1. Stoichiometry

2. Stoichiometry- mass and quantity relationships among reactants and products in a chemical reaction Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction

3. Stoichiometry How is a balanced equation like a recipe? -A balanced chemical equation provides the same kind of quantitative information

4. Solving Stoichiometry Problems 1.List what you know 2.Set up the problem 3.Estimate and calculate

5. Example An equation can represent the manufacturing of a single tricycle:

6. Example In a five-day workweek, Tiny Tyke is scheduled to make 640 tricycles. How many wheels should be in the plant on Monday morning to make these tricycles? Knowns: Number of tricycles = 640 tricycles = 640 FSW 3 HP 2 F + S + 3W + H + 2P = FSW 3 HP 2

7. Example Two conversion factors: 3 W1FSW 3 HP 2 1FSW 3 HP 2 3 W Calculate: 640 FSW 3 HP 2 X 3W = 1920 W 1 FSW 3 HP 2

8. A balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume. Stoichiometry

9. Number of Atoms

10. Stoichiometry Number of Molecules

11. Stoichiometry Moles

12. Stoichiometry Mass

13. Stoichiometry Volume

14. Stoichiometry Mass and atoms are conserved in every chemical reaction

15. Interpreting a Balanced Chemical Equation Example: Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is: 2H 2 S(g) + 3O 2 (g) --- > 2SO 2 (g) + 2H 2 O(g) Looking at the coefficients, you can determine the relative amounts of molecules or moles of reactants and products

16. Interpreting a Balanced Chemical Equation Example Continued: 2 molecules H 2 S + 3 molecules O 2 -  2 molecules SO 2 + 2 molecules H 2 O 2 mol H 2 S + 3 mol O 2 -  2 mol SO 2 + 2 mol H 2 O

17. Interpreting a Balanced Chemical Equation Example Continued: 2 mol H 2 S + 3 mol O 2  2 mol SO 2 + 2 mol H 2 0 (2 mol X 34.1 g/mol) + (3 mol X 32 g/mol) -> (2 mol X 64.1 g/mol) + (2 mol X 18.0 g/mol) 68.2 g H 2 S + 96.0 g O 2 -> 128.2 g SO 2 + 36.0 g H 2 O 164.2 g = 164.2 g

18. Writing and Using Mole Ratios In chemical calculations, mole ratios are used to convert between: - moles of reactant and moles of product, between moles of reactants, or between moles of products

19. Writing and Using Mole Ratios Mole-Mole Calculations – Mole ratio- is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles

20. Writing and Using Mole Ratios To determine the number of moles in a sample of a compound, first measure the mass of the sample Then use the molar mass to calculate the number of moles in that mass

21. Calculating Moles of a Product How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? Known: moles of nitrogen = 0.60 mol N 2 0.60 mol N 2 X 2 mol NH 3 = 1.2 mol NH 3 1 mol N 2 Ratio of 1.2 mol NH 3 to 0.60 mol H 2 is 2:1 N 2 + 3H 2 ---  2NH 3

22. Calculating the Mass of a Product –Mass-Mass Calculations

23. Calculating the Mass of a Product Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is: N 2 (g) + 3H 2 (g) -  2NH 3 (g) Known: Mass of hydrogen = 5.40 g H 2 3 mol H 2 = 2 mol NH 3 (from equation) 1 mol H 2 = 2.0 g H 2 (molar mass) 1 mol NH 3 = 17.0g NH 3 (molar mass)

24. Calculating the Mass of a Product Example Continued: Steps: g H 2  mol H 2  mol NH 3  g NH 3 5.40 g H 2 X 1 mol H 2 X 2 mol NH 3 2.0 g H 2 3 mol H 2 X 17.0g NH 3 = 31 g NH 3 1 mol NH 3

25. Other Stoichiometric Calculations In a typical stoichiometric problem: - the given quantity is first converted to moles - then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance - finally the moles are converted to any other unit of measurement related to the unit mole

26. Other Stoichiometric Calculations Solution Diagram:

27. Other Stoichiometric Calculations Problem-Solving Approach:

28. Calculating Molecules of a Product How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2H 2 O(l)2H 2 (g) + O 2 (g) electricity

29. Calculating Molecules of a Product Example Continued: Known: Mass of water = 29.2 g H 2 O 2 mol H 2 O = 1 mol O 2 (from equation) 1 mol H 2 O = 18.0 g H 2 O (molar mass) 1 mol O 2 = 6.02 X 10 23 molecules O 2 Follow: g H 2 O  mol H 2 O  mol O 2  molecules O 2

30. Calculating Molecules of a Product Example Continued: 29.2 g H 2 O X 1 mol H 2 O X 1 mol O 2 18.0 g H 2 O 2 mol H 2 O X 6.02 X 10 23 molecules O 2 1 mol O 2 = 4.88 X 10 23 molecules O 2

