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Outline 6.1 Chemical Equations ► 6.1 Chemical Equations ► 6.2 Balancing Chemical Equations ► 6.3 Avogadro’s Number and the Mole ► 6.4 Gram–Mole Conversions ► 6.5 Mole Relationships and Chemical Equations ► 6.6 Mass Relationships and Chemical Equations ► 6.7 Percent Yield ► 6.8 Classes of Chemical Reactions ► 6.9 Precipitation Reactions and Solubility Guidelines ► 6.10 Acids, Bases, and Neutralization Reactions ► 6.11 Redox Reactions ► 6.12 Recognizing Redox Reactions ► 6.13 Net Ionic Equations Prentice Hall © 2007 Chapter Six ► Chemical equation: An expression in which symbols are used to represent a chemical reaction. ► Reactant: A substance that undergoes change in a chemical reaction and is written on the left side of the reaction arrow in a chemical equation. ► Product: A substance that is formed in a chemical reaction and is written on the right side of the reaction arrow in a chemical equation. 1 ► The numbers and kinds of atoms must be the same on both sides of the reaction arrow. ► Numbers in front of formulas are called coefficients; they multiply all the atoms in a formula. ► The symbol 2 NaHCO3 indicates two units of sodium bicarbonate, which contains 2 Na, 2 H, 2 C, and 6 O. ► Substances involved in chemical reactions may be solids, liquids, gases, or they may be in solution. ► This information is added to an equation by placing the appropriate symbols after the formulas: ► Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq) Prentice Hall © 2007 Chapter Six 3 ► A polyatomic ion appearing on both sides of an equation can be treated as a single unit. Chapter Six Chapter Six 2 6.2 Balancing Chemical Equations ► Balancing chemical equations can be done using four basic steps: ► STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. ► STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Prentice Hall © 2007 Chapter Six 4 ► STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. ► The equation: 2 H2SO4 + 4 NaOH → 2 Na2SO4 + 4 H2O is balanced, but can be simplified by dividing all coefficients by 2: H2SO4 + 2 NaOH → Na2SO4 + 2 H2O ► Hint: If an equation contains a pure element as a product or reactant it helps to assign that element’s coefficient last. ► STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. Prentice Hall © 2007 Prentice Hall © 2007 5 Prentice Hall © 2007 Chapter Six 6 1 Symbols Used in Equations Symbols used in chemical equations show -TABLE Chemical Equations are Balanced In a balanced chemical reaction 5.2 • atoms are not gained or lost. • the states of the reactants. • the number of reactant atoms is equal to the number of product atoms. • the states of the products. • the reaction conditions. 7 8 A Balanced Chemical Equation Learning Check In a balanced chemical equation, Determine if each equation is balanced or not. • there must be the same number of each type of atom on the reactant side and on the product side of a balanced equation. A. Na(s) + N2(g) Na3N(s) • numbers called coefficients are used in front of one or more formulas. B. C2H4(g) + H2O(l) C2H5OH(l) Al + 2Al S Al2S3 Not Balanced + 3S Al2S3 2 Al 3S Balanced = 2 Al = 3S 9 10 -Guide to Balancing a Chemical Equation Solution Determine if each equation is balanced or not. A. Na(s) + N2(g) Na3N(s) No. 2 N on reactant side, 1 N on product side. 1 Na on reactant side, 3 Na on product side. B. C2H4(g) + H2O(l) C2H5OH(l) Yes. 2 C = 2C 6H = 6H 1O = 1O -Copyright © 2005 by Pearson Education, Inc. -Publishing as Benjamin Cummings 11 12 2 Steps in Balancing an Equation Balancing Chemical Equations 1. Write the equation with the correct formulas. NH3(g) + O2(g) NO(g) + H2O(g) To balance the following equation, Fe3O4(s) + H2(g) Fe(s) + H2O(l) 2. Determine if the equation is balanced. No, not all atoms are balanced. • work on one element at a time. • use only coefficients in front of formulas. • do not change any subscripts. Fe: Fe3O4(s) + H2(g) 3Fe(s) + H2O(l) O: Fe3O4(s) + H2(g) 3Fe(s) + 4H2O(l) H: Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l) 3. Balance with coefficients in front of formulas. