Chapter 3: Stoichiometry

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. Chapter 3: Stoichiometry 2 3 2 C2H4 + O2→ CO2 + H2O 2 Al + 6 HCl → 2 AlCl3 + 3 H2 2 NH4NO3→ 2 N2 + O2 + 4 H2O

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

5. 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)

6. Chapter 3: Stoichiometry Combustions in Air = reactions with oxygen Write the balanced reaction equation for the combustion of magnesium to magnesium oxide:

7. 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)

8. 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

9. Chapter 3: Stoichiometry Molecular weight / Formula weight: => sum of all atomic weights in molecular formula for molecular compounds MW of C2H4 FW of Mg(OH)2 for ionic compounds

10. Chapter 3: Stoichiometry Ca(NO3)2 Type of compound: Ions: Total number of oxygen atoms: Name:

11. Chapter 3: Stoichiometry Ca(NO3)2 Percentage of oxygen, by mass: (1) total mass of Ca(NO3 )2 in amu (2) mass of oxygen in compound, in amu (3) percentage of oxygen

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 soda sixpack = 6 individual bottles of soda . . . . . . . 1 mole of molecules = 6.02 x 1023 individual molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Avogadro's Number

13. Chapter 3: Stoichiometry 1 dozen eggs = 12 individual eggs How many moles of eggs are in an egg carton that holds a dozen eggs?

14. Chapter 3: Stoichiometry 1 dozen eggs = 12 individual eggs How many moles of eggs are in an egg carton that holds a dozen eggs?

15. Chapter 3: Stoichiometry Ca(NO3)2 How many moles of oxygen are in 2.4 moles of Ca(NO3)2 ?

16. 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

17. Chapter 3: Stoichiometry What is the mass in grams of 0.527 moles of CH3OH? (1) determine molar mass of CH3OH 32 g/mol (2) use MM to convert moles into grams g CH3OH 16.9

18. Chapter 3: Stoichiometry mol → gram ← Molar Mass How many hydrogen atoms are in 4.5 g of CH3OH? 0.14 = mol CH3OH 0.56 = mol H atoms 3.4 x 1023 = H atoms

19. 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

20. 2 CO + O2→ CO2 2 1 mol O2 1 mol O2 2mol CO 2 mol CO2 2mol CO 1 mol O2 Chapter 3: Stoichiometry Quantitative Information from Balanced Equations You can write a series of stoichiometric factors for this reaction:

21. Chapter 3: Stoichiometry How many grams of H2O are formed from the complete combustion of 2 g of C2H4? 18 g/mol 28 g/mol C2H4 (g) + O2 (g) 3 → CO2 (g) + H2O (g) 2 2 1 mol 2 mol = 2.6 g H2O

22. 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

23. + Limiting Reactant Chapter 3: Stoichiometry Limiting Reactants + 2 of these will be left over “in excess”

24. 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

25. 3 mol N2, 6 mol H2 Available: Chapter 3: Stoichiometry Limiting Reactants N2 + 3 H2→ 2 NH3 How much H2 would we need to completely react 3 mol N2: = 9 mol H2 How much NH3 can we form with the available reagents? = 4 mol NH3

26. 3 mol N2, 6 mol H2 Available: Chapter 3: Stoichiometry Limiting Reactants N2 + 3 H2→ 2 NH3 How much N2 is left over (in excess)? = 1 mol N2

27. Chapter 3: Stoichiometry Limiting Reactants 2 Al + 3 Cl2→ 2 AlCl3 Available: 0.5 mol Al , 2.5 mol Cl2 How much Cl2 would we need to completely react 0.5 mol Al: = 0.75 mol Cl2 How much AlCl3 can we form with the available reagents? = 0.5 mol AlCl3

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

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

30. 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)

31. Chapter 3: Stoichiometry Consider the combustion of methanol: 2 CH3OH + 3 O2→ 2 CO2 + 4 H2O What is the theoretical yield of water if 44 g of methanol are reacted with 128 g of oxygen?

32. Chapter 3: Stoichiometry 2 CH3OH + 3 O2→ 2 CO2 + 4 H2O available (initial) moles 1.4 mol 4.0 mol O2 = 2.1 mol O2

33. Chapter 3: Stoichiometry 2 CH3OH + 3 O2→ 2 CO2 + 4 H2O (3) how many product moles can be formed with limiting reactant ? = 1.4 mol CO2 = 2.8 mol H2O

34. Chapter 3: Stoichiometry 2 CH3OH + 3 O2→ 2 CO2 + 4 H2O (4) What is the mass of H2O formed (theoretical yield)?