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Chapter 3: Stoichiometry

→. +. Chapter 3: Stoichiometry. “ measuring elements” Must account for ALL atoms in a chemical reaction. 2. 2. H 2 + O 2 → H 2 O. →. +. +. +. →. Chapter 3: Stoichiometry. 2. CO + O 2 → CO 2. 2. CH 4 + Cl 2 → CCl 4 + HCl. 4. 4. 2 3. 2 3. x 3.

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Chapter 3: Stoichiometry

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  1. → + Chapter 3: Stoichiometry • “measuring elements” • Must account for ALL atoms in a chemical reaction 2 2 H2 + O2 → H2O

  2. → + + + → Chapter 3: Stoichiometry 2 CO + O2 → CO2 2 CH4 + Cl2 → CCl4 + HCl 4 4

  3. 2 3 2 3 x 3 Chapter 3: Stoichiometry 2 3 2 C2H4 + O2 → CO2 + H2O 2 Al + HCl → AlCl3 + H2 2 2 3 6 or: 2 Al + HCl → AlCl3 + H2 2 Al + 6 HCl → 2 AlCl3 + 3 H2

  4. 1 2 x 2 Chapter 3: Stoichiometry 2 NH4NO3 → N2 + O2 + H2O 2 NH4NO3 → 2 N2 + O2 + 4 H2O

  5. Chapter 3: Stoichiometry Three basic reaction types: • Combination Reactions • Decomposition Reactions • Combustions (in air)

  6. Decomposition Reactions A single reactant breaks into two or more products 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) Chapter 3: Stoichiometry Combination Reactions Two or more reactants combine to form a single product C (s) + O2 (g) → CO2 (g) 2 Na (s) + Cl2 (g) → 2 NaCl (s)

  7. metal + nonmetal = ionic compound: Chapter 3: Stoichiometry Combustions in Air = reactions with oxygen Write the balanced reaction equation for the combustion of magnesium to magnesium oxide: Mg2+ O2- => MgO 2 Mg (s) + O2 (g) → 2 MgO (s)

  8. Chapter 3: Stoichiometry Combustions of Hydrocarbons in Air = reactions with oxygen to form carbon dioxide and water (complete combustion) Write the balanced reaction equation for the combustion of C2H4 gas C2H4 (g) + O2 (g) 3 → CO2 (g) + H2O (g) 2 2

  9. Chapter 3: Stoichiometry C2H4 (g) + O2 (g) 3 → CO2 (g)+ H2O (g) 2 2 + → + How many C2H4 molecules are in the flask? • If you know the weight of one molecule of C2H4 • and the total weight of gas in the flask, you can • calculate the number of molecules in the flask

  10. Chapter 3: Stoichiometry Molecular weight / Formula weight: => sum of all atomic weights in molecular formula MW of C2H4 = 2 x 12.0 amu + 4 x 1.0 amu = 28.0 amu 1 x 24.3 amu + (16.0 amu + 1.0 amu) x 2 FW of Mg(OH)2 = = 24.3 amu + 34.0 amu = 58.3 amu

  11. Chapter 3: Stoichiometry Molar Mass = mass of one mole of a substance in grams FW or MW of substance in amu's = mass of 1mole of substance in grams FW of Ca(NO3)2 = 164.1 amu Molar Mass of Ca(NO3)2 = 164.1 g/mol MW of O2 = 2 x 16.0 amu = 32 amu Molar Mass of O2 = 32 g/mol

  12. Chapter 3: Stoichiometry Number of individual molecules are difficult to deal with => definition of a “package” of molecules or particles 1 dozen eggs = 12 individual eggs 1 sixpack of cans = 6 cans . . . . . . . 1 mole of molecules = 6.02 x 1023 individual molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Avogadro's Number

