1. Chapter 3: Stoichiometry ● “measuring elements” ● Must account for ALL atoms in a chemical reaction + → H2+O2→H2OH2+O2→H2O 2 2

2. Chapter 3: Stoichiometry CO+O 2 →CO 2 22 + → CH 4 +Cl 2 →CCl 4 +HCl → + + 44

3. Chapter 3: Stoichiometry C 2 H 4 +O 2 →CO 2 +H 2 O 2 2 3 Al + HCl →AlCl 3 +H 2 22 6 2 3 2 2323 2323 x 3 2 Al + 6 HCl→2 AlCl 3 +3 H 2 or:

4. Chapter 3: Stoichiometry NH 4 NO 3 → N 2 + O 2 + H 2 O 2 1212 x 2 2 NH 4 NO 3 →2 N 2 + O 2 + 4 H 2 O

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

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

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

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 C 2 H 4 gas C 2 H 4 (g)+ O 2 (g)→ CO 2 (g)+ H 2 O (g)223

9. Chapter 3: Stoichiometry C 2 H 4 (g)+ O 2 (g)→ CO 2 (g)+ H 2 O (g)223 +→+ How many C 2 H 4 molecules are in the flask? ● If you know the weight of one molecule of C 2 H 4 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 C 2 H 4 = 2 x 12.0 amu + 4 x 1.0 amu = 28.0 amu FW of Mg(OH) 2 = 1 x 24.3 amu + (16.0 amu + 1.0 amu) x 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(NO 3 ) 2 = 164.1 amu Molar Mass of Ca(NO 3 ) 2 = 164.1 g/mol MW of O 2 = 2 x 16.0 amu = 32 amu Molar Mass of O 2 = 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 10 23 individual molecules Avogadro's Number......................................

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

14. Chapter 3: Stoichiometry Ca(NO 3 ) 2 Type of compound: ionic Ions: Ca 2+ NO 3 - Total number of oxygen atoms: 6 Name: Calcium nitrate

15. Chapter 3: Stoichiometry Ca(NO 3 ) 2 How many moles of calcium nitrate are in 394g of Ca(NO 3 ) 2 ? = moles Ca(NO 3 ) 2 2.4 Molar Mass of Ca(NO 3 ) 2 = 164.1 g/mol 394 g Ca(NO 3 ) 2 x 164.1 g 1 mol mol →gram MM

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

17. Chapter 3: Stoichiometry Ca(NO 3 ) 2 How many moles of oxygen are in 2.4 moles of Ca(NO 3 ) 2 ? 1 formula unit of Ca(NO 3 ) 2 contains 6 oxygen atoms 1 mole of Ca(NO 3 ) 2 contains 6 moles of oxygen atoms = moles oxygen 14.4

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

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

20. Chapter 3: Stoichiometry Ca(NO 3 ) 2 What is the percentage of oxygen, by mass, in calcium nitrate? FW of Ca(NO 3 ) 2 = 40.1 amu + (14.0 amu + 3 x 16.0 amu) x 2 =164.1 amu (1) total mass of Ca(NO 3 ) 2 in 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. Chapter 3: Stoichiometry mol→gram ← Molar Mass How many hydrogen atoms are in 4.5 g of CH 3 OH? gram CH 3 OH → = mol CH 3 OH = mol H atoms (1)(2)(3) (1) (2) (3) = H atoms 0.14 0.56 3.4 x 10 23 mol CH 3 OH →mol hydrogen → hydrogen atoms

22. Chapter 3: Stoichiometry Quantitative Information from Balanced Equations CO+O 2 →CO 2 22 How many grams of CO 2 would be produced by the combustion of 2 moles of CO? + → 2 moles CO+ 1 mole O 2 →2 moles CO 2 2 x 28 g+32 g→2 x 44 g = 88 g

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

24. Chapter 3: Stoichiometry C 2 H 4 (g)+ O 2 (g)→ CO 2 (g)+ H 2 O (g)223 How many grams of H 2 O are formed from the complete combustion of 2.0 g of C 2 H 4 ? 28 g/mol 18 g/mol 2 g C 2 H 4 = 2.6 g H 2 O from balanced equation grams C 2 H 4 → moles C 2 H 4 → moles H 2 O→ grams H 2 O 1 need balanced reaction equation !!

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

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, 2 cups of lime soda a bottle of Grenadine syrup, and 10 maraschino cherries? a. 2b. 4c. 6d.8

27. 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!

28. Chapter 3: Stoichiometry Limiting Reactants +

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

30. 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

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

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

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

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

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

36. 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 Convert into grams of product (theoretical yield) Compare with actual amount of product recovered (actual yield) Determine % yield of the reaction