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Newport High School Academic Chemistry Mrs. Teates

Stoichiometry. Newport High School Academic Chemistry Mrs. Teates. Lesson 1 – Introduction to Stoichiometry. Lesson Essential Questions: What is stoichiometry and how is it used to describe reactions?

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Newport High School Academic Chemistry Mrs. Teates

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  1. Stoichiometry Newport High SchoolAcademic ChemistryMrs. Teates

  2. Lesson 1 – Introduction to Stoichiometry • Lesson Essential Questions: • What is stoichiometry and how is it used to describe reactions? Vocabulary: stoichiometry, composition stoichiometry, reaction stoichiometry, mole, Avogadro’s number, molar mass

  3. Stoichiometry Definition • Stoichiometry • mass relationships between substances in a chemical reaction • Composition stoichiometry deals with the mass relationships of elements in compounds. • Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

  4. What is the Mole? VERY A largeamount!!!! • A counting number (like a dozen) • Can be used to measure anything (DVDs, cars, eggs, molecules) • 1 mol = 6.02  1023 items

  5. What is the Mole? HOW LARGE IS IT??? • 1 mole of pennies would cover the Earth 1/4 mile deep! • 1 mole of hockey pucks would equal the mass of the moon! • 1 mole of basketballs would fill a bag the size of the earth!

  6. Molar Mass • Mass of 1 mole of an element or compound. • Atomic mass tells the... • atomic mass units per atom (amu) • grams per mole (g/mol) • Round to 2 decimal places

  7. Molar Mass Examples 12.01 g/mol 26.98 g/mol 65.39 g/mol • carbon • aluminum • zinc

  8. Molar Mass Examples • water • sodium chloride • H2O • 2(1.01) + 16.00 = 18.02 g/mol • NaCl • 22.99 + 35.45 = 58.44 g/mol

  9. Molar Mass Examples • sodium bicarbonate • sucrose • NaHCO3 • 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol • C12H22O11 • 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

  10. Molar Conversions MASS IN GRAMS MOLES NUMBER OF PARTICLES molar mass 6.02  1023 (g/mol) (particles/mol)

  11. Molar Conversion Examples • How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

  12. Molar Conversion Examples • How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules 1 mol 2.50 mol = 1.51  1024 molecules C12H22O11

  13. Molar Conversion Examples • Find the mass of 2.1  1024 molecules of NaHCO3. 2.1  1024 molecules 1 mol 6.02  1023 molecules 84.01 g 1 mol = 290 g NaHCO3

  14. Percentage Composition • Calculate because it is useful to know the percentage by mass of an element in a compound or

  15. Sample Problems • Sample Problem J on page 243 • Practice problem #1

  16. Mole Ratio • A mole ratiois a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example:2Al2O3(l) → 4Al(s) + 3O2(g) Mole Ratios:2 mol Al2O3 2 mol Al2O3 4 mol Al 4 mol Al 3 mol O2 3 mol O2 , ,

  17. Converting Between Amounts in Moles

  18. Stoichiometry Calculations

  19. Lesson 2 – Ideal Stoichiometric Calculations • Lesson Essential Questions: • How are calculations used to describe chemical reactions?

  20. STOICH MAP 23 6.02X10 PARTICLES PARTICLES 23 6.02X10 A B MASS MOLE MOLE MASS MOLE MOLAR MASS MOLAR MASS RATIO A A B B VOLUME VOLUME 22.4 L A B 22.4 L

  21. Proportional Relationships Ratio of eggs to cookies • I have 5 eggs. How many cookies can I make? 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

  22. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio - moles  moles • Molar mass - moles  grams • Molarity - moles  liters soln • Molar volume - moles  liters gas • Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

  23. Stoichiometry Calculations

  24. Conversion of Quantities in Moles

  25. Conversions of Quantities in Moles Sample Problem A In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation. CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l) How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?

  26. Conversions of Quantities in Moles Sample Problem A Solution CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l) Given:amount of CO2 = 20 mol Unknown:amount of LiOH (mol) Solution:

  27. Conversions of Amounts in Moles to Mass

  28. Conversions of Amounts in Moles to Mass Sample Problem B In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?

  29. Conversions of Amounts in Moles to Mass Sample Problem B Solution Given:amount of H2O = 3.00 mol Unknown:mass of C6H12O6 produced (g) Solution: Balanced Equation: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) mol ratio molar mass factor 90.1 g C6H12O6

  30. Conversions of Mass to Amounts in Moles

  31. Conversions of Mass to Amounts in Moles Sample Problem D The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced) The reaction is run using 824 g NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed?

  32. Conversions of Mass to Amounts in Moles Sample Problem D Solution Given:mass of NH3 = 824 g Unknown:a. amount of NO produced (mol) b. amount of H2O produced (mol) Solution: Balanced Equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) molar mass factor mol ratio a. b.

  33. Conversions of Mass to Amounts in Moles Sample Problem D Solution molar mass factor mol ratio a. b.

  34. Mass-Mass to Calculations

  35. Mass-Mass to Calculations Sample Problem E Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation. Sn(s) + 2HF(g) → SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?

  36. Mass-Mass to Calculations Sample Problem E Solution Given:amount of HF = 30.00 g Unknown:mass of SnF2 produced (g) Solution: molar mass factor mol ratio molar mass factor = 117.5 g SnF2

  37. More Practice for Homework • Page 320 #4-16

  38. Lesson 3 – Limiting Reactants and Percent Yield • Lesson Essential Questions: • How does the limiting reactant affect the percentage yield and actual yield in a chemical reaction? Vocabulary: limiting reactant, excess reactant, theoretical yield, actual yield, percent yield

  39. Limiting Reactants • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant • bread • Excess Reactants • peanut butter and jelly

  40. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  41. Limiting Reactants 1. Write a balanced equation. 2. Write your known and unknown. 3. For each reactant, calculate the amount of product formed. 4. Smaller answer indicates: • limiting reactant • amount of product

  42. Limiting Reactant Practice • Sample Problem F on page 313. • Practice problem #1 on page 313. • Sample Problem G on pages 314-315. • Practice problems #1 & 2 on page 315. • Homework pg. 321 #22-25

  43. Percent Yield measured in lab calculated on paper

  44. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  45. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  46. Percent Yield 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

  47. Percent Yield Practice • Sample problem H on page 317. • Practice problem #1 & 2 on page 318.

  48. Homework • Page 321 #22-30, 31-36, 38 • Review for the test – Worksheet titled Stoichiometry Review Problems

  49. Works Cited • Modern Chemistry Textbook • www.nclark.net • http://mrsj.exofire.net/chem/

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