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Chapter 4 Chemical Quantities and Aqueous Reactions

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

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Chapter 4 Chemical Quantities and Aqueous Reactions

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

  2. 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 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) • 2 molecules of C8H18 react with 25 molecules of O2 • to form 16 molecules of CO2 and 18 molecules of H2O • 2 moles of C8H18 react with 25 moles of O2 • to form 16 moles of CO2 and 18 moles of H2O • 2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O Tro, Chemistry: A Molecular Approach

  3. 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 CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18? 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) 2 moles C8H18 : 16 moles CO2 Tro, Chemistry: A Molecular Approach

  4. g C8H18 mol C8H18 mol CO2 g CO2 Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline • assuming that gasoline is octane, C8H18, the equation for the reaction is: 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) • the equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the Concept Plan will be: Tro, Chemistry: A Molecular Approach

  5. g C8H18 mol C8H18 mol CO2 g CO2 Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline (C8H18) Given: Find: 3.4 x 1015 g C8H18 g CO2 Concept Plan: Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2 Solution: Check: since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

  6. Practice • According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l) • convert 9.0 g of glucose into moles (MM 180) • convert moles of glucose into moles of water • convert moles of water into grams (MM 18.02) • convert grams of water into mL • How? what is the relationship between mass and volume? density of water = 1.00 g/mL Tro, Chemistry: A Molecular Approach

  7. Practice According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C6H12O6(s) + 6 O2(g)® 6 CO2(g) + 6 H2O(l) Tro, Chemistry: A Molecular Approach

  8. 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 Approach

  9. 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 Approach

  10. O O O O H H H H + + + C C + H H O O O O H H Limiting and Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) • Our balanced equation for the combustion of methane implies that every 1 molecule of CH4 reacts with 2 molecules of O2 Tro, Chemistry: A Molecular Approach

  11. O O O O O O O O H H H H H H H H H H C C C C C ? H H H H H + H H H H H O O O O O O O O Limiting and Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) • If we have 5 molecules of CH4 and 8 molecules of O2, which is the limiting reactant?

  12. + O O O O O O O O H H H H H H H H H H C C C C C H H H H H H H H H H O O O O O O O O Limiting and Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) since less CO2 can be made from the O2 than the CH4, the O2 is the limiting reactant

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

  14. Example: When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield. Tro, Chemistry: A Molecular Approach

  15. Write down the given quantity and its units. Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced Example:When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Tro, Chemistry: A Molecular Approach

  16. Write down the quantity to find and/or its units. Find: limiting reactant theoretical yield percent yield Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Tro, Chemistry: A Molecular Approach

  17. Write a Concept Plan: Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. } kg C g C mol C mol Ti smallest amount is from limiting reactant kg TiO2 g TiO2 mol TiO2 mol Ti g Ti kg Ti T.Y. % Yield Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) smallest mol Ti

  18. Collect Needed Relationships: 1000 g = 1 kg Molar Mass TiO2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO2: 1 mol Ti (from the chem. equation) 2 mole C  1 mol Ti (from the chem. equation) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)

  19. Apply the Concept Plan: Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Limiting Reactant smallest moles of Ti Tro, Chemistry: A Molecular Approach

  20. Apply the Concept Plan: Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Theoretical Yield Tro, Chemistry: A Molecular Approach

  21. Apply the Concept Plan: Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Tro, Chemistry: A Molecular Approach

  22. Check the Solutions: Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Limiting Reactant = TiO2 Theoretical Yield = 52.9 kg Percent Yield = 80.9% Since Ti has lower molar mass than TiO2, the T.Y. makes sense The Percent Yield makes sense as it is less than 100%. Tro, Chemistry: A Molecular Approach

  23. Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) Tro, Chemistry: A Molecular Approach

  24. g NH3 g CuO mol NH3 mol CuO mol N2 mol N2 smallest moles N2 g N2 Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Concept Plan: Relationships: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g 2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2

  25. Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) Check: There are fewer mols of N2 produced from CuO than from NH3, therefore CuO is the limiting reagent.

  26. 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 Approach

  27. 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 Approach

  28. 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 Approach

  29. Solution ConcentrationMolarity • 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 Approach

  30. Preparing 1 L of a 1.00 M NaCl Solution Tro, Chemistry: A Molecular Approach

  31. g KBr mol KBr M L sol’n Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution • Sort Information Given: Find: 25.5 g KBr, 1.75 L solution Molarity, M • Strategize Concept Plan: Relationships: 1 mol KBr = 119.00 g, M = moles/L • Follow the Concept Plan to Solve the problem Solution: • Check since most solutions are between 0 and 18 M, the answer makes sense Check:

  32. 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 Tro, Chemistry: A Molecular Approach

  33. mol NaOH L sol’n Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH? • Sort Information Given: Find: 0.125 M NaOH, 0.255 mol NaOH liters, L • Strategize Concept Plan: Relationships: 0.125 mol NaOH = 1 L solution • Follow the Concept Plan to Solve the problem Solution: • Check since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L Check:

  34. 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 M1∙V1 = M2∙V2 Tro, Chemistry: A Molecular Approach

  35. V1, M1, M2 V2 Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH? • Sort Information Given: Find: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M V2, L • Strategize Concept Plan: Relationships: M1V1 = M2V2 • Follow the Concept Plan to Solve the problem Solution: • Check since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does Check:

  36. 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 Tro, Chemistry: A Molecular Approach

  37. L Pb(NO3)2 mol Pb(NO3)2 mol KCl L KCl Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq) • Sort Information Given: Find: 0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 L KCl • Strategize Concept Plan: Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 = 2 mol KCl • Follow the Concept Plan to Solve the problem Solution: • Check since need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x volume Pb(NO3)2 Check:

  38. 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

  39. 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 Approach

  40. 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 Approach

  41. 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 Approach

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

  43. 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 Approach

  44. 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 Na2SO4(aq) 2 Na+(aq) + SO42-(aq) HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) Tro, Chemistry: A Molecular Approach

  45. Classes of Dissolved Materials Tro, Chemistry: A Molecular Approach

  46. 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 Approach

  47. 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 Approach

  48. Solubility RulesCompounds that Are Generally Soluble in Water Tro, Chemistry: A Molecular Approach

  49. Solubility RulesCompounds that Are Generally Insoluble Tro, Chemistry: A Molecular Approach

  50. 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 Tro, Chemistry: A Molecular Approach

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