# Chapter 4 Chemical Quantities and Aqueous Reactions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts.

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Chapter 4 Chemical Quantities and Aqueous Reactions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Tro, Chemistry: A Molecular Approach2 Reaction Stoichiometry the numerical relationships between chemical amounts in a reaction is called stoichiometry the coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 molecules of C 8 H 18 react with 25 molecules of O 2 to form 16 molecules of CO 2 and 18 molecules of H 2 O 2 moles of C 8 H 18 react with 25 moles of O 2 to form 16 moles of CO 2 and 18 moles of H 2 O 2 mol C 8 H 18 : 25 mol O 2 : 16 mol CO 2 : 18 mol H 2 O

Tro, Chemistry: A Molecular Approach3 Predicting Amounts from Stoichiometry the amounts of any other substance in a chemical reaction can be determined from the amount of just one substance How much CO 2 can be made from 22.0 moles of C 8 H 18 in the combustion of C 8 H 18 ? 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 moles C 8 H 18 : 16 moles CO 2

Tro, Chemistry: A Molecular Approach4 Example – Estimate the mass of CO 2 produced in 2004 by the combustion of 3.4 x 10 15 g gasoline assuming that gasoline is octane, C 8 H 18, the equation for the reaction is: 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) the equation for the reaction gives the mole relationship between amount of C 8 H 18 and CO 2, but we need to know the mass relationship, so the Concept Plan will be: g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18

Tro, Chemistry: A Molecular Approach5 Practice According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O(l) 1.convert 9.0 g of glucose into moles (MM 180) 2.convert moles of glucose into moles of water 3.convert moles of water into grams (MM 18.02) 4.convert grams of water into mL a)How? what is the relationship between mass and volume? density of water = 1.00 g/mL

Tro, Chemistry: A Molecular Approach6 Limiting Reactant for reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others when this reactant is used up, the reaction stops and no more product is made the reactant that limits the amount of product is called the limiting reactant sometimes called the limiting reagent the limiting reactant gets completely consumed reactants not completely consumed are called excess reactants the amount of product that can be made from the limiting reactant is called the theoretical yield

Tro, Chemistry: A Molecular Approach7 Things Don’t Always Go as Planned! many things can happen during the course of an experiment that cause the loss of product the amount of product that is made in a reaction is called the actual yield generally less than the theoretical yield, never more! the efficiency of product recovery is generally given as the percent yield

Tro, Chemistry: A Molecular Approach8 Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) Our balanced equation for the combustion of methane implies that every 1 molecule of CH 4 reacts with 2 molecules of O 2 H H C H H + O O C + OO OO + O HH O HH +

9 Limiting and Excess Reactants in the Combustion of Methane If we have 5 molecules of CH 4 and 8 molecules of O 2, which is the limiting reactant? H H C H H + OO OO OO OO OO OO OO OO ? H H C H H H H C H H H H C H H H H C H H CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)

Example 4.4 Finding Limiting Reactant, Theoretical Yield, and Percent Yield

Tro, Chemistry: A Molecular Approach11 Example: When 28.6 kg of C are allowed to react with 88.2 kg of TiO 2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield.

Tro, Chemistry: A Molecular Approach12 Practice – How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l)

Tro, Chemistry: A Molecular Approach13 Solutions when table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can tell from the taste, or simply boiling away the water homogeneous mixtures are called solutions the component of the solution that changes state is called the solute the component that keeps its state is called the solvent if both components start in the same state, the major component is the solvent

Tro, Chemistry: A Molecular Approach14 Describing Solutions since solutions are mixtures, the composition can vary from one sample to another pure substances have constant composition salt water samples from different seas or lakes have different amounts of salt so to describe solutions accurately, we must describe how much of each component is present we saw that with pure substances, we can describe them with a single name because all samples identical

Tro, Chemistry: A Molecular Approach15 Solution Concentration qualitatively, solutions are often described as dilute or concentrated dilute solutions have a small amount of solute compared to solvent concentrated solutions have a large amount of solute compared to solvent quantitatively, the relative amount of solute in the solution is called the concentration

Tro, Chemistry: A Molecular Approach16 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution

Tro, Chemistry: A Molecular Approach17 Preparing 1 L of a 1.00 M NaCl Solution

Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution

Tro, Chemistry: A Molecular Approach19 Using Molarity in Calculations molarity shows the relationship between the moles of solute and liters of solution If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar 1 L solution : 2 moles sugar

Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH?