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 13 4. Check that atoms of each element are equal in reactants and products. 4 N (4 x 1 N) = 4 N (4 x 1 N) 12 H (4 x 3 H) = 12 H (6 x 2 H) 10 O (5 x 2 O) = 10 O (4 O + 6 O) Learning Check Learning Check Check the balance of atoms in the following. Balance each equation and list the coefficients in the balanced equation going from reactants to products: Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l) A. __Mg(s) + __N2(g) A. Number of H atoms in products. 1) 2 2) 4 14 3) 8 1) 1, 3, 2 __Mg3N2(s) 2) 3, 1, 2 3) 3, 1, 1 + __Cl2(g) __AlCl3(s) B. Number of O atoms in reactants. 1) 2 2) 4 3) 8 B. __Al(s) C. Number of Fe atoms in reactants. 1) 1 2) 3 3) 4 1) 3, 3, 22) 1, 3, 1 3) 2, 3, 2 15 6.3 Avogadro’s Number and the Mole Solution ► Molecular weight: The sum of atomic weights of all atoms in a molecule. ► Formula weight: The sum of atomic weights of all atoms in one formula unit of any compound. ► Mole: One mole of any substance is the amount whose mass in grams (molar mass) is numerically equal to its molecular or formula weight. ► Avogadro’s number: The number of molecules or formula units in a mole. NA = 6.022 x 1023 A. 3) 3, 1, 1 3Mg(s) + 1N2(g) 1Mg3N2(s) B. 3) 2, 3, 2 2Al(s) 16 + 3Cl2(g) 2AlCl3(s) 17 Prentice Hall © 2007 Chapter Six 18 3 6.4 Gram – Mole Conversions ► Molar mass = Mass of 1 mole of a substance. = Mass of 6.022 x 1023 molecules of a substance. = Molecular (formula) weight of substance in grams. ► Molar mass serves as a conversion factor between numbers of moles and mass. If you know how many moles you have, you can calculate their mass; if you know the mass of a sample, you can calculate the number of moles. Prentice Hall © 2007 19 Chapter Six The molar mass of water is 18.0 g. The conversion factor between moles of water and mass of water is 18.0 g/mol and the conversion factor between mass of water and moles of water is 1 mol/18.0 g: Prentice Hall © 2007 Collection Terms Chapter Six 20 A Mole of Atoms A collection term states a specific number of items. A mole is a collection that contains • the same number of particles as there are carbon atoms in 12.0 g of carbon. • 1 dozen donuts • 6.02 x 1023 atoms of an element (Avogadro’s number). = 12 donuts • 1 ream of paper = 500 sheets • 1 case = 24 cans -Copyright -Publishing © 2005 by Pearson Education, Inc. as Benjamin Cummings 1 mole element Number of Atoms 1 mole C = 6.02 x 1023 C atoms 1 mole Na = 6.02 x 1023 Na atoms 1 mole Au = 6.02 x 1023 Au atoms 22 21 A Mole of a Compound Avogadro’s Number Avogadro’s number 6.02 x 1023 can be written as an equality and two conversion factors. A mole • of a covalent compound has Avogadro’s number of molecules. 1 mole CO2 = 6.02 x 1023 CO2 molecules Equality: 1 mole = 6.02 x 1023 particles 1 mole H2O = 6.02 x 1023 H2O molecules Conversion Factors: 6.02 x 1023 particles and 1 mole • of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 1023 NaCl formula units 1 mole 6.02 x 1023 particles 1 mole K2SO4 = 6.02 x 1023 K2SO4 formula units 23 24 4 Using Avogadro’s Number Using Avogadro’s Number Avogadro’s number is used to convert moles of a substance to particles. Avogadro’s number is used to convert particles of a substance to moles. How many Cu atoms are in 0.50 mole Cu? How many moles of CO2 are in 2.50 x 1024 molecules CO2? 