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Notes One Unit Four

Notes One Unit Four. Making Cookies Making Chemicals. Pages 344-345. Making Cookies. Making Cookies. What does the following recipe tell us?

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Notes One Unit Four

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  1. Notes One Unit Four Making Cookies Making Chemicals Pages 344-345

  2. Making Cookies

  3. Making Cookies What does the following recipe tell us? 1 C butter +1 C sugar +2 egg yolks +1/4 C vegetable oil +1/2 C cocoa + 2 tsp vanilla + 2-1/4 C flour + 1/4 tsp salt +1 C Chocolate Powder + 4 Oz semisweet chocolate, 2 TBS cream 32 Mud Puddle Cookies How many cookies can be made with… 2 cups of butter and 10 eggs? 160 cookies 64 cookies 4 cups of sugar and 1 Egg? 128 cookies 16 cookies 1-1/8 of C flour and 8 cups of oil? 16 cookies 1024 cookies 0.25 8 Cookies_____ cups butter? 1.5 24 Cookies_____ tsp vanilla? 4 64 Cookies_____ egg yolks?

  4. Making Chemicals

  5. Making Chemicals 1) Combustion: hydrocarbon + oxygen  carbon dioxide + water 2m 4m 3m 10m 1 8 C5H12(l)+ O2(g)  CO2(g) + H2O(l) 5 6 0.67m 2.5m 4m 4m

  6. Making Chemicals 2) Synthesis: A + B  AB 2m 4m 2m 4 1 Sn(s) + S8(c)  SnS2(s) 4 16m 2m 8m

  7. Moles to Moles 3) Single displacement: A + BC AC + B 4m 1m 0.5m 4m 2 6 Al(s)+ H2O(l) Al(OH)3(aq) + H2(g) 2 3 0.67m 1.33m 1m 4m

  8. Moles to Moles 4) Acid Base: Acid + Base Salt+ Water 2.4m 1m 1m 2.4m 6 2 6 2 HCl(aq)+ Al(OH)3(s) H2O(l)+ AlCl3(aq) 4m 12m 4.5m 1.5m

  9. Moles to Moles Standard Format • How many moles of product can be made using 4.0 moles of each reactant? • 1) Balanced Equation • H2(g)+ N2(g) NH3(g) • 2) Before: • 3) Limiting? • 4) Changes: • 5) After: 3 1 2 4.0 m 4.0 m 0 m 4.0 4.0 1 3 - - + 4.0 4.0 x2 x1 3 3 0 m 2.7 m 2.7 m

  10. Moles to Moles Standard Format • How many moles of product can be made using 4.0 moles of P4(g) and 5.0 moles of F2(g). • 1) Balanced Equation • P4(g)+ F2(g) PF3(g) • 2) Before: • 3) Limiting? • 4) Changes: • 5) After: 1 6 4 4.0 m 5.0 m 0 m 4.0 5.0 1 6 - - + 5.0 5.0 x4 x1 6 6 3.2 m 0 m 3.3 m

  11. Moles to Moles Standard Format • How many moles of product can be made using 5.0 moles of Fe(s) and 7.0 moles of O2(g). • 1) Balanced Equation • Fe(s)+ O2(g) Fe3O4(s) • 2) Before: • 3) Limiting? • 4) Changes: • 5) After: 3 2 1 5.0 m 7.0 m 0 m 5.0 7.0 3 2 - - + 5.0 5.0 x1 x2 3 3 0 m 3.7 m 1.7 m

  12. Notes Two Unit Four Quiz Mass-Mass Calculation White Board Assignment Pages 344-345

  13. Stoichiometry Quantitative study of reactants and products in chemical reactions. Mg(s) + O2 (g)→ MgO(s) Reactants → Products Answers the question: • How much product will we get from a given amount of reactant?

