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Chapter 4 Chemical Quantities and Aqueous Reactions

Chemistry: A Molecular Approach , 2nd Ed. Nivaldo Tro. Chapter 4 Chemical Quantities and Aqueous Reactions. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. Global Warming. Scientists have measured an average 0.6 °C rise in atmospheric temperature since 1860

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Chapter 4 Chemical Quantities and Aqueous Reactions

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  1. Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro Chapter 4Chemical Quantities and Aqueous Reactions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

  2. Global Warming Scientists have measured an average 0.6 °C rise in atmospheric temperature since 1860 During the same period atmospheric CO2 levels have risen 25% Are the two trends causal? Tro: Chemistry: A Molecular Approach, 2/e

  3. The Sources of Increased CO2 One source of CO2 is the combustion reactions of fossil fuels we use to get energy Another source of CO2 is volcanic action How can we judge whether global warming is natural or due to our use of fossil fuels? Tro: Chemistry: A Molecular Approach, 2/e

  4. Quantities in Chemical Reactions The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction Law of Conservation of Mass Balancing equations by balancing atoms The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry Tro: Chemistry: A Molecular Approach, 2/e

  5. Reaction 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, 2/e

  6. Making Pizza The number of pizzas you can make depends on the amount of the ingredients you use 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza • This relationship can be expressed mathematically • 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza • If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make • assuming you have enough crusts and tomato sauce Tro: Chemistry: A Molecular Approach, 2/e

  7. 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, 2/e

  8. Practice According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C6H12O6 + 6 O2® 6 CO2 + 6 H2O Tro: Chemistry: A Molecular Approach, 2/e

  9. Practice  How many moles of water are made in the combustion of 0.10 moles of glucose? Given: Find: 0.10 moles C6H12O6 moles H2O Conceptual Plan: Relationships: C6H12O6 + 6 O2→ 6 CO2 + 6 H2O 1 mol C6H12O6: 6 mol H2O mol C6H12O6 mol H2O Solution: Check: 0.6 mol H2O = 0.60 mol H2O because 6x moles of H2O as C6H12O6, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

  10. g C8H18 mol C8H18 mol CO2 g CO2 Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasolne • 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 conceptual plan will be Tro: Chemistry: A Molecular Approach, 2/e

  11. g C8H18 mol C8H18 mol CO2 g CO2 Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasoline Given: Find: 3.4 x 1015 g C8H18 g CO2 Conceptual Plan: Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2 Solution: Check: because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

  12. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion? • Our calculation just showed that the world produced 1.1 x 1016 g of CO2 just from petroleum combustion in 2007 • 1.1 x 1013 kg CO2 • Estimates of volcanic CO2 production are 2 x 1011 kg/year • This means that volcanoes produce less than 2% of the CO2 added to the air annually Tro: Chemistry: A Molecular Approach, 2/e

  13. Example 4.1: How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis? g CO2 mol CO2 mol C6H12O6 g C6H12O6 1 mol 180.2 g 1 mol C H O 6 12 6 44.01 g 1 mol 6 mol CO 2 1 mol C H O 180.2 g C H O 1 mol CO    2 6 12 6 6 12 6 3 7.8 g CO 2 44.01 g CO 6 mol CO 1 mol C H O 2 2 6 12 6 = 2 5.8 g C H O 6 12 6 Given: Find: 37.8 g CO2, 6 CO2 + 6 H2O  C6H12O6+ 6 O2 g C6H12O6 Conceptual Plan: Relationships: 1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6: 6 mol CO2 Solution: Check: because 6x moles of CO2 as C6H12O6, but the molar mass of C6H12O6 is 4x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

  14. Practice — How many grams of O2 can be made from the decomposition of 100.0 g of PbO2?2 PbO2(s)→2 PbO(s) + O2(g)(PbO2 =239.2, O2 = 32.00) Tro: Chemistry: A Molecular Approach, 2/e

  15. Practice — How many grams of O2 can be made from the decomposition of 100.0 g of PbO2?2 PbO2(s)→2 PbO(s) + O2(g) g PbO2 mol PbO2 mol O2 g O2 Given: Find: 100.0 g PbO2, 2 PbO2→ 2 PbO + O2 g O2 Conceptual Plan: Relationships: 1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2 Solution: Check: because ½ moles of O2 as PbO2, and the molar mass of PbO2 is 7x O2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

  16. % A(aq) ppm A(aq) % B(aq) ppm B(aq) moles L M = Moles A Moles B Stoichiometry Road Map a A ® b B M A(aq) M B(aq) Solution MM equation MM 22.4 L 22.4 L mass A mass B Pure Substance equation Volume A(g) Volume B(g) density density volume A (l) volume B(l) Tro: Chemistry: A Molecular Approach, 2/e

  17. More Making Pizzas • We know that 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza • But what would happen if we had 4 crusts, • 15 oz. tomato sauce, and 10 cu cheese? Tro: Chemistry: A Molecular Approach, 2/e