0.50 mole Cu x 6.02 x 1023 Cu atoms 1 mole Cu = 3.0 x 1023 2.50 x 1024 molecules CO2 x 1 mole CO2 6.02 x 1023 molecules CO2 Cu atoms -Copyright -Publishing = 4.15 mole CO2 © 2005 by Pearson Education, Inc. as Benjamin Cummings 25 26 Learning Check Solution 1. The number of atoms in 2.0 mole Al is C. 1.2 x 1024 Al atoms A. 2.0 Al atoms. 2.0 mole Al x 6.02 x 1023 Al atoms 1 mole Al B. 3.0 x 1023 Al atoms. C. 1.2 x 1024 Al atoms. B. 3.0 mole S atoms 2. The number of moles of S in 1.8 x 1024 atoms S is 1.8 x 1024 S atoms x A. 1.0 mole S atoms. B. 3.0 mole S atoms. 1 mole S 6.02 x 1023 S atoms C. 1.1 x 1048 mole S atoms. 27 6.5 Mole Relationships and Chemical Equations The coefficients in a balanced chemical equation tell how many molecules, and thus how many moles, of each reactant are needed and how many molecules, and thus moles, of each product are formed. See the example below: Prentice Hall © 2007 Chapter Six 29 28 ► The coefficients can be put in the form of mole ratios, which act as conversion factors when setting up factor-label calculations. ► In the ammonia synthesis, the mole ratio of H2 to N2 is 3:1, the mole ratio of H2 to NH3 is 3:2, and the mole ratio of N2 to NH3 is 1:2 leading to the following conversion factors: (3 mol H2)/(1 mol N2) (3 mol H2)/(2 mol NH3) (1 mol N2)/(2 mol NH3) Prentice Hall © 2007 Chapter Six 30 5 Subscripts and Moles Factors from Subscripts The subscripts in a formula show • the relationship of atoms in the formula. • the moles of each element in 1 mole of compound. The subscripts are used to write conversion factors for moles of each element in 1 mole compound. For aspirin C9H8O4, the following factors can be written: Glucose C6H12O6 In 1 molecule: 6 atoms C 12 atoms H In 1 mole: 6 mole C 12 mole H 9 mole C 1 mole C9H8O4 and 1 mole C9H8O4 9 mole C 8 mole H 1 mole C9H8O4 4 mole O 1 mole C9H8O4 1 mole C9H8O4 8 mole H 1 mole C9H8O4 4 mole O 6 atoms O 6 mole O 31 32 Learning Check Solution A. How many mole O are in 0.150 mole aspirin C9H8O4? A. How many mole O are in 0.150 mole aspirin C9H8O4? 0.150 mole C9H8O4 x 4 mole O = 0.600 mole O 1 mole C9H8O4 subscript factor B. How many O atoms are in 0.150 mole aspirin C9H8O4? B. How many O atoms are in 0.150 mole aspirin C9H8O4? 0.150 mole C9H8O4 x 4 mole O x 6.02 x 1023 O atoms 1 mole C9H8O4 1 mole O subscript factor 33 = 3.61 x 1023 O atoms 6.6 Mass Relationships and Chemical Equations Avogadro’s Number 34 Molar Mass Mole to mole conversions are carried out using mole ratios as conversion factors. The molar mass • is the mass of one mole of an element or compound. • is the atomic mass expressed in grams. -Copyright -Publishing Prentice Hall © 2007 Chapter Six 35 © 2005 by Pearson Education, Inc. as Benjamin Cummings 36 6 Molar Mass from Periodic Table Learning Check Molar mass is the atomic mass expressed in grams. Give the molar mass for each (to the tenths decimal place). 1 mole Ag 1 mole S - = 107.9 g g = 32.07 g - A. 1 mole K atoms = ________ B. 1 mol Sn atoms = ________ 1 mole C = 12.01 38 37 Solution Molar Mass of K3PO4 Calculate the molar mass of K3PO4. Give the molar mass for each (to the tenths decimal place). A. B. 1 mole K atoms = Element Number of Moles K 3 39.1 g 1 mole Sn atoms = 118.7 g Atomic Mass 39.1 g/mole Total Mass in K3PO4 117.3 g P 1 31.0 g/mole 31.0 g O 4 16.0 g/mole 64.0 g K3PO4 212.3 g 39 40 Some One-mole Quantities -One One--Mole Learning Check Quantities What is the molar mass of each of the following? A. K2O B. Al(OH)3 32.1 g 55.9 g 58.5 g 294.2 g 342.2 g 41 42 7 Solution Molar Mass Factors A. K2O 94.2 g/mole Molar mass conversion factors • are written from molar mass. • relate grams and moles of an element or compound. 