31. Volume-Volume Stoichiometric Calculations Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of Standard Temperature and Pressure. 2NO(g) + O 2 (g) -----  2NO 2 (g)

32. Volume-Volume Stoichiometric Calculations Example Continued: Known: Volume of oxygen = 34 L O 2 2 mol NO 2 / 1 mol O 2 (mol ratio) 1 mol O 2 = 22.4 L O 2 (at STP) 1 mol NO 2 = 22.4 L NO 2 (at STP)

33. Volume-Volume Stoichiometric Calculations Example Continued: 34 L O 2 X 1 mol O 2 X 2 mol NO 2 22.4 L O 2 1 mol O 2 X 22.4 L NO 2 = 68 L NO 2 1 mol NO 2

34. Finding the Volume of a Gas Needed for a Reaction Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO 3 according to this balanced equation? 2SO 2 (g) + O 2 (g) ---  2SO 3 (g) Known: Volume of sulfur trioxide = 20.4 mL 2 mL SO 3 / 1 mL O 2 (volume ratio)

35. Volume-Volume Stoichiometric Calculations Example Continued: 20.4 mL SO 3 X 1 mL O 2 2 mL SO 3 = 10.2 mL O 2

36. Limiting and Excess Reagents In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. – limiting reagent- is the reagent that determines the amount of product that can be formed by a reaction – Excess Reagent- the reagent that is not used up

37. Limiting and Excess Reagents The Chemical Equation for the Preparation of Ammonia:

38. Determining the Limiting Reagent in a Reaction Copper reacts with sulfur to form copper(I) sulfide according to the following equation. 2Cu(s) + S(s) -  Cu 2 S(s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? Known: Mass of copper = 80.0 g Cu Mass of sulfur = 25.0 g S

39. Determining the Limiting Reagent in a Reaction Example Continued: Steps: g Cu --  mol Cu g S ---  mol S mol Cu --  mol S

40. Determining the Limiting Reagent in a Reaction Example Continued: 80.0 g Cu X 1 mol Cu = 1.26 mol Cu 63.5 g Cu 25.0 g S X 1 mol S = 0.779 mol S 32.1 g S 1.26 mol Cu X 1 mol S = 0.630 mol S 2 mol Cu

41. Determining the Limiting Reagent in a Reaction Example Continued: You can compare the amount of sulfur needed (0.630 mol S) with the given amount (0.779 mol S). - indicates sulfur is in excess - copper is the limiting reagent

42. Using a Limiting Reagent to Find the Quantity of a Product What is the maximum number of grams of Cu 2 S that can be formed when 80.0 g Cu reacts with 25.0 g S? 2 Cu(s) + S(s) ---  Cu 2 S(s) Known: Limiting reagent = 1.26 mol Cu 1 mol Cu 2 S = 159.1 g Cu 2 S (molar mass)

43. Using a Limiting Reagent to Find the Quantity of a Product Example Continued: Steps: mol Cu  mol Cu 2 S  g Cu 2 S 1.26 mol Cu X 1 mol Cu 2 S X 159.1 g Cu 2 S 2 mol Cu1 mol Cu 2 S = 1.00 X 10 2 g Cu 2 S = mole ratio = 1 mol Cu 2 S / 2 mol Cu

44. Percent Yield Percent yield is a measure of the efficiency of a reaction carried out in the laboratory – Example: A batting average is actually a percent yield.

45. Percent Yield Theoretical yield- maximum amount of product that could be formed from given amounts of reactants Actual yield- the amount of product that actually forms when the reaction is carried out in the laboratory

46. Percent Yield Percent yield- is the ratio of the actual yield to the theoretical yield expressed as a percent

47. Calculating the Theoretical Yield of a Reaction Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is: CaCO 3 (s)CaO(s) + CO 2 (g) Known: Mass of calcium carbonate=24.8 g CaCO 3 1 mol CaCO 3 = 100.1 g CaCO 3 1 mol CaO = 56.1 g CaO Δ

48. Calculating the Theoretical Yield of a Reaction Example Continued: Steps: g CaCO 3  mol CaCO 3  mol CaO  g CaO 24.8 g CaCO 3 X 1 mol CaCO 3 X 1 mol CaO 100.1 g CaCO 3 1 mol CaCO 3 X 56.1 g CaO= 13.9 g CaO 1 mol CaO

49. Calculating the Percent Yield of a Reaction What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? CaCO 3 (s) CaO(s) + CO 2 (g) Known: Actual yield = 13.1 g CaO Theoretical yield = 13.9 g CaO Δ

50. Calculating the Percent Yield of a Reaction Example Continued: percent yield = actual yieldX 100% theoretical yield Percent yield = 13.1 g CaO X 100% = 94.2% 13.9 g CaO