 Step 2: Divide MM by Empirical Formula Mass 238.88 g = 2 141.94g

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? Which substances have the same empirical 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: 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 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 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 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?