  14. Necessary Information to remember 1. Writing balanced Equations • Use coefficients to balance the atoms in the reactants with the atoms in the products magnesium plus oxygen gas creates magnesium oxide 2Mg(s) + O2 (g) → 2MgO(s)

  15. Necessary Information to remember 2. Convert Grams to Moles • 1 mole of any substance = molar mass of substance in grams • You have 8.45 g of each substance convert to moles. • Molar Mass of Mg = 24.31 g • Molar Mass of O2 = 16.00g x 2 = 32.00 g • Molar Mass of MgO = 24.31g + 16.00g = 40.31 g

  16. 3 Types of Stoichiometry Conversions moles of reactants  moles of products  mass of reactants  moles of reactants  moles of products mass of reactants  moles of reactants  moles of products  mass of products

  17. Steps to Follow for Stoichiometry

  18. Balanced Equations Have Ratios • the coefficients in a chemical equation can be interpreted as the ratio of molecules of each substance. 2 CO(g) + O2 (g)2 CO2 (g) The image and the Balance equation mean: 2 molecules of CO reacts with 1 molecule of oxygen gas to generate 2 molecules of CO2

  19. What does the ratio mean?2 CO(g) + O2 (g) 2 CO2 (g) • If I have 6 molecules of CO and 6 molecules of O2 can I make 6 molecules of CO2? • (HINT… think how many CO must reach with each O2? • 6 molecules CO(g) + 6 molecules O2 (g)

  20. Ratio 2 CO(g) + O2 (g) 2 CO2 (g) • I can’t make 6 molecules of CO2 because for every one O2 molecule I need 2 CO molecules • That means I will run out of CO molecules after I react with 3 O2 molecules. 1 3 2

  21. Results • 6 molecules of CO create 3 molecules CO2 2 to 1 ratio • 6 molecules of O2 create 6 molecules CO2 1 to 1 ratio

  22. Ratio Using Moles vs. Molecules • Stoichiometry answers the question… • How much product will we get from a given amount of reactant? • Stoichiometry (stoich) uses the ratios in a balanced chemical equation • Stoichiometry is used in chemistry labs with large amounts of substances, therefore Stoich uses the ratios in MOLES instead of molecules

  23. Molecules Changes to Moles 2 CO(g) + O2 (g)2 CO2 (g) Molecule Version 2 molecules of carbon monoxide reacts with 1 molecule of oxygen gas to yield 2 molecules of carbon dioxide Mole Version 2 moles of carbon monoxide molecules reacts with 1 mole of oxygen gas molecules to yield 2 moles of carbon dioxide molecules

  24. Using Coefficients to get Mole Ratios 2 CO(g) + O2 (g)2 CO2 (g) Conversion factors from equation: 2 moles CO = 1 mole O2 2 moles CO = 2 moles of CO2  1 mole O2 = 2 moles of CO2

  25. Stoichiometry Example #1moles of reactants  moles of products  A reaction takes place between Lithium and water: Li(s) + H2O (l) LiOH(aq) + H2 (g) Question: How many moles of H2 will be formed if 6.23 moles of Li combine with water?

  26. Stoichiometry Example #1 Cont’dQuestion: How many moles of H2 will be formed if 6.23 moles of Li combine with water? 1. Balance the equation using coefficients 2Li(s) + 2 H2O (l) 2 LiOH(aq) + H2 (g) 2. Convert the required amount to moles *do not need this step for this question* 6.23 moles of Li (from question)

  27. Stoichiometry Example #1 Cont’dQuestion: How many moles of H2 will be formed if 6.23 moles of Li combine with water? 3. Find the mole ratio using the coefficients. 2Li(s) + 2 H2O (l) 2 LiOH(aq) + H2 (g) 2 moles of Li = 1 mole H2 4. Convert to desired unit The question asked for moles so 3.12 moles H2 doesn’t need to be converted and is the final answer.

