Chapter One: CHEMICAL FOUNDATIONS

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 2 Chemistry: An Overview A main challenge of chemistry is to understand the connection between the macroscopic world that we experience and the microscopic world of atoms and molecules. You must learn to think on the atomic level. 1.1

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 3 Atoms vs. Molecules 1.1

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 4 The Fundamental SI Units Physical QuantityName of UnitAbbreviation Masskilogramkg Lengthmeterm Timeseconds TemperaturekelvinK Electric currentampereA Amount of substancemolemol Luminous intensitycandelacd 1.3

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 5 Uncertainty in Measurement A digit that must be estimated is called uncertain. A measurement always has some degree of uncertainty. Record the certain digits and the first uncertain digit (the estimated number). 1.4

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 6 Rules for Counting Significant Figures (continued) Leading zeros do not count as significant figures: has 2 sig figs 1.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 7 Rules for Counting Significant Figures (continued) Trailing zeros are significant only if the number contains a decimal point: has 4 sig figs 150 has 2 sig figs 1.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 8 Rules for Counting Significant Figures (continued) Exact numbers have an infinite number of significant figures:  1 inch = 2.54 cm, exactly  9 pencils (obtained by counting) 1.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 9 Sig Figs in Mathematical Operations For multiplication or division, the number of significant figures in the result is the same as the number in the calculation that has the fewest significant figures: × 5.5 = 

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 10 Sig Figs in Mathematical Operations (continued) For addition or subtraction, the result has the same number of decimal places as the measurement with the fewest number of decimal places: = 

Chapter Two: ATOMS, MOLECULES, AND IONS

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 12 Avogadro’s Hypothesis (1811) At the same temperature and pressure, equal volumes of different gases contain the same number of particles: –5 liters of oxygen –5 liters of nitrogen –Same number of particles! 2.3

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 13 Nuclear Atom Viewed in Cross Section

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 14 The Modern View of Atomic Structure Isotopes: –Atoms with the same number of protons but different numbers of neutrons –Show almost identical chemical properties; chemistry of atom is due to its electrons –In nature most elements contain mixtures of isotopes 2.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 15 Two Isotopes of Sodium 2.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 16 Chemical Bonds Covalent Bonds: –Bonds form between atoms by sharing electrons –Resulting collection of atoms is called a molecule 2.6

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 17 Chemical Bonds Ionic Bonds:  Bonds form due to force of attraction between oppositely charged ions  Ion – atom or group of atoms that has a net positive or negative charge  Cation – positive ion; lost electron(s)  Anion – negative ion; gained electron(s) 2.6

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 18 The Periodic Table Metals vs. Nonmetals Groups or Families – elements in the same vertical columns –Alkali metals, alkaline earth metals, halogens, and noble gases Periods – horizontal rows of elements 2.7

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 19 Polyatomic Ions Must be memorized Examples of compounds containing polyatomic ions: NaOHSodium hydroxide Mg(NO 3 ) 2 Magnesium nitrate (NH 4 ) 2 SO 4 Ammonium sulfate 2.8

Chapter Three: STOICHIOMETRY

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 21 Percent Composition Mass percent of an element: For iron in iron(III) oxide, (Fe 2 O 3 ): 3.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 22 Analyzing for Carbon and Hydrogen Device used to determine the mass percent of each element in a compound. 3.6

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 23 Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals. 3.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 24 Calculating Masses in Reactions Balance the equation. Convert mass to moles. Set up mole ratios from the balanced equation. Calculate number of moles of desired reactant or product. Convert back to grams. 3.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 25 Limiting Reactants Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. 3.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 26 Percent Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. 3.10

Chapter Four: TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 28 Nature of Aqueous Solutions Solute – substance being dissolved Solvent – liquid water Electrolyte – substance that when dissolved in water produces a solution that can conduct electricity 4.2

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 29 Electrolytes Strong Electrolytes – conduct current very efficiently (bulb shines brightly) Weak Electrolytes – conduct only a small current (bulb glows dimly) Nonelectrolytes – no current flows (bulb remains unlit) 4.2

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 30 Molarity Molarity (M) = moles of solute per volume of solution in liters: 4.3

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 31 Dilution The process of adding water to a stock solution to achieve the molarity desired for a particular solution. Dilution with water does not alter the numbers of moles of solute present. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2 4.3

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 32 Precipitation Reaction A double displacement reaction in which a solid forms and separates from the solution. –When ionic compounds dissolve in water, the resulting solution contains the separated ions –Precipitate – the solid that forms 4.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 33 Precipitates Soluble – solid dissolves in solution; (aq) is used in reaction Insoluble – solid does not dissolve in solution; (s) is used in reaction Insoluble and slightly soluble are often used interchangeably 4.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 34 Two Solubility Rules to Memorize 1.nitrate (NO 3  ) salts are soluble. 2.alkali (group 1A) and NH 4 + salts are soluble. 4.5

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 35 Describing Reactions in Solution Formula Equation (Molecular Equation) –Reactants and products as compounds AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) Net Ionic Equation –Show only components that actually react Ag + (aq) + Cl  (aq)  AgCl(s) –Na + and NO 3  are spectator ions 4.6

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 36 Oxidation-Reduction Reactions (Redox Reactions) Reactions in which one or more electrons are transferred. 4.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 37 Rules for Assigning Oxidation States 1.Oxidation state of an atom in an element = 0 2.Oxidation state of monatomic element = charge 3.Oxygen =  2 in covalent compounds (except in peroxides where it =  1) 4.Hydrogen = +1 in covalent compounds 5.Fluorine =  1 in compounds 6.Sum of oxidation states = 0 in compounds 7.Sum of oxidation states = charge of the ion 4.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 38 Exercise Find the oxidation states for each of the elements in each of the following compounds: K 2 Cr 2 O 7 CO 3 2- MnO 2 PCl 5 SF 4 4.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 39 Redox Characteristics Transfer of Electrons Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent 4.9

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 40 Balancing Oxidation-Reduction Reactions Cr 2 O 7 2- (aq) + SO 3 2- (aq)  Cr 3+ (aq) + SO 4 2- (aq) –In acidic solution First Steps: –Separate into half-reactions –Balance elements except H and O 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 41 Method of Half Reactions Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  SO 4 2- (aq) 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 42 Method of Half Reactions Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  SO 4 2- (aq) Now assign oxidation numbers to non- oxygen, non-hydrogen atoms. Add enough electrons to the appropriate side to account for the electrons gained or lost by that element 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 43 Method of Half Reactions (continued) 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) + 2e

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 44 Method of Half Reactions (continued) 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) + 2e - Now balance the charge of both sides of each half reaction by adding the appropriate number of: H + ions in acidic solution or OH- ions in basic solution 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 45 Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) + 7H 2 O H 2 O + SO 3 2- (aq)  SO 4 2- (aq) + 2e - + 2H

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 46 Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) + 7H 2 O H 2 O + SO 3 2- (aq)  SO 4 2- (aq) + 2e - + 2H + Balance “electrons in / electrons out” by multiplying each half reaction by a factor that results in the LCM of the 2 electron values. 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 47 Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2-  2Cr H 2 O 3[H 2 O +SO 3 2-  SO e - + 2H + ] 4.10

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 1 | Slide 48 Method of Half Reactions (continued) 14H + + 6e - + Cr 2 O 7 2-  2Cr H 2 O 3[H 2 O +SO 3 2-  SO e - + 2H + ] After distributing the “3”, recombining ½ reactions and simplifying the H 2 O terms: Final Balanced Equation: Cr 2 O SO H +  2Cr SO H 2 O 4.10