Midterm II Material/Topics Autumn 2010

1 Midterm II Material/Topics Autumn 2010 Supplemental Material: Resonance Structures Ch 5.8 Molecular Geometry Ch 5.9 Electronegativity Ch 5.10 Bond Polarity Ch 5.11 Molecular Polarity Ch 5.12 Naming Binary Molecular Compounds Writing and Balancing Chemical Equations (Ch 6.6) A. Combination Reactions (Ch 9.1) B. Decomposition Reactions (Ch 9.1) C. Single-Replacement Reactions (Ch 9.1) D. Double-Replacement Reactions (Ch 9.1) with Solubility Rules (Ch 8.4) E. Combustion Reactions (Ch 9.1) F. Oxidation - Reduction Reactions (Ch. 9.2, 9.3) Supplemental Material: Comparing Strength of Agents H. Acid-Base Neutralization Reactions (Ch 10.1, 10.6 & 10.7) Ch 6.1 Formula Masses Supplemental material: Percent Composition Ch 6.2 The Mole & Avogadro's Number Ch 6.3 Molar Mass Ch 6.4 & 6.5 Chemical Formula & Mole Calculations Supplemental material & Lab Exp 8: Empirical & Molecular Formula Ch 6.6, 6.7, 6.8 Chemical Equations & Stoichiometric Calculations Supplemental material: Heat of Reaction Supplemental material: Percent Yield Supplemental material: Limiting Reagent

2 Ch 7.1 & 7.2 The Kinetic Molecular Theory and States of Matter Ch 7.3 Gas Law Variables Ch 7.4 Boyle s Law Ch 7.5 Charles s Law Supplemental material: Amontons's Law Ch 7.6 The Combined Gas Law Ch 7.7 The Ideal Gas Law Supplemental material: Avogadro s Hypothesis and the Molar Volume Ch 7.8 Dalton s Law of Partial Pressures Ch 7.9 Changes of Physical State Ch 7.10 Evaporation of Liquids Ch 7.11Vapor Pressure of Liquids Ch. 7.12 Boiling and Boiling Point Supplemental material: Heat of Vaporization, Viscosity, and Surface Tension Ch 7.13 Intermolecular Forces in Liquids A. London Forces B. Dipole-Dipole Interactions C. Hydrogen Bonds Ch 8.1 Characteristics of Solutions Ch 8.2 Solubility Ch 8.3 Solution Formation Supplemental material: Entropy Ch 8.4 Solubility Rules

3 Resonance Structures Resonance structures are two or more Lewis structures that represent the same ion or molecule equally well. Examples: nitrite ion NO 2 - ozone O 3 carbonate ion CO 3 2- benzene

4 Ch 5.8 Molecular Geometry Valence Shell Electron-Pair Repulsion theory (VSEPR theory) Electrons repel each other Electrons tend to be as far apart as possible Electron pairs control geometry 2 electron pairs or VSEPR groups 3 electron pairs or VSEPR groups 4 electron pairs or VSEPR groups Figure 5.8

5 A. Two VSEPR groups = linear.... : O = C = O : 2 VSEPR groups around the central atom bond angle 180 o B. Three VSEPR groups = trigonal planar or angular : O : H C H 3 VSEPR groups Trigonal planar bond angle ~120 o...... - : O N O : 3 VSEPR groups around N 2 bonding, 1 nonbonding bent or angular shape bond angle ~120 o

6 C. Four VSEPR groups: Tetrahedral (no nonbonding e pairs) Trigonal pyramidal (1 nonbonding e pair) Angular or bent (2 nonbonding e pairs) 109.5 o <109.5 o <109.5 o

7 Ch 5.10 Bond Polarity pure covalent polar covalent ionic Differences in Electronegativity Nonpolar covalent bonds (similar values) < 0.4 0.4 < Polar covalent bonds < 1.5 1.5 < Borderline area < 2.0 (ionic or polar covalent) Ionic bonds > 2.0

8 Ch 5.11 Molecular Polarity Bond polarity Molecular geometry A. Polar bond + unsymmetrical distribution of electronic charge = polar molecule

9 B. Polar bond + symmetrical distribution of electronic charge = nonpolar molecule Polarity cancels Polarity cancels in a trigonal planar molecule with 3 identical atoms or groups attached. Polarity cancels in a tetrahedral molecule with 4 identical atoms or groups attached.

Examples of polar and nonpolar molecules (all contain polar bonds) NH 3 is trigonal pyramidal, not planar 10

11 Ch 5.12 Naming Binary Molecular Compounds Different compounds exist for most pairs of nonmetals. Examples of N-O compounds: NO NO 2 N 2 O 3 N 2 O 4 N 2 O 5 N 2 O 3 dinitrogen trioxide 1. prefix + full name of least electronegative nonmetal 2. prefix + stem name of more electronegative nonmetal + ide

12 Chapter Medley: Chemical Equations and Reactions I. Balancing Equations Law of Conservation of Mass NEVER change subscripts Eliminate fractions Check the final answer II. Types of Chemical Reactions A. Combination Reactions X + Y XY B. Decomposition Reactions XY X + Y C. Single-Replacement Reactions A + BY B + AY Order of reactivity: F > Cl > Br > I

13 D. Double-Replacement Reactions AX + BY AY + BX Complete chemical equation Co(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) CoCO 3 (s) + 2 NaNO 3 (aq) Ionic equation Co 2+ + 2NO 3 - + 2Na + + CO 3 2- CoCO 3 (s) + 2Na + + 2NO 3 - Net ionic equation Co 2+ + CO 3 2- CoCO 3 (s) Na + and NO 3 - are spectator ions.