Tro, Chemistry: A Molecular Approach21 Dilution often, solutions are stored as concentrated stock solutions to make solutions of lower concentrations from these stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 the concentrations and volumes of the stock and new solutions are inversely proportional M 1 ∙ V 1 = M 2 ∙ V 2

Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?

Tro, Chemistry: A Molecular Approach23 Solution Stoichiometry since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO 3 ) 2 in the reaction 2 KCl(aq) + Pb(NO 3 ) 2 (aq)  PbCl 2 (s) + 2 KNO 3 (aq)

25 What Happens When a Solute Dissolves? there are attractive forces between the solute particles holding them together there are also attractive forces between the solvent molecules when we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules if the attractions between solute and solvent are strong enough, the solute will dissolve

Tro, Chemistry: A Molecular Approach26 Table Salt Dissolving in Water Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity

Tro, Chemistry: A Molecular Approach27 Electrolytes and Nonelectrolytes materials that dissolve in water to form a solution that will conduct electricity are called electrolytes materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

Tro, Chemistry: A Molecular Approach28 Molecular View of Electrolytes and Nonelectrolytes in order to conduct electricity, a material must have charged particles that are able to flow electrolyte solutions all contain ions dissolved in the water ionic compounds are electrolytes because they all dissociate into their ions when they dissolve nonelectrolyte solutions contain whole molecules dissolved in the water generally, molecular compounds do not ionize when they dissolve in water  the notable exception being molecular acids

Tro, Chemistry: A Molecular Approach29 Salt vs. Sugar Dissolved in Water ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve

Tro, Chemistry: A Molecular Approach30 Acids acids are molecular compounds that ionize when they dissolve in water the molecules are pulled apart by their attraction for the water when acids ionize, they form H + cations and anions the percentage of molecules that ionize varies from one acid to another acids that ionize virtually 100% are called strong acids HCl(aq)  H + (aq) + Cl - (aq) acids that only ionize a small percentage are called weak acids HF(aq)  H + (aq) + F - (aq)

Tro, Chemistry: A Molecular Approach31 Strong and Weak Electrolytes strong electrolytes are materials that dissolve completely as ions ionic compounds and strong acids their solutions conduct electricity well weak electrolytes are materials that dissolve mostly as molecules, but partially as ions weak acids their solutions conduct electricity, but not well when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together Na 2 SO 4 (aq)  2 Na + (aq) + SO 4 2- (aq) HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq)

Tro, Chemistry: A Molecular Approach32 Classes of Dissolved Materials

Tro, Chemistry: A Molecular Approach33 Solubility of Ionic Compounds some ionic compounds, like NaCl, dissolve very well in water at room temperature other ionic compounds, like AgCl, dissolve hardly at all in water at room temperature compounds that dissolve in a solvent are said to be soluble, while those that do not are said to be insoluble NaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be meaningful

Tro, Chemistry: A Molecular Approach34 When Will a Salt Dissolve? Predicting whether a compound will dissolve in water is not easy The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results we call this method the empirical method

Tro, Chemistry: A Molecular Approach35 Compounds Containing the Following Ions are Generally Soluble Exceptions (when combined with ions on the left the compound is insoluble) Li +, Na +, K +, NH 4 + none NO 3 –, C 2 H 3 O 2 – none Cl –, Br –, I – Ag +, Hg 2 2+, Pb 2+ SO 4 2– Ag +, Ca 2+, Sr 2+, Ba 2+, Pb 2+ Solubility Rules Compounds that Are Generally Soluble in Water