  13. 1 sixpack 6 individual eggs Chapter 3: Stoichiometry 1 dozen eggs = 12 individual eggs How many sixpacks of eggs are in an egg carton that holds 12 eggs? sixpacks = 2 How many moles of eggs are in an egg carton that holds 12 eggs? 2.0 x 10-23 = moles of eggs

  14. Chapter 3: Stoichiometry Ca(NO3)2 Type of compound: ionic Ca2+ Ions: NO3- Total number of oxygen atoms: 6 Name: Calcium nitrate

  15. Chapter 3: Stoichiometry Ca(NO3)2 How many moles of calcium nitrate are in 394g of Ca(NO3)2 ? MM mol → gram Molar Mass of Ca(NO3)2 = 164.1 g/mol 1 mol = moles Ca(NO3)2 394 g Ca(NO3)2 x 2.4 164.1 g

  16. Chapter 3: Stoichiometry What is the mass in grams of 0.527 moles of calcium sulfate dihydrate (gypsum), CaSO4·2 H2O ? MM mol → gram (1) determine molar mass (MM) of CaSO4·2 H2O MW = (40+32+4*16)+2*(2*1+16) amu = 172 amu MM = 172 g/mol (2) use MM to convert moles into grams 0.527 moles CaSO4·2 H2O  = g CaSO4·2 H2O 90.6

  17. Chapter 3: Stoichiometry Ca(NO3)2 How many moles of oxygen are in 2.4 moles of Ca(NO3)2 ? 1 formula unit of Ca(NO3)2 contains 6 oxygen atoms 1 mole of Ca(NO3)2 contains 6 moles of oxygen atoms = moles oxygen 14.4

  18. Chapter 3: Stoichiometry Gretel(shoes)2 1 unit of Gretel(shoes)2 contains 2 shoes 1 mole ofGretel(shoes)2contains 2 moles of shoes How many moles of shoes are in 2.4 moles of Gretel(shoes)2 ? 2 moles of shoes 2.4 moles of Gretel(shoes)2 = moles shoes x 4.8 1 mol of Gretel(shoes)2

  19. Chapter 3: Stoichiometry Gretel(shoes)2 What is the percentage of shoes, by mass, in Gretel(shoes)2? (1) total mass of Gretel(shoes)2 mass of Gretel(shoes)2 = 55.2 kg + 0.5kg x 2 = 56.2 kg (2) mass of shoes inGretel(shoes)2 2 x 0.5kg = 1.0 kg (3) percentage of shoe mass

  20. Chapter 3: Stoichiometry Ca(NO3)2 What is the percentage of oxygen, by mass, in calcium nitrate? (1) total mass of Ca(NO3 )2 in amu FW of Ca(NO3)2 = 40.1 amu + (14.0 amu + 3 x 16.0 amu) x 2 = 164.1 amu (2) mass of oxygen in compound, in amu 6 x 16.0 amu = 96.0 amu (3) percentage of oxygen, by mass mass

  21. 2 CO + O2 → CO2 2 Chapter 3: Stoichiometry Quantitative Information from Balanced Equations How many grams of CO2 would be produced by the combustion of 2 moles of CO? → + 2 moles CO + 1 mole O2 → 2 moles CO2 2 x 28 g + 32 g → 2 x 44 g = 88 g

  22. 1 mol O2 1 mol O2 2 mol CO 2 mol CO2 2 mol CO 1 mol O2 Chapter 3: Stoichiometry Quantitative Information from Balanced Equations 2 CO + O2 → CO2 2 You can write a series of stoichiometric factors for this reaction:

  23. Chapter 3: Stoichiometry How many grams of H2O are formed from the complete combustion of 2.0 g of C2H4? need balanced reaction equation !! 18 g/mol 28 g/mol C2H4 (g) + O2 (g) 3 → CO2 (g) + H2O (g) 2 2 1 →moles C2H4 →moles H2O → grams H2O grams C2H4 2 g C2H4 = 2.6 g H2O from balanced equation