2 mole K (39.1 g/mole) + 1 mole O (16.0 g/mole) 78.2 g + 16.0 g = 94.2 g Example: Write molar mass factors for methane, CH4, used in gas stoves and gas heaters. Molar mass: 1 mol CH4 = 16.0 g B. Al(OH)3 78.0 g/mole 1 mole Al (27.0 g/mole) + 3 mol O (16.0 g/mole) + 3 mol H (1.01 g/mole) 27.0 g + 48.0 g + 3.03 g = 78.0 g Conversion factors: 16.0 g CH4 1 mole CH4 and 1 mole CH4 16.0 g CH4 43 44 Learning Check Calculations Using Molar Mass Molar mass factors are used to convert between the grams of a substance and the number of moles. - Grams -Molar mass factor Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g? -Moles 45 46 Solution Grams, Moles, and Particles A molar mass factor and Avogadro’s number convert • grams to particles. Calculate the molar mass of C6H10S. (6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole molar mass Set up the calculation using a mole factor. 225 g C6H10S x 1 mole C6H10S 114.2 g C6H10S g Avogadro’s number mole particles • particles to grams. molar mass factor (inverted) Avogadro’s number = 1.97 mole C6H10S particles 47 molar mass mole g 48 8 Learning Check Solution How many H2O molecules are in 24.0 g H2O? How many H2O molecules are in 24.0 g H2O? 1) 4.52 x 1023 3) 8.02 x 1023 2) 1.44 x 1025 24.0 g H2O x 1 mole H2O x 6.02 x 1023 H2O molecules 18.0 g H2O 1 mole H2O 3) 8.03 x 1023 = 8.03 x 1023 H2O molecules 49 ► Mole to mass and mass to mole conversions are carried out using molar mass as a conversion factor. ► Mass to mass conversions are frequently needed, but cannot be carried out directly. ► Overall, there are four steps for determining mass relationships among reactants and products. Prentice Hall © 2007 Mass to mass conversions: ►STEP 1: Write the balanced chemical equation. ►STEP 2: Choose molar masses and mole ratios to convert known information into needed information. ►STEP 3: Set up the factorlabel expression, and calculate the answer. ►STEP 4: Estimate or check the answer using a ballpark solution. 51 Chapter Six 50 Prentice Hall © 2007 Reading Equations In Moles Consider the following equation: 4 Fe(s) + 3 O2(g) 52 Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for any two substances in an equation. 2 Fe2O3(s) 4Fe(s) + 3O2(g) This equation can be read in “moles” by placing the word “moles” between each coefficient and formula. 4 moles Fe + 3 moles O2 Chapter Six 2Fe2O3(s) Fe and O2 4 moles Fe and 3 moles O2 3 moles O2 4 moles Fe Fe and Fe2O3 4 moles Fe and 2 moles Fe2O3 2 moles Fe2O3 4 moles Fe O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3 2 moles Fe2O3 3 moles O2 2 moles Fe2O3 53 54 9 Learning Check Solution 3H2(g) + N2(g) Consider the following equation: 3 H2(g) + N2(g) 2 NH3(g) A. A mole-mole factor for H2 and N2 is 1) 3 moles N2 2) 1 mole N2 1 mole H2 3 moles H2 B. A mole-mole factor for NH3 and H2 is 1) 1 mole H2 2) 2 moles NH3 2 moles NH3 3 moles H2 2NH3(g) A. A mole-mole factor for H2 and N2 is 2) 1 mole N2 3 moles H2 3) 1 mole N2 2 moles H2 B. A mole-mole factor for NH3 and H2 is 2) 2 moles NH3 3 moles H2 3) 3 moles N2 2 moles NH3 55 56 -Guide Calculations with Mole Factors to Using Mole Factors How many moles of Fe2O3 can form from 6.0 mole O2? 4Fe(s) + 3O2(g) Relationship: 2Fe2O3(s) 3 mole O2 = 2 mole Fe2O3 Write a mole-mole factor to determine the moles of Fe2O3. 6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3 3 mole O2 57 58 Learning Check Solution How many moles of Fe are needed for the reaction of 12.0 moles O2? 3) 16.0 moles Fe 4 Fe(s) 12.0 moles O2 x + 3 O2(g) 2 Fe2O3(s) 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe 59 4 moles Fe = 16.0 moles Fe 3 moles O2 60 10 Steps in Finding the Moles and Masses in a Chemical Reaction Moles to Grams Suppose we want to determine the mass (g) of NH3 that can form from 2.50 moles N2. N2(g) + 3 H2(g) 2 NH3(g) The plan needed would be moles N2 moles NH3 grams NH3 The factors needed would be: mole factor NH3/N2 and the molar mass NH3 61 62 Moles to Grams Learning Check The setup for the solution would be: How many grams of O2 are needed to produce 0.400 mole Fe2O3 in the following reaction? 