Tro, Chemistry: A Molecular Approach36 Compounds Containing the Following Ions are Generally Insoluble Exceptions (when combined with ions on the left the compound is soluble or slightly soluble) OH – Li +, Na +, K +, NH 4 +, Ca 2+, Sr 2+, Ba 2+ S 2– Li +, Na +, K +, NH 4 +, Ca 2+, Sr 2+, Ba 2+ CO 3 2–, PO 4 3– Li +, Na +, K +, NH 4 + Solubility Rules Compounds that Are Generally Insoluble

Tro, Chemistry: A Molecular Approach37 Precipitation Reactions reactions between aqueous solutions of ionic compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate

38 2 KI(aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq)

Tro, Chemistry: A Molecular Approach39 No Precipitate Formation = No Reaction KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction

Tro, Chemistry: A Molecular Approach40 Process for Predicting the Products of a Precipitation Reaction 1. Determine what ions each aqueous reactant has 2. Determine formulas of possible products Exchange ions  (+) ion from one reactant with (-) ion from other Balance charges of combined ions to get formula of each product 3. Determine Solubility of Each Product in Water Use the solubility rules If product is insoluble or slightly soluble, it will precipitate 4. If neither product will precipitate, write no reaction after the arrow

Tro, Chemistry: A Molecular Approach41 Process for Predicting the Products of a Precipitation Reaction 5. If either product is insoluble, write the formulas for the products after the arrow – writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate 6. Balance the equation

Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 1. Write the formulas of the reactants K 2 CO 3 (aq) + NiCl 2 (aq)  2. Determine the possible products a)Determine the ions present (K + + CO 3 2- ) + (Ni 2+ + Cl - )  b)Exchange the Ions (K + + CO 3 2- ) + (Ni 2+ + Cl - )  (K + + Cl - ) + (Ni 2+ + CO 3 2- ) c)Write the formulas of the products  cross charges and reduce K 2 CO 3 (aq) + NiCl 2 (aq)  KCl + NiCO 3

Tro, Chemistry: A Molecular Approach43 3. Determine the solubility of each product KCl is soluble NiCO 3 is insoluble 4. If both products soluble, write no reaction does not apply since NiCO 3 is insoluble Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride

Tro, Chemistry: A Molecular Approach44 5. Write (aq) next to soluble products and (s) next to insoluble products K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) 6. Balance the Equation K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride

Tro, Chemistry: A Molecular Approach45 Ionic Equations equations which describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO 3 ) 2 (aq)  2 KNO 3 (aq) + Mg(OH) 2 (s) equations which describe the actual dissolved species are called complete ionic equations aqueous strong electrolytes are written as ions  soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes written in molecule form  solids, liquids, and gases are not dissolved, therefore molecule form 2K +1 (aq) + 2OH -1 (aq) + Mg +2 (aq) + 2NO 3 -1 (aq)  K +1 (aq) + 2NO 3 -1 (aq) + Mg(OH) 2(s)

Tro, Chemistry: A Molecular Approach46 Ionic Equations ions that are both reactants and products are called spectator ions 2K +1 (aq) + 2OH -1 (aq) + Mg +2 (aq) + 2NO 3 -1 (aq)  K +1 (aq) + 2NO 3 -1 (aq) + Mg(OH) 2(s) an ionic equation in which the spectator ions are removed is called a net ionic equation 2OH -1 (aq) + Mg +2 (aq)  Mg(OH) 2(s)

Tro, Chemistry: A Molecular Approach47 Acid-Base Reactions also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) the net ionic equation for an acid-base reaction is H + (aq) + OH  (aq)  H 2 O(l) as long as the salt that forms is soluble in water

Tro, Chemistry: A Molecular Approach48 Acids and Bases in Solution acids ionize in water to form H + ions more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H 3 O +  most chemists use H + and H 3 O + interchangeably bases dissociate in water to form OH  ions bases, like NH 3, that do not contain OH  ions, produce OH  by pulling H off water molecules in the reaction of an acid with a base, the H + from the acid combines with the OH  from the base to make water the cation from the base combines with the anion from the acid to make the salt acid + base  salt + water

Common Acids

Tro, Chemistry: A Molecular Approach50 Common Bases

Tro, Chemistry: A Molecular Approach51 HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

