Chapter 7 Stoichiometry of Chemical Reactions Chapter 7 Equations and Stoichiometry 1 Learning Objectives 7.1 •Derive chemical equations from narrative descriptions of chemical reactions. •Write and balance chemical equations in molecular, total ionic, and net ionic formats. 7.2 •Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction) •Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations •Identify common acids and bases •Predict the solubility of common inorganic compounds by using solubility rules •Compute the oxidation states for elements in compounds Chapter 7 Equations and Stoichiometry 2 7.3 •Explain the concept of stoichiometry as it pertains to chemical reactions •Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products •Perform stoichiometric calculations involving mass, moles, and solution molarity 7.4 •Explain the concepts of theoretical yield and limiting reactants/reagents. •Derive the theoretical yield for a reaction under specified conditions. •Calculate the percent yield for a reaction. 7.5 •Describe the fundamental aspects of titrations and gravimetric analysis. •Perform stoichiometric calculations using typical titration and gravimetric data. Chapter 7 Equations and Stoichiometry 3 Chemical Equations reactants + "reacts with" products → "to produce” coefficients - indicate amount of substance Chapter 7 Equations and Stoichiometry 4 Balancing Equations Equations must be balanced. Equal amounts of each element on each side of the equation. 4H 2O 4H, 2O • When balancing an equation, subscripts should never change, as this changes the chemical identity. • Changing coefficients only changes amount of a substance, not the identity. Chapter 7 Equations and Stoichiometry 5 Chapter 7 Equations and Stoichiometry 6 Guidelines: • write unbalanced equation • use coefficients to indicate how many formula units are required to balance equation • balance those species that occur in the fewest formulas on each side. • reduce coefficients to smallest whole number values • When balancing reactions involving organic compounds, balance in the order: C, H, O KClO3 ® KCl + Chapter 7 Equations and Stoichiometry O2 7 Chapter 7 Equations and Stoichiometry 8 Chapter 7 Equations and Stoichiometry 9 There are three major types of chemical processes occurring in aqueous solutions: Precipitation reactions Acid-base reactions Redox reactions Chapter 7 Equations and Stoichiometry 10 Precipitation Reactions (a type of double-displacement reaction) Reactions that result in the formation of an insoluble product are known as precipitation reactions. A precipitate is an insoluble solid formed by a reaction in solution. AgNO3 (aq) + KBr (aq) → AgBr(s) + KNO3 (aq) This is an example of a metathesis reaction (also called at double replacement), a reaction that involves the exchange of parts between the two compounds. They exchanged anions. Chapter 7 Equations and Stoichiometry 11 Ionic Equations Molecular equation – shows complete chemical formulas of the reactants and products. AgNO3(aq) + KCl(aq)→ AgCl(s) + KNO3(aq) Complete ionic equation or ionic equation Shows dissolved species as free ions. Ag+(aq) + NO3–(aq) + K+ (aq) + Cl– (aq) → AgCl(s) + K+ (aq) + NO3– (aq) Spectator ions – present but play no role in the reaction. NO3– , K+. Chapter 7 Equations and Stoichiometry 12 Net ionic equation Shows only the species that actually take part in the reaction. Ag+ (aq) + Cl–(aq) → AgCl(s) 1. Write a balanced molecular equation for the reaction. 