  28. Stoichiometry Example #2mass of reactants  moles of reactants  moles of products How many moles of H2 will be formed by a reaction of 80.57g of Li with water? 1. Balance the equation using coefficients  2 Li(s) + 2 H2O (l) 2 LiOH(aq) + H2 (g)

  29. Stoichiometry Example #2 Cont’dHow many moles of H2 will be formed by a reaction of 80.57g of Li with water? • Convert given value into moles • Find mole ratios from equation and use in factor label method for the unknown value (H2) 2 Li(s) + 2 H2O (l) 2 LiOH(aq) + H2 (g)

  30. Stoichiometry Example #3mass of reactants  moles of reactants  moles of products  mass of products How many grams of H2 will be formed by a reaction of 80.57g of Li with water? 1. Write reaction again  2 Li(s) + 2 H2O (l) 2 LiOH(aq) + H2 (g) • Convert given value into moles • Find mole ratios from equation and use in factor label method for the unknown value (H2) Example #2 work

  31. Stoichiometry Example #3 Cont’dHow many grams of H2 will be formed by a reaction of 80.57g of Li with water? 4. Convert the moles into grams

  32. Additional Example #3mass of reactants  moles of reactants  moles of products  mass of products How many grams of NO2 are formed by reacting 1.44 grams of NO with oxygen gas? • Write the balanced chemical equation 2NO(g) + O 2(g) → 2NO2(g) • Convert the given value into moles

  33. Additional Example #3How many grams of NO2 are formed by reacting 1.44 grams of NO with oxygen gas? 3. Find mole ratios from equation and use in factor label method for the unknown value (NO2) 2NO(g) + O 2(g) → 2NO2(g) 4. Convert the moles into grams

  34. Let’s Put all the steps together! • Factor label Method: it uses UNITS to cancel and work your way through the problem ( ) ( ) ( ) = Given Info (g) x 1 Mole Given (mol)Moles RequestedFM Requested (g) g of FM Given (g) Moles Given 1 Mole Requested Requested Given Bracket Mole to Mole Bracket Requested Bracket The given bracket converts grams of given to moles of given. The middle bracket converts moles of given to moles of requested. The requested bracket converts moles of requested to grams of requested. All in one multistep process !!! THIS IS A MASS MASS PROBLEM.

  35. Moles to Moles Standard Format • How many moles of product can be made using 4.50 moles of C(s) and 6.50 moles of H2(g). • 1) Balanced Equation • C(s)+ H2(g) C6H6(g) • 2) Before: • 3) Limiting? • 4) Changes: • 5) After: 6 3 1 4.50 m 6.50 m 0 m 4.50 6.50 6 3 - - + 4.50 4.50 x1 x3 6 6 0 m 4.25 m 0.75 m

  36. Mass-Mass Calculation One 1 N2(g) + H2(g) NH3(l) 3 2 How many grams of ammonia will be made reacting with 14.0 grams of nitrogen reaction with hydrogen?

  37. Grams to Grams 28.0g ____g 34.0 ÷28.0g/m 3 1 2 H2(g)+ N2(g)  NH3(g) ___m 2.0 x17.0g/m ___m 1.0 # E Mass # E Mass H 3.0 N 2 x 14.0= 28.0g/m 3 x 1.0 = N 1 x 14.0 = 14.0 17.0g/m

  38. 17.0 ____g 14.0g ÷28.0g/m 1N2(g) + 3H2(g)  2NH3(l) ____m x17.0g/m 1.00 _____m 0.500 # E Mass N 28.0g/m 2 x 14.0 = Mass # E 28.0gN2/m= 0.500m N2 1) grams N2 to moles N2 14.0g÷ 2) moles N2to moles NH3 0.500m N2x 3) moles NH3 to grams NH3 1.00mNH3 x H 3.0 3x 1.0 = N 1x 14.0 = 14.0 (2m NH3) 17.0g/m = 1.00 m NH3 (1mN2) 17.0g/m = 17.0g NH3

  39. 2 H2(g) + O2(g) H2O(l) 1 2 Mass-Mass Calculation Two How many grams of water will be made reacting with 4.0grams of hydrogen with oxygen?