Tro, Chemistry: A Molecular Approach52 Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 1. Write the formulas of the reactants HNO 3 (aq) + Ca(OH) 2 (aq)  2. Determine the possible products a)Determine the ions present when each reactant dissociates (H + + NO 3 - ) + (Ca +2 + OH - )  b)Exchange the ions, H +1 combines with OH -1 to make H 2 O(l) (H + + NO 3 - ) + (Ca +2 + OH - )  (Ca +2 + NO 3 - ) + H 2 O(l) c)Write the formula of the salt cross the charges (H + + NO 3 - ) + (Ca +2 + OH - )  Ca(NO 3 ) 2 + H 2 O(l)

Tro, Chemistry: A Molecular Approach53 3. Determine the solubility of the salt Ca(NO 3 ) 2 is soluble 4. Write an (s) after the insoluble products and a (aq) after the soluble products HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + H 2 O(l) 5. Balance the equation 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide

Tro, Chemistry: A Molecular Approach54 Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6. Dissociate all aqueous strong electrolytes to get complete ionic equation not H 2 O 2 H + (aq) + 2 NO 3 - (aq) + Ca +2 (aq) + 2 OH - (aq)  Ca +2 (aq) + 2 NO 3 - (aq) + H 2 O(l) 7. Eliminate spectator ions to get net-ionic equation 2 H +1 (aq) + 2 OH -1 (aq)  H 2 O(l) H +1 (aq) + OH -1 (aq)  H 2 O(l)

Tro, Chemistry: A Molecular Approach55 Titration often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio the unknown solution is added slowly from an instrument called a burette  a long glass tube with precise volume markings that allows small additions of solution

Tro, Chemistry: A Molecular Approach56 Acid-Base Titrations the difficulty is determining when there has been just enough titrant added to complete the reaction the titrant is the solution in the burette in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity the chemical is called an indicator at the endpoint of an acid-base titration, the number of moles of H + equals the number of moles of OH  aka the equivalence point

Tro, Chemistry: A Molecular Approach57 Titration

Tro, Chemistry: A Molecular Approach58 Titration The base solution is the titrant in the burette. As the base is added to the acid, the H + reacts with the OH – to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.

Tro, Chemistry: A Molecular Approach59 Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Tro, Chemistry: A Molecular Approach60 Gas Evolving Reactions Some reactions form a gas directly from the ion exchange K 2 S(aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K 2 SO 3 (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 SO 3 (aq) H 2 SO 3  H 2 O(l) + SO 2 (g)

Tro, Chemistry: A Molecular Approach61 NaHCO 3 (aq) + HCl(aq)  NaCl(aq) + CO 2 (g) + H 2 O(l)

Tro, Chemistry: A Molecular Approach62 Compounds that Undergo Gas Evolving Reactions Reactant Type Reacting With Ion Exchange Product Decom- pose? Gas Formed Example metal n S, metal HS acidH2SH2SnoH2SH2S K 2 S(aq) + 2HCl(aq)  2KCl(aq) + H 2 S(g) metal n CO 3, metal HCO 3 acidH 2 CO 3 yesCO 2 K 2 CO 3 (aq) + 2HCl(aq)  2KCl(aq) + CO 2 (g) + H 2 O(l) metal n SO 3 metal HSO 3 acidH 2 SO 3 yesSO 2 K 2 SO 3 (aq) + 2HCl(aq)  2KCl(aq) + SO 2 (g) + H 2 O(l) (NH 4 ) n anionbaseNH 4 OHyesNH 3 KOH(aq) + NH 4 Cl(aq)  KCl(aq) + NH 3 (g) + H 2 O(l)

Tro, Chemistry: A Molecular Approach63 Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 1. Write the formulas of the reactants Na 2 CO 3 (aq) + HNO 3 (aq)  2. Determine the possible products a)Determine the ions present when each reactant dissociates (Na +1 + CO 3 -2 ) + (H +1 + NO 3 -1 )  b)Exchange the anions (Na +1 + CO 3 -2 ) + (H +1 + NO 3 -1 )  (Na +1 + NO 3 -1 ) + (H +1 + CO 3 -2 ) c)Write the formula of compounds cross the charges Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + H 2 CO 3