2. Rewrite the equation to show the ions that form in solution when each soluble strong electrolyte dissociates or ionizes into its component ions. • Only dissolved strong electrolytes are written in ionic form. Na+(aq) + Cl–(aq) NH3(aq) HF(aq) 3. Identify and cancel spectator ions that occur on both sides of the equation. Chapter 7 Equations and Stoichiometry 13 Solubility Guidelines for Ionic Compounds The solubility of a substance is the amount of that substance that can be dissolved in a given quantity of solvent. Solubility Guidelines A compound is probably soluble if it contains one of the following cations: • Group 1A cation: Li+, Na+, K+, Rb+, Cs+ • Ammonium ion: NH4+ A compound is probably soluble if it contains one of the following anions: • Halides: Cl–, Br–, I– • Except Ag+, Hg22+, and Pb2+ compounds. • Nitrate (NO3–), perchlorate (ClO4–), acetate (CH3CO2–), • Sulfate (SO42–) • Except Ba2+, Hg22+, and Pb2+. A compound is probably insoluble if it contains one of the following anions: • Carbonate (CO32–), phosphate (PO43–), chromate (CrO42–), sulfides (S2–) • Except compounds containing alkali metal (Group 1A) ions and the ammonium ion. • Hydroxides (OH–) • Except compounds containing alkali metal (Group 1A) ions and the Ba2+ ion. Chapter 7 Equations and Stoichiometry 14 Chapter 7 Equations and Stoichiometry 15 From left to right: AgI, AgBr, AgCl, and AgF Chapter 7 Equations and Stoichiometry 16 Acid-Base Reactions Acids • Have a sour taste • Cause litmus to change from blue to red • React with certain metals to produce hydrogen gas • React with carbonates and bicarbonates to produce carbon dioxide. Bases • Have a bitter taste • Fell slippery. E.g.: soap • Cause litmus to change from red to blue Chapter 7 Equations and Stoichiometry 17 Acids are substances that are able to ionize in aqueous solutions to form a hydrogen ion (H+) and thereby increase the concentration of H+(aq) ions. (Arrhenius definition) Bronsted acid - proton donor. Monoprotic acid - each unit of acid yields one hydrogen ion. HCl → H3O+(aq) + Cl–(aq) Chapter 7 Equations and Stoichiometry 18 Polyprotic acids Diprotic acid - each unit of acid gives up two H+ ions, in two separate steps. H2SO4 → H3O+(aq) + HSO4–(aq) HSO4–(aq) → H3O+(aq) + SO42–(aq) Triprotic acids - yield three H+ ions H3PO4 → H3O+(aq) + H2PO4–(aq) H2PO4–(aq) → H3O+(aq) + HPO42–(aq) HPO42–(aq) → H3O+(aq) + PO43–(aq) Chapter 7 Equations and Stoichiometry 19 Chapter 7 Equations and Stoichiometry 20 Bases Bases increase the concentration of OH–(aq) ions in water. (Arrhenius definition) Bases are substances that accept (react with) H+ ions. (Brønsted) LiOH(s) + H2O(l) → Li+(aq) + OH–(aq) NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq) Chapter 7 Equations and Stoichiometry 21 Strong and Weak Acids and Bases Acids and bases that are strong electrolytes are called strong acids and strong bases. Those that are weak electrolytes are weak acids and weak bases. Common Strong Acids Weak Acids are incompletely deprotonated in solution. Ex: HF, HNO2, H2SO3, H3PO4, CH3COOH (acetic acid) Chapter 7 Equations and Stoichiometry 22 Bases Strong bases: all Group 1A hydroxides, (LiOH, NaOH, KOH, RbOH, CsOH) and Ba(OH)2. Ca(OH)2 and Sr(OH)2 are slightly soluble. Be(OH)2 and Mg(OH)2 are insoluble. Weak bases: The most common weak base is NH3. Most other weak bases are derivatives of ammonia called amines. Ex: (CH3)3N, C5H5N, C6H5NH2. NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq) Chapter 7 Equations and Stoichiometry 23 Neutralization Reactions and Salts A neutralization reaction occurs when a solution of an acid and that of a base are mixed to produce a salt (and water if the base is strong). Another type of double displacement reaction. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) acid base water salt Chapter 7 Equations and Stoichiometry 24 Pay attention to stoichiometry HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) 2 HCl(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaCl2(aq) H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq) Weak acid and strong base HCN(aq) + NaOH(aq) → H2O(l) +NaCN(aq) Weak base and strong acid HNO3(aq) + NH3(aq) → NH4NO3(aq) In solution, NH3 reacts with water to produce NH4+ and OH–. HNO3(aq) + NH4+(aq) + OH–(aq) → H2O(l) + NH4NO3(aq) Chapter 7 Equations and Stoichiometry 25 Oxidation –Reduction Reactions • Loss of electrons by a substance is called oxidation. • The gain of electrons by a substance is reduction. • Oxidation of one substance is always accompanied by the