Stoichiometry

1. Stoichiometry Assignments: 11.1 362/38cd, 39cd, 43acde, 40cd 11.2 363/46-48; 363/49,50ab,51 11.3 364/59-62 11.4 365/64 and 66 Chapter 11 Natural Approach to Chemistry

3. Learning Objectives • Apply the mole concept and the law of conservation of mass to calculate quantities of chemicals participating in reactions. • Important terms: stoichiometry, percent yield, actual yield, theoretical yield, limiting reactant

4. Chemical equations tell stories… 2CO(g) + O2(g) → 2CO2(g) … and stories can be put into different categories Nonfiction Science fiction Adventure Romance History Psychology Children’s literature … Synthesis / Decomposition Single / Double replacement Precipitate reaction Polymerization reaction

5. Chemical equations tell stories… But what exactly do they tell us? 2CO(g) + O2(g) 2CO2(g) They tell us what compounds we start with: Carbon monoxide (CO) gas Oxygen (O2) gas what compounds are formed: Carbon dioxide (CO2) gas

6. 2CO(g) + O2(g) 2CO2(g) Chemical equations tell stories… What else do they tell us? 2 CO molecules 1 O2 molecules 2 CO2 molecules They tell us how much of each compound is involved stoichiometry: the study of the amounts of substances involved in a chemical reaction.

7. 2CO(g) + O2(g) 2CO2(g) 2 CO molecules 2 dozen CO molecules 2 moles CO molecules 2 x (6.023 x 1023) CO molecules 2 CO2 molecules 2 dozen CO2 molecules 2 moles CO2 molecules 2 x (6.023 x 1023) CO2 molecules 1 O2 molecules 1 dozen O2 molecules 1 mole O2 molecules (1 x) 6.023 x 1023 O2 molecules

8. 2CO(g) + O2(g) 2CO2(g) Is that okay? Number of moles is notconserved + ≠ 2 moles CO molecules 1 mole O2 molecules 2 moles CO2 molecules Yes, as long as the chemical equation is balanced! The coefficients are important!!!

9. 2CO(g) + O2(g) 2CO2(g) These are important! Coefficients 2 moles CO molecules 1 mole O2 molecules 2 moles CO2 molecules This chemical equation isbalanced The coefficients are correct

10. Coefficients are important 1 bag cake mix + 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes Write as a ratio:

11. Coefficients are important 1 bag cake mix + 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes Write as a ratio:

12. Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide Write as a ratio: These are stoichiometric equivalents

13. Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide Write as a ratio: mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

14. 11.1 Analyzing a Chemical Reaction Mole ratios Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide mole ratios for this chemical equation

15. Mole ratios Consider the following equation: CO(g) + 2H2(g) CH3OH(l) carbon monoxide hydrogen methanol If the reaction produces 5 moles of CH3OH, how many moles of H2 are consumed? Asked: moles of H2 5 moles CH3OH x 2 moles H2= 10 moles H2 1 mole CH3OH

16. A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react with excess aluminum metal? Asked:moles AlCl3 Given:moles Cl2 5 mole Cl2 x 2 mole AlCl3 = 3.3 mole AlCl3 3 mole Cl2

17. Potassium + Hydrogen Phosphate  Finish the reaction in symbols and balance… If 14.72 moles of hydrogen phosphate are consumed in the above reaction, how many moles of hydrogen are produced?