2.50 mole N2 x 2 moles NH3 x 17.0 g NH3 1 mole N2 1 mole NH3 given mole-mole factor 4 Fe(s) + 3 O2(g) molar mass 2 Fe2O3(s) 1) 38.4 g O2 2) 19.2 g O2 = 85.0 g NH3 3) 1.90 g O2 63 64 6.7 Percent Yield Solution ► The amount of product actually formed in a chemical reaction is somewhat less than the amount predicted by theory. ► Unwanted side reactions and loss of product during handling prevent one from obtaining a perfect conversion of all the reactants to desired products. ► The amount of product actually obtained in a chemical reaction is usually expressed as a percent yield. 2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O2 2 mole Fe2O3 1 mole O2 mole factor molar mass 65 Prentice Hall © 2007 Chapter Six 66 11 ► Percent yield is defined as: (Actual yield ÷ Theoretical yield) x 100% 6.8 Classes of Chemical Reactions ► When learning about chemical reactions it is helpful to group the reactions of ionic compounds into three general classes: precipitation reactions, acid–base neutralization reactions, and oxidation–reduction reactions. ► Precipitation reactions are processes in which an insoluble solid called a precipitate forms when reactants are combined in aqueous solution. ► The actual yield is found by weighing the product obtained. ► The theoretical yield is found by a mass-to-mass calculation. Prentice Hall © 2007 Chapter Six 67 Prentice Hall © 2007 Chapter Six 6.9 Precipitation Reactions and Solubility Guidelines ► Acid–base neutralization reactions are processes in which H+ ions from an acid react with OH- ions from a base to yield water. An ionic compound called a salt is also produced. The “salt” produced need not be common table salt. Any ionic compound produced in an acid–base reaction is called a salt. ► Oxidation–reduction reactions, or redox reactions, are processes in which one or more electrons are transferred between reaction partners (atoms, molecules, or ions). As a result of this transfer, the charges on atoms in the various reactants change. ► Reaction of aqueous Pb(NO3)2 with aqueous KI gives a yellow precipitate of PbI2. ► To predict whether a precipitation reaction will occur on mixing aqueous solutions of two ionic compounds, you must know the solubility of the potential products. Prentice Hall © 2007 Prentice Hall © 2007 Chapter Six 69 If a potential product does not contain at least one of the ions listed below, it is probably not soluble and will precipitate from solution when formed. 68 Chapter Six 70 6.10 Acids, Bases, and Neutralization Reactions ► When acids and bases are mixed together in correct proportion, acidic and basic properties disappear. ► A neutralization reaction produces water and a salt. HA(aq) + MOH(aq) → H2O(l) + MA(aq) acid + base → water + salt ► The reaction of hydrochloric acid with potassium hydroxide to produce potassium chloride is an example: ► HCl(aq) + KOH(aq) → H2O(l) + KCl(aq) Prentice Hall © 2007 Chapter Six 71 Prentice Hall © 2007 Chapter Six 72 12 6.11 Redox Reactions ► Oxidation–reduction (redox) reaction: A reaction in which electrons transfer from one atom to another. ► Oxidation: Loss of one or more electrons by an atom. ► Reduction: Gain of one or more electrons by an atom. Prentice Hall © 2007 Chapter Six 73 Chapter Six Prentice Hall © 2007 Chapter Six 74 6.12 Recognizing Redox Reactions Reducing agent: ► Loses one or more electrons ► Causes reduction ► Undergoes oxidation ► Becomes