reduction of another as electrons are transferred between them. LEO-GER Loss of Electrons-Oxidation Gain of Electrons-Reduction OIL-RIG Oxidation Is Loss of electrons Reduction Is Gain of electrons Chapter 7 Equations and Stoichiometry 26 Oxidation Numbers The oxidation number of an atom in a substance is the charge an atom would have in a molecule (or an ionic compound) if the electrons were transferred completely. • For an atom in its elemental form, the oxidation number is always zero. • For any monatomic ion, the oxidation number equals the charge on the ion. • Nonmetals usually have negative oxidation numbers, although they can sometimes be positive. • The oxidation number of oxygen is usually –2 in both ionic and molecular compounds. (The major exception is in compounds called peroxides, which contain O22- ion, giving each oxygen an oxidation number of –1.) • The Oxidation number of hydrogen is +1 when bonded to nonmetals and –1 when bonded to metals. • The oxidation number of fluorine is –1 in all compounds. The other halogens have an oxidation number of –1 in most binary compounds. When combined with oxygen, as in oxyanions, they have positive oxidation states. Chapter 7 Equations and Stoichiometry 27 • The sum of the oxidation numbers of all atoms in a neutral compound is zero. • The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion. Oxidation numbers can be determined by setting the sum of the oxidation numbers to equal the charge on the molecule or ion. Chapter 7 Equations and Stoichiometry 28 Chapter 7 Equations and Stoichiometry 29 Maximum Oxidation number = Group number Minimum Oxidation number = Group number – 8 – NO3 Oxidation number of N is +5. NH3 Oxidation number of N is –3. N in NO3– – Cl in ClO3 N in NH3 Mn in KMnO4 Chapter 7 Equations and Stoichiometry 30 Types of Redox Reactions • Combination reactions A + B → AB C(s) + O2(g) → CO2(g) 3Mg(s) + N2(g) → Mg3N2(g) • Decomposition Reactions AB → A + B 2KClO3(s) → 2KCl(s) + 3O2(g) Substances that are oxidized are reducing agents. Substances that are reduced are oxidizing agents. Chapter 7 Equations and Stoichiometry 31 Chapter 7 Equations and Stoichiometry 32 Method of Half-Reactions Treat oxidation and reduction as separate reactions. C(s) + O2(g) → CO2(g) 3Mg(s) + N2(g) → Mg3N2(g) Chapter 7 Equations and Stoichiometry 33 • Disproportionation Reaction When one substance is oxidized and reduced in the same reaction. Cl2 + H2O → HCl + HOCl Hg2S → Hg(l) + HgS(s) Chapter 7 Equations and Stoichiometry 34 Displacement Reactions Metal displacement (Single Displacement) Oxidation of Metals by Acids and Salts A + BX → AX + B Zn(s) + 2HBr(aq) → ZnBr2(aq) + H2(g) Mn(s) + Pb(NO3)2(aq) → Mn(NO3)2(aq) + Pb(s) These reactions are called displacement reactions because the ion in solution is displaced or replaced through oxidation of an element. Whenever one substance is oxidized, some other substance must be reduced. Chapter 7 Equations and Stoichiometry 35 Chapter 7 Equations and Stoichiometry 36 Chapter 7 Equations and Stoichiometry 37 Half-Reactions of Displacement Reactions Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) Chapter 7 Equations and Stoichiometry 38 Hydrogen Displacement Alkali metals and some alkaline earth metals (Ca, Sr, and Ba) react with water to produce hydrogen gas. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Some metals that don’t react with water (e.g. Fe, Zn, Mg) will react with acid to produce hydrogen gas. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Halogen Displacement (the higher element ends up as the anion). F2 > Cl2 > Br2 > I2 Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l) Cl2(g) + 2KF(aq) → no reaction Chapter 7 Equations and Stoichiometry 39 Chemical Symbols on Different Levels • Chemical symbols represent both a microscopic and macroscopic level. 