19. mass  moles…. There is no scale that measures in moles! How do you convert from moles to grams? By using the molar mass (g/mole) The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2. How do you convert from grams of Al to grams of Cl2? By using the molar mass (g/mole) and mole ratios

20. Process for calculating grams from grams given

21. If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced? Asked:grams of CO2 Given:grams of CaCO3 Relationships: mole ratios molar mass of CaCO3 = 40.078 + 12.011 + (15.999 x 3) = 100.0 g/mole molar mass of CO2 = 12.011 + (15.999 x 2) = 44.01 g/mole Strategy:

22. If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced? B B • 0.45 mole CaC03 x 1 mole CO2 x 44.01 g CO2 = 19.8 g CO2 • 1 mole CaC031 mole C02

23. CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration

24. In theory, all 100 kernels should have popped. Did you do something wrong? + 100 kernels 82 popped 18 unpopped

25. In theory, all 100 kernels should have popped. Did you do something wrong? No In real life (and in the lab) things are often not perfect + 100 kernels 82 popped 18 unpopped

26. Percent yield What you get to eat! + 100 kernels 82 popped 18 unpopped

27. actual yield: the amount obtained in the lab in an actual experiment. theoretical yield: the expected amount produced if everything reacted completely.

28. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Heating

29. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 4.87 g 10.00 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) • There is usually some human error, like not measuring exact amounts • carefully • Maybe the heating time was not long enough; not all the Na2HCO3 • reacted • - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too • - CO2 is a gas and does not get measured

30. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield obtained in experiment calculated

31. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield calculated

32. Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield calculated

33. 10.00 g 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) This is a gram-to-gram conversion: 10.00g NaHCO3 1 mole NaHCO3 1 mole Na2CO3 1mole Na2CO3 = 84.01 g NaHCO3 2 mole NaHCO3 105.99 g Na2CO3 Answer: Mass of D = 6.306 g Na2CO3

34. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g 10.00 g 6.306 g 0.1190 moles 0.05950 moles For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g. The actual yield (measured) is 4.87 g.

35. 2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g. The actual yield (measured) is 4.87 g.

36. Stoichiometry with solutions Decomposition of baking soda: (We just looked at this.) 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g Convert to moles Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) 50.0 mL of a 3.0 M solution Reactions in solution Convert to moles

37. A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = 1.0079 x 2 = 2.02 g/mole Asked: grams of H2produced Given:50.0 mL of 3.0 M HCl Reacting with excess zinc Solve: 0.0500L HCl x 3.0mole HCl x 1 mol H2 x 2.02 g H2= 0.15 g H2 L HCl 2 mol HCl 1 mol H2 Answer:0.15 grams of H2 are produced

39. Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) 50.0 mL of a 3.0 M solution Convert molarity to moles Sometimes the concentration is written in mass percent Vinegar is 5% acetic acid by mass

40. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve: Answer: 6.0 g of acetic acid.

41. Let’s Review: Obtained from the experiment Calculate using molar masses and mole ratios

42. Assignments 11.2:

43. Limiting Reactants Ch 11.3

44. Suppose you want to make 2 ham & cheese sandwiches Can you still make 2 ham & cheese sandwiches if you have 4 slices of bread, 4 slices of ham, and 1 slice of cheese? Limiting factor No, you are limited by the cheese! You can only get 1 ham & cheese sandwich.

45. Limiting reactant Excess reactant limiting reactant: the reactant that “runs out” first in a chemical reaction. excess reactant: the reactant that is remaining after the reaction is complete.

47. Sample Problem: 365/58. Iron can be produced from the following reaction: Fe2O3 + 2Al  2Fe + Al2O3 a. If 100.0 g of Fe2O3 reacts with 30.0 g Al, which one will be used up first?

48. Sample Problem: 365/58. Iron can be produced from the following reaction: Fe2O3 + 2Al  2Fe + Al2O3 b. For this next step, use the smaller mole answer from a. to find the needed amount

49. How much Fe can be produced? 1.12 mole Al 2 mole Fe 55.85 g Fe = 62.10 g Fe 2 mole Al 1 mole Fe How much of the excessive reactant is remaining? Fe2O3 is the excessive reactant. mole Al  mole Fe2O3  mass Fe2O3 Have – needed = excess 1.112 mole Al 1 mole Fe2O3 159.70 g Fe2O3 = 88.795 Fe2O3 2 mole Al 1 mole Fe2O3 100.0g – 88.80 g = 11.20 g remaining (excess)

50. Asgn: 364/59, 60