14 E. Combustion Reactions C 2 H 6 O + 3 O 2 2 CO 2 + 3 H 2 O + Heat F. Oxidation Reduction Reactions = Redox 2 Mg + O 2 2 MgO 0 0 +2-2 oxidation = loss of electrons reduction = gain of electrons Mg is oxidized from 0 to +2 O 2 accepts the electrons from Mg O 2 oxidizes Mg to Mg 2+ O 2 is the oxidizing agent O 2 is reduced from 0 to -2 Mg transfers electrons to O 2 Mg reduces O 2 to O 2- Mg is the reducing agent

15 Assignments of Oxidation Numbers KMnO 4 (+1) + (ox# Mn) + 4(-2) = 0 (ox# Mn) = 8-1 = +7 K = +1 Mn = +7 O = -2 G. Acid-Base Neutralization Reactions Acid + Base Salt + Water HX + BOH BX + HOH H 3 PO 4 (aq) + 3 NaOH(aq) Na 3 PO 4 (aq) + 3 H 2 O(l) 3H + + PO 4 3- + 3Na + + 3OH - 3Na + + PO 4 3- + 3H 2 O Net ionic equation: H + + OH - H 2 O

16 Chapter 6 Chemical Calculations 1. Formula Masses C 2 H 6 O M.W. = 46.0 amu (NH 4 ) 2 S F.W. = 68.1 amu 2. Percent Composition % Composition = mass of element x 100 total mass Ammonium sulfide (NH 4 ) 2 S F.W. = 68.1 amu % N = 2 x 14.0 amu N x 100 = 41.1 % N 68.1 amu (NH 4 ) 2 S 3. The Mole & Avogadro s Number 1 mole = 6.02 x 10 23 objects How many He atoms are in 2.55 moles of He? 2.55 mole He x 6.02 x 10 23 He atoms 1 mole He = 1.54 x 10 24 He atoms

17 4. Molar Mass Ammonium sulfide (NH 4 ) 2 S formula mass = 68.1 amu 68.1 g (NH 4 ) 2 S = 1 mole F.W. = 68.1 g/mol Carbon dioxide CO 2 formula mass = 44.01 amu 44.01 g CO 2 = 1 mole M.W. = 44.01 g/mol If 7.50 moles of ammonia, NH 3, are required for a certain experiment, what mass of ammonia is needed? M.W. = 17.0 g/mol 7.50 moles NH 3 x 17.0 g NH 3 = 128 g NH 3 1 mole NH 3 5. Chemical Formula & Mole Calculations How many g of (NH 4 ) 2 S are required to obtain 0.50 moles of NH 4 +? (NH 4 ) 2 S = 68.1 g/mol (molar mass) 0.50 moles NH 4 + = 17 g (NH 4 ) 2 S x 1 mole (NH 4 ) 2 S + 2 moles NH 4 x 68.1 g (NH 4 ) 2 S 1 mol (NH 4 ) 2 S

18 6. Empirical & Molecular Formula Composition of Borazole = 40.28%B, 52.20%N, 7.52%H Molar mass = 80.5 amu Calculated the molecular formula B: 40.28g x 1 mole = 3.726 moles = 1.0 10.81 g 3.726 N: 52.20g x 1 mole = 3.726 moles = 1.0 14.01 3.726 H: 7.52g x 1 mole = 7.45 moles = 2.0 1.01 g 3.726 E.F.= BNH 2 E.F. mass: 10.81 + 14.01 + (2x1.01) = 26.84 Molar mass = 80.5 amu = 3 E.F. mass 26.84 amu Molecular Formula = 3 x (BNH 2 ) = B 3 N 3 H 6 Borazole

19 7. Chemical Equations & Stoichiometric Calculations How many g O 2 are needed to convert 45 g glucose (C 6 H 12 O 6 ) into CO 2 and H 2 O? C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O M.W. 180 amu + 32 amu 45 g glu x 1 mol glu x 6 mol O 2 x 32 g O 2 180 g glu 1 mol glu 1 mol O 2 = 48 g O 2

20 8. Heat of Reaction combustion of propane is exothermic. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) + 530 kcal C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H = - 530 kcal photosynthesis is endothermic 6 CO 2 (g) + 6 H 2 O(g) + 678 kcal C 6 H 12 O 6 (aq) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (aq) + 6 O 2 (g) H = + 678 kcal How much energy is produced when 0.50 g of butane, C 4 H 10, is burned in a butane lighter? C 4 H 10 = 58.1 g/mole 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) + 1365 kcal 0.50g C 4 H 10 x 1 mol C 4 H 10 x 1365 kcal = 5.87 kcal 58.1 g C 4 H 10 2 moles C 4 H 10 Is this reaction exothermic or endothermic? What is the sign for H?