more positive (or less negative) Oxidizing agent: ► Gains one or more electrons ► Causes oxidation ► Undergoes reduction ► Becomes more negative (or less positive) Prentice Hall © 2007 ► Oxidation and reduction always occur together. ► A substance that is oxidized gives up an electron, causes reduction, and is called a reducing agent. ► A substance that is reduced gains an electron, causes oxidation, and is called an oxidizing agent. ► The charge on the reducing agent increases during the reaction, and the charge on the oxidizing agent decreases. ► One can determine whether atoms are oxidized or reduced in a reaction by keeping track of changes in electron sharing by the atoms. Each atom in a substance is assigned a value called an oxidation number or oxidation state. ► The oxidation number indicates whether the atom is neutral, electron-rich, or electron-poor. ► By comparing the oxidation state of an atom before and after reaction, we can tell whether the atom has gained or lost electrons. 75 ► Rules for assigning oxidation numbers: ► An atom in its elemental state has an oxidation number of zero. Prentice Hall © 2007 Chapter Six 76 ► In a molecular compound, an atom usually has the same oxidation number it would have if it were a monatomic ion. ► Examples: H often has an oxidation number of +1, oxygen often has an oxidation number of -2, halogens often have an oxidation number of -1. ► A monatomic ion has an oxidation number equal to its charge. Prentice Hall © 2007 Chapter Six 77 Prentice Hall © 2007 Chapter Six 78 13 6.13 Net Ionic Equations ► For compounds with more than one nonmetal element, such as SO2, NO, and CO2, the more electronegative element—oxygen in these examples—has its preferred negative oxidation number. ► The less electronegative element is assigned a positive oxidation number so that the sum of the oxidation numbers in a neutral compound is 0. Prentice Hall © 2007 Chapter Six ► Ionic equation: An equation in which ions are explicitly shown. ► Spectator ion: An ion that appears unchanged on both sides of a reaction arrow. ► Net ionic equation: An equation that does not include spectator ions. 79 80 ► The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction. ► Mole ratios relate amounts of reactants and/or products. Using molar masses and mole ratios in factor-label calculations relates unknown masses to known masses or molar amounts. ► The yield is the amount of product obtained. ► The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%. ► Chemical equations must be balanced; the numbers and kinds of atoms must be the same in both the reactants and the products. ► To balance an equation, coefficients are placed before formulas but the formulas themselves cannot be changed. ► A mole refers to Avogadro’s number of formula units of a substance. One mole of any substance has a mass equal to its formula weight in grams. ► Molar masses act as conversion factors between numbers of molecules and masses in grams. Chapter Six Chapter Six Chapter Summary Cont. Chapter Summary Prentice Hall © 2007 Prentice Hall © 2007 81 Prentice Hall © 2007 Chapter Six 82 Chapter Summary Cont. ► Precipitation reactions are processes in which an insoluble solid called a precipitate is formed. ► In acid–base neutralization reactions an acid reacts with a base to yield water plus a salt. ► Oxidation–reduction (redox) reactions are processes in which one or more electrons are transferred between reaction partners. ► Oxidation is the loss of electrons by an atom, and reduction is the gain of electrons by an atom. ► Oxidation numbers are assigned to provide a measure of whether an atom is neutral, electron-rich, or electron-poor. Prentice Hall © 2007 Chapter Six 83 14