  18. More Making Pizzas, Continued Each ingredient could potentially make a different number of pizzas But all the ingredients have to work together! We only have enough tomato sauce to make three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have. Tro: Chemistry: A Molecular Approach, 2/e

  19. More Making Pizzas, Continued The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant. also known as the limiting reagent The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. it also determines the amounts of the other ingredients we will use! Tro: Chemistry: A Molecular Approach, 2/e

  20. The 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, 2/e

  21. 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 one molecule of CH4 reacts with two molecules of O2 Tro: Chemistry: A Molecular Approach, 2/e

  22. Limiting and Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) • If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant? since less CO2 can be made from the O2 than the CH4, sothe O2 is the limiting reactant Tro: Chemistry: A Molecular Approach, 2/e

  23. Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4? Tro: Chemistry: A Molecular Approach, 2/e

  24. Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4? Pick least amount mol N2 mol Si mol Si3N4 mol Si3N4 Limiting reactant and theoretical yield Given: Find: 1.20 mol Si, 1.00 mol N2 mol Si3N4 Conceptual Plan: Relationships: 2 mol N2: 1 Si3N4; 3 mol Si : 1 Si3N4 Solution: Limiting reactant Theoretical yield Tro: Chemistry: A Molecular Approach, 2/e

  25. More Making Pizzas Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield. Tro: Chemistry: A Molecular Approach, 2/e

  26. Theoretical and Actual Yield As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make The theoretical yield will always be the least possible amount of product the theoretical yield will always come from the limiting reactant Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield Tro: Chemistry: A Molecular Approach, 2/e

  27. Example 4.4:Finding limiting reactant, theoretical yield, and percent yield Tro: Chemistry: A Molecular Approach, 2/e

  28. 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, 2/e

  29. 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) • Write down the given quantity and its units Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced Tro: Chemistry: A Molecular Approach, 2/e

  30. 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) • Write down the quantity to find and/or its units Find: limiting reactant theoretical yield percent yield Tro: Chemistry: A Molecular Approach, 2/e

  31. } smallest amount is from limiting reactant kg C kg TiO2 Example:Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) • Write a conceptual plan Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. smallest mol Ti Tro: Chemistry: A Molecular Approach, 2/e

  32. Example:Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) • 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 Tro: Chemistry: A Molecular Approach, 2/e

  33. 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 • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e

  34. 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 • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e

  35. 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 • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e

  36. 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 • Check the solutions limiting reactant = TiO2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the percent yield makes sense as it is less than 100% Tro: Chemistry: A Molecular Approach, 2/e

  37. 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)If 4.61 g of N2 are made, what is the percent yield? Tro: Chemistry: A Molecular Approach, 2/e

  38. g NH3 g CuO mol NH3 mol CuO mol N2 mol 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) If 4.61 g of N2 are made, what is the percent yield? Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Conceptual 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 Choose smallest g N2 Tro: Chemistry: A Molecular Approach, 2/e

  39. 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) If 4.61 g of N2 are made, what is the percent yield? Solution: Theoretical yield because the percent yield is less than 100, the answer makes sense Check: Tro: Chemistry: A Molecular Approach, 2/e

  40. 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, 2/e

  41. Describing Solutions • Because solutions are mixtures, the composition can vary from one sample to another • pure substances have constant composition • saltwater 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 are identical Tro: Chemistry: A Molecular Approach, 2/e

  42. Solution Concentration • Qualitatively, solutions are often described as dilute or concentrated • Dilute solutionshave a small amount of solute compared to solvent • Concentrated solutionshave a large amount of solute compared to solvent Tro: Chemistry: A Molecular Approach, 2/e

  43. Concentrations—Quantitative Descriptions of Solutions • A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution • Concentration = amount of solute in a given amount of solution • occasionally amount of solvent Tro: Chemistry: A Molecular Approach, 2/e

  44. 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, 2/e

  45. Preparing 1 L of a 1.00 M NaCl Solution Tro: Chemistry: A Molecular Approach, 2/e

  46. 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 Given: Find: 25.5 g KBr, 1.75 L solution molarity, M Conceptual Plan: Relationships: 1 mol KBr = 119.00 g, M = moles/L Solution: because most solutions are between 0 and 18 M, the answer makes sense Check: Tro: Chemistry: A Molecular Approach, 2/e

  47. Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution? Tro: Chemistry: A Molecular Approach, 2/e

  48. Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution? Given: Find: 3.4 g NH3, 200.0 mL solution M 0.20 mol NH3, 0.2000 L solution M Conceptual Plan: Relationships: M = mol/L, 1 mol NH3 = 17.03 g, 1 mL = 0.001 L g NH3 mol NH3 M mL sol’n L sol’n Solve: Check: the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters Tro: Chemistry: A Molecular Approach, 2/e

  49. 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, 2/e

  50. mol NaOH L sol’n Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH? Given: Find: 0.125 M NaOH, 0.255 mol NaOH liters, L Conceptual Plan: Relationships: 0.125 mol NaOH = 1 L solution Solution: because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L Check: Tro: Chemistry: A Molecular Approach, 2/e

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