  40. 36.0 ____g 4.0g ÷2.0 g/m 2H2(g) + 1O2(g)  2H2O(l) 2.0 2.0 ___m x18.0g/m ___m # E Mass H 2.0g/m 2 x 1.0 = # E 2.0gH2 /m= 2.0m H2 Mass 1) grams H2 to moles H2 4.0g÷ 2) moles H2to moles H2O 2.0m H2x 3) moles H2O to grams H2O 2.0m H2O x H 2.0 2x 1.0 = O 1x 16.0 = 16.0 (2m H2O) = 2.0 mH2O 18.0g/m (2mH2) 18.0g/m = 36.0g H2O

  41. 2 Ag2O(s) O2(g)+ Ag(s) 1 4 Mass-Mass Calculation Three How many grams of silver will be made by decomposing 8.00grams of Silver oxide?

  42. ____g 7.45 8.00g ÷231.8 g/m # E Mass Ag 215.8 2x 107.9 = O 1x 16.0 = 16.0 2Ag2O(g)  1O2(g)+4Ag(s) 231.8g/m # E Mass Ag 107.9g/m 1 x 107.9 = ______m X107.9g/m 0.0690 0.0345 ______m 231.8g Ag2O /m= 0.0345m Ag2O 1) grams Ag2Oto moles Ag2O 8.00g÷ 2) moles Ag2Oto moles Ag 0.0345m Ag2Ox 3) moles Ag to grams Ag 0.0690 m Ag x (4m Ag) = 0.0690 m Ag (2 Ag2O) 107.9g/m = 7.45g Ag

  43. Notes Three Unit Four Quiz Lab

  44. 1 C8H8(l)+ O2(g) CO2(g) + H2O(g) 10 8 4 Mass-Mass Calculation Four How many grams of carbon dioxide will be made by burning 6.00grams of Isovanillin?

  45. 20.3 ____g # E Mass 6.00g ÷104.0 g/m C 96.0 8x 12.0 = H 8x 1.0 = 8.0 1C8H8(l)+10O2(g)8CO2(g)+4H2O(g) 104.0 g/m # E Mass C 12.0 1 x 12.0 = O 32.0 2 x 16.0 = 0.0577 ______m 44.0g/m _____m X44.0g/m 0.462 104.0g C8H8 /m= 0.0577m C8H8 1) grams C8H8 to moles C8H8 6.00g÷ 2) moles C8H8to moles CO2 0.0577m C8H8 x 3) moles CO2 to grams CO2 0.462 m CO2 x (8m CO2) = 0.462 m CO2 (1mC8H8) 20.3g CO2 44.0g/m =

  46. Notes Four Unit Four Limiting Reagent Mass-Mass Problem Molar Volume Mass-Volume Problem Lab Results

  47. # E Mass Determining Limiting Reactant N 2 x 14.0= 28.0g/m 1 N2(g) + H2(g) NH3(l) 3 2 14.0g 7.0g ÷2.0g/m ÷28.0g/m # E Mass 3.5m 0.500m H 2 x 1.0= 2.0g/m How many grams of ammonia will be made reacting with 14.0 grams of nitrogen with 7.0 g of hydrogen? 3.5m 0.500m 3 1

  48. 17.0 ____g 14.0g ÷28.0g/m 1N2(g) + 3H2(g)  2NH3(l) ____m x17.0g/m 1.00 _____m 0.500 # E Mass N 28.0g/m 2 x 14.0 = Mass # E 28.0gN2/m= 0.500m N2 1) grams N2 to moles N2 14.0g÷ 2) moles N2to moles NH3 0.500m N2x 3) moles NH3 to grams NH3 1.00mNH3 x H 3.0 3x 1.0 = N 1x 14.0 = 14.0 (2m NH3) 17.0g/m = 1.00 m NH3 (1mN2) 17.0g/m = 17.0g NH3

  49. Molar Volume @STP 1mN2(g) 28.0g/m 22.4L/m 1mO2(g) 32.0g/m 22.4L/m 1mAr(g) 39.9g/m 22.4L/m

  50. 1 C8H8(l)+ O2(g) CO2(g) + H2O(g) 10 8 4 Mass-Volume Calculation One How many liters of carbon dioxide will be made by burning 6.00grams of Isovanillin?

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