21 9. Percent Yield % yield = actual yield x 100 theoretical yield When 884 g of Fe 2 O 3 was reduced with excess carbon, 507 g of Fe were obtained. What was the percent yield? Fe 2 O 3 + 3 C 2 Fe + 3 CO 884g Fe 2 O 3 x 1 mol Fe 2 O 3 x 2 moles Fe x 55.85g Fe = 618g Fe 159.7g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe % yield = actual yield x 100 = 507 g x 100 = 82.0 % theoretical yield 618 g

22 10. Limiting Reagent Yield calculations must be based on the limiting reagent. If 3.33 g CaCl 2 are reacted with 8.50 g AgNO 3 and 5.63 g AgCl are obtained, what is the % yield? CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) 3.33 g 8.50 g 5.63 g 111.0 amu 169.9 amu 143.3 amu 3.33 g CaCl 2 x 1 mol = 0.0300 mol CaCl 2 111.0 g 8.50 g AgNO 3 x 1 mol = 0.0500 mol AgNO 3 169.9 g Mole-to-coefficient ratios: 0.0300 CaCl 2 = 0.0300 0.0500 AgNO 3 = 0.0250 limiting reagent 0.0500 mol AgNO 3 x 2 mol AgCl x 143.3 g AgCl = 7.16 g AgCl 2 mol AgNO 3 1 mol AgCl actual yield x 100 = 5.63 g x 100 = 78.6 % theoretical yield 7.16 g

23 Chapter 7 Solids, Liquids, Gases Boyle s Law: Pressure-Volume P 1 x V 1 = P 2 x V 2 Charles s Law: Temperature-Volume V 1 = V 2 T 1 T 2 T must be in kelvin Amontons s Law: Pressure-Temperature P 1 = P 2 T 1 T 2 T must be in kelvin Dalton s Law of Partial Pressures P total = P 1 + P 2 + P 3 +.

24 The Combined Gas Law P 1 V 1 P 2 V 2 T 1 T 2 The Ideal Gas Law PV = nrt R = universal gas constant = 0.0821 L atm mol K (will be given) Avogadro s Hypothesis and the Molar Volume V a n a V b n b At STP (0 o C & 1 atm) Molar volume of any gas = 22.4 L/mol 1 atm = 760 mm Hg = 760 torr

25 Assuming a diver has a lung capacity of 5.0 L and is at a depth of 66 ft (with water temperature of 15 o C), what would be the volume of air in her lung if she returns to the surface while holding her breath? Surface temperature is 27 o C and the atmospheric pressure is 1.0 atm. Every 33 ft water depth = 1 atm. V 1 = 5.0 L V 2 =? P 1 = (66 ft x 1 atm/33 ft) + 1.0 atm = 3.0 atm P 2 = 1 atm T 1 = 15 o C T 2 = 27 o C P 1 V 1 P 2 V 2 T 1 T 2 5.0 L x 3.0 atm = V 2 x 1.0 atm 15 + 273 K 27 + 273 K V 2 = 16 L

26 A flask has V = 125 ml and contains 0.911 g of gas. T = 20.0 o C; P = 75.0 cm Hg; R = 0.0821 L atm mol -1 K -1 What is the molecular weight of the gas? M.W. =? g/mol PV = nrt P = 75.0 cm Hg/76 cm Hg = 0.987 atm V = 125 ml = 0.125 L T = 20 + 273 K = 293 K n = PV = 0.987 atm x 0.125 L RT 0.0821 L atm x 293 K mol K = 0.00513 mole M.W. = 0.911 g = 178 g/mol 0.00513 mol

27 Indicators of Strong Intermolecular Forces in Liquids Low Vapor Pressure High Heat of Vaporization High Viscosity High Surface Tension Intermolecular Forces in Liquids = Van der Waals Forces London Forces Dipole-Dipole Interactions Hydrogen Bonds Account for the following differences in boiling points: CH 4 NH 3 H 2 O HF b.p. -161-33 100-19.5 o C

28 Arrange these molecules in order of: 1. Increasing attractive forces 2. Increasing boiling point 3. Decreasing melting point 4. Decreasing vapor pressure 5. Increasing molar heat of vaporization

29 Chapter 8 Solutions homogeneous mixture of solvent and one or more solutes saturated supersaturated unsaturated concentrated dilute Factors affecting solubility Nature of solute & solvent Temperature Pressure (in case of gases) When salts dissolve in water, disorder/entropy increases.

30 Like Dissolves Like Molecules that mix in all proportions are miscible. CH 3 CH 2 -OH + H 2 O are miscible because of hydrogen-bonding. CH 3 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -OH Hydrophobic portion (nonpolar), no interaction with water Predict the solubilities in water: Hydrophilic polar & attracted to water NH 3 CH 4 CCl 4 LiNO 3