2 H2(g) + O2(g) ® 2 H2O(g) 2 molecules 1 molecule 2 molecules 2(6x1023 molecules) 6x1023 molecules 2(6x1023 molecules) 2 moles 1 mole 2 moles The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules (or formula units) involved in the reaction and as the relative numbers of moles. Chapter 7 Equations and Stoichiometry 40 Stoichiometry: Chemical Arithmetic • 2 mol H2 ≏ 1 mol O2 ≏ 2 mol H2O • They are stoichiometrically equivalent. Chapter 7 Equations and Stoichiometry 41 Examples How many grams of H2O can be produced from 40.0 g O2? Chapter 7 Equations and Stoichiometry 42 Combustion of butane, C4H10. If we burn 1.00 g of butane, what mass of CO2 is produced? Chapter 7 Equations and Stoichiometry 43 Chapter 7 Equations and Stoichiometry 44 Reactions with Limiting Amounts of Reactants • The reagent that is completely consumed in a reaction is called the limiting reactant or limiting reagent, because it determines, or limits, the amount of product formed. • The reagent that gives the lesser amount of product is the limiting reagent. • The amount it produces is the maximum amount of product. • The other reactants are sometimes called excess reactants or excess reagents. Chapter 7 Equations and Stoichiometry 45 A simpler explanation. Suppose we’re making cheese sandwiches. The rule is one slice of cheese for two slices of bread to make one sandwich. 1 slice cheese + 2 slices Bread → 1 Sandwich We have six slices of cheese and twenty slices of bread. How many sandwiches can we make? Chapter 7 Equations and Stoichiometry 46 Example Assume that 5.52 g of sodium reacts with 5.10 g of aluminum oxide. a) Which reactant is limiting, and which reactant is in excess? b) How many grams of aluminum are produced? c) What mass of excess reactant remains at the end of the reaction? Always make sure the chemical equation is balanced! Na(l) + Al2O3(s) ® Al(l) + Chapter 7 Equations and Stoichiometry Na2O(s) 47 Chapter 7 Equations and Stoichiometry 48 Chapter 7 Equations and Stoichiometry 49 Yields of Chemical Reactions • The quantity of product that is calculated to form when all of the limiting reactant reacts is called the theoretical yield. It is the maximum quantity that can be expected on the basis of the stoichiometry of a chemical equation. • The amount actually obtained in a reaction is called the actual yield. • The percent yield of a reaction relates the actual yield to the theoretical (calculated) yield: • The percent yield is the percentage of the theoretical yield that is actually achieved. Chapter 7 Equations and Stoichiometry 50 Example: Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by controlled reaction between cyclohexane (C6H12; mw = 84.14) and O2. 2 C6H12 + 5 O2 ® 2 H2C6H8O4 + 2 H2O Assume you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reagent. a) What is the theoretical yield of adipic acid (mw = 146.14)? b) If you obtain 37.6 g of adipic acid from your reaction, what is the percent yield of adipic acid? Chapter 7 Equations and Stoichiometry 51 Chapter 7 Equations and Stoichiometry 52 How many moles of O₂ would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 83.1% yield? 2 NO (g) + O₂ (g) → 2 NO₂ (g) Chapter 7 Equations and Stoichiometry 53 Learning Objectives By the end of this section, you will be able to: • Describe the fundamental aspects of titrations and gravimetric analysis. • Perform stoichiometric calculations using typical titration and gravimetric data. Chapter 7 Equations and Stoichiometry 54 Gravimetric Analysis - an analytical technique based on the measurement of mass. Ex: Calculate the concentration of an aqueous KCl solution if 25.00 mL of the solution gives 0.430 g of AgCl when treated with excess AgNO3. Chapter 7 Equations and Stoichiometry Acid-Base Titrations Titration is a procedure for determining the concentration of a solution (analyte) using another solution of known concentration, called a standard solution. Chapter 7 Equations and Stoichiometry Example: 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfuric acid solution? Write a balanced equation: Chapter 7 Equations and Stoichiometry NaOH + HCl à NaOH + H2SO4 à Ba(OH)2 + HBr à KOH + H3PO4 à Chapter 7 Equations and Stoichiometry