  24. Chapter 3: Stoichiometry Summary 1) determine equation for the reaction 2) balance equation • 3) formulate problem: • how much of A => gets converted into how much of B 4) determine MW/FW of substances involved 5) determine stoichiometric factors from balanced equation

  25. Chapter 3: Stoichiometry Limiting Reactants ½ cup Ginger ale + ½ cup lime soda + 1 Tbsp. Grenadine syrup + 1 Maraschino cherry Shirley Temple Cocktail How many ST’s can you make from 4 cups Ginger ale, 2 cups of lime soda a bottle of Grenadine syrup, and 10 maraschino cherries? a. 2 b. 4 c. 6 d.8

  26. Chapter 3: Stoichiometry Limiting Reactants ½ cup Ginger ale + ½ cup lime soda + 1 Tbsp. Grenadine syrup + 1 Maraschino cherry Shirley Temple Cocktail How many ST’s can you make from 4 cups Ginger ale, 4 cups of lime soda a bottle of Grenadine syrup, and 100 grams of maraschino cherries? You need to know how many maraschino cherries are in 100 grams!

  27. is the limiting reactant! The amount of limits the amount of product that can be formed Chapter 3: Stoichiometry Limiting Reactants +

  28. Chapter 3: Stoichiometry Limiting Reactants Limiting Reactant - limits the amount of product that can be formed - reacts completely(disappears during the reaction) - other reactants will be left over, i.e. in excess

  29. 3 mol N2, 6 mol H2 Available (given): Chapter 3: Stoichiometry Limiting Reactants: how much NH3 can be formed from 3 moles N2 and 6 moles H2? N2 + 3 H2 → 2 NH3 1 How much H2 would we need to completely react 3 mol N2: = 9 mol H2 compare with what is available 3 mol N2 We only have 6 mol H2 available – it is limiting ! How much NH3 can we form with the available reagents? 6 mol H2 = 4 mol NH3 continue with limiting reagent

  30. 3 mol N2, 6 mol H2 Available (given): Chapter 3: Stoichiometry Limiting Reactants N2 + 3 H2 → 2 NH3 1 How much N2 is left over (in excess)? H2 is limiting, there is plenty of N2 How much N2 is actually reacting? 6 mol H2 = 2 mol N2 this is the amount that reacts continue with limiting reagent Available – amount reacted = left over 3 mol – 2 mol = 1 mol N2 left over

  31. Chapter 3: Stoichiometry Limiting Reactants 2 Al + 3 Cl2 → 2 AlCl3 0.5 mol Al , 2.5 mol Cl2 Available (given): How much Cl2 would we need to completely react 0.5 mol Al: = 0.75 mol Cl2 compare with what is available 0.5 mol Al We have more than enough Cl2 available – it is in excess ! If Cl2 is in excess, Al must be limiting ! How much AlCl3 can we form with the available reagents? = 0.5 mol AlCl3 0.5 mol Al continue with limiting reagent

  32. Chapter 3: Stoichiometry Theoretical Yield 2 Al + 3 Cl2 → 2 AlCl3 0.5 mol Al , 2.5 mol Cl2 Available (given): What mass do 0.5 mol AlCl3 correspond to? = 67 g AlCl3 0.5 mol AlCl3 The maximum mass of product that can be formed is the theoretical yield

  33. Chapter 3: Stoichiometry Theoretical Yield 2 Al + 3 Cl2 → 2 AlCl3 0.5 mol Al , 2.5 mol Cl2 Available (given): Fritz does the reaction with the available reagents he only ends up with 34g. What is the % yield of the reaction? = 51 %

  34. Chapter 3: Stoichiometry Summary Determine availabe quantity of reactants in moles Determine if one of the reactants is a limiting reactant Determine the maximum # of moles of product that can be formed Determine % yield of the reaction Compare with actual amount of product recovered (actual yield) Convert into grams of product (theoretical yield)

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