Tro, Chemistry: A Molecular Approach64 3. Check to see either product H 2 S - No 4. Check to see if either product decomposes – Yes H 2 CO 3 decomposes into CO 2 (g) + H 2 O(l) Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + CO 2 (g) + H 2 O(l) Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves

Tro, Chemistry: A Molecular Approach65 5. Determine the solubility of other product NaNO 3 is soluble 6. Write an (s) after the insoluble products and a (aq) after the soluble products Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 (aq) + CO 2 (g) + H 2 O(l) 7. Balance the equation Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 + CO 2 (g) + H 2 O(l) Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves

Tro, Chemistry: A Molecular Approach66 Other Patterns in Reactions the precipitation, acid-base, and gas evolving reactions all involved exchanging the ions in the solution other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions also known as redox reactions many involve the reaction of a substance with O 2 (g) 4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s)

Tro, Chemistry: A Molecular Approach67 Combustion as Redox 2 H 2 (g) + O 2 (g)  2 H 2 O(g)

Tro, Chemistry: A Molecular Approach68 Redox without Combustion 2 Na(s) + Cl 2 (g)  2 NaCl(s) 2 Na  2 Na + + 2 e  Cl 2 + 2 e   2 Cl 

Tro, Chemistry: A Molecular Approach69 Reactions of Metals with Nonmetals consider the following reactions: 4 Na(s) + O 2 (g) → 2 Na 2 O(s) 2 Na(s) + Cl 2 (g) → 2 NaCl(s) the reaction involves a metal reacting with a nonmetal in addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O 2 (g) → 2 Na + 2 O – (s) 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s)

Tro, Chemistry: A Molecular Approach70 Oxidation and Reduction in order to convert a free element into an ion, the atoms must gain or lose electrons of course, if one atom loses electrons, another must accept them reactions where electrons are transferred from one atom to another are redox reactions atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na → Na + + 1 e – oxidation Cl 2 + 2 e – → 2 Cl – reduction Leo Ger

Tro, Chemistry: A Molecular Approach71 Electron Bookkeeping for reactions that are not metal + nonmetal, or do not involve O 2, we need a method for determining how the electrons are transferred chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction even though they look like them, oxidation states are not ion charges!  oxidation states are imaginary charges assigned based on a set of rules  ion charges are real, measurable charges

Tro, Chemistry: A Molecular Approach72 Rules for Assigning Oxidation States rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

Tro, Chemistry: A Molecular Approach73 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3 –, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2

Tro, Chemistry: A Molecular Approach74 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority NonmetalOxidation StateExample FCF 4 H+1CH 4 O-2CO 2 Group 7ACCl 4 Group 6A-2CS 2 Group 5A-3NH 3

Tro, Chemistry: A Molecular Approach75 Practice – Assign an Oxidation State to Each Element in the following Br 2 K + LiF CO 2 SO 4 2- Na 2 O 2

Tro, Chemistry: A Molecular Approach76 Oxidation and Reduction Oxidation and Reduction Another Definition oxidation occurs when an atom’s oxidation state increases during a reaction reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction

Tro, Chemistry: A Molecular Approach77 Oxidation–Reduction oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent

Tro, Chemistry: A Molecular Approach78 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

Tro, Chemistry: A Molecular Approach79 Combustion Reactions Reactions in which O 2 (g) is a reactant are called combustion reactions Combustion reactions release lots of energy Combustion reactions are a subclass of oxidation- reduction reactions 2 C 8 H 18 (g) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g)

Tro, Chemistry: A Molecular Approach80 Combustion Products to predict the products of a combustion reaction, combine each element in the other reactant with oxygen ReactantCombustion Product contains CCO 2 (g) contains HH 2 O(g) contains SSO 2 (g) contains NNO(g) or NO 2 (g) contains metalM2On(s)M2On(s)

Tro, Chemistry: A Molecular Approach81 Practice – Complete the Reactions combustion of C 3 H 7 OH(l) combustion of CH 3 NH 2 (g)

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