1. Stoichiometry Calculations with Chemical Formulas and Equations

2. Stoichiometry Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789

3. Stoichiometry Chemical Equations Concise representations of chemical reactions

4. Stoichiometry Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

5. Stoichiometry Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

6. Stoichiometry Anatomy of a Chemical Equation Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

7. Stoichiometry Anatomy of a Chemical Equation The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

8. Stoichiometry Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

9. Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule

10. Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules

11. Stoichiometry Balancing Equations

12. Stoichiometry Balancing Equations Law of Conservation of Matter:  In a chemical reaction, matter can be neither created nor destroyed.  In a chemical reaction, the amount of reactants equal the amount of products.

13. Stoichiometry Balancing Equations Paraphrase: Law of Conservation of Atoms:  The number of atoms of each type of element must be the same on each side of the equation.

14. Stoichiometry Balancing Equations Hydrogen and oxygen are diatomic elements. Their subscripts cannot be changed. The subscripts on water cannot be changed. Hydrogen + oxygen water Hydrogen + oxygen water H 2 + O 2 H 2 O

15. Stoichiometry Balancing Equation Count the atoms on each side.  Reactant side:  2 atoms H  2 atoms O  Product side:  2 atoms H  1 atom O H 2 + O 2 H 2 O

16. Stoichiometry Balancing Equations H 2 + O 2 H 2 O If the subscripts cannot be altered, how can the atoms be made equal? Adjust the number of molecules by changing the coefficients.

17. Stoichiometry Balancing Equations Reactants:  2 atoms of H  2 atoms of O Products:  4 atoms of H  2 atoms of O H is no longer balanced! H 2 + O 2 2H 2 O

18. Stoichiometry Balancing Equations Reactant side:  4 atoms of H  2 atoms of O Product side:  4 atoms of H  2 atoms of O It’s Balanced! 2H 2 + O 2 2H 2 O

19. Stoichiometry Balancing Equations Count atoms. –Reactants: 2 atoms N and 2 atoms H –Products: 1 atom N and 3 atoms of NH 3 N 2 + H 2 NH 3 Nitrogen + hydrogen ammonia Nitrogen + hydrogen ammonia

20. Stoichiometry Balancing Equations Nothing is balanced. Balance the nitrogen first by placing a coefficient of 2 in front of the NH 3. N 2 + H 2 2NH 3

21. Stoichiometry Balancing Equations Hydrogen is not balanced. Place a 3 in front of H 2. Reactant side: 2 atoms N, 6 atoms H Product side: 2 atoms N, 6 atoms H N 2 + 3H 2 2NH 3

22. Stoichiometry Balancing Equations In this equation, the ion groups do not break up. Instead of counting individual atoms, ion groups may be counted. Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4

23. Stoichiometry Balancing Equations Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Reactants:  Ca 2+ – 3  PO 4 3- - 2  H + – 2  SO 4 2+ - 1 Products:  Ca 2+ - 1  SO 4 2- - 1  H + - 3,  PO 4 3- - 1

24. Stoichiometry Balancing Equations Balance the metal first by placing a coefficient of 3 in front of CaSO 4. Products:  Ca – 3 atoms  SO 4 2- - 3 groups Ca 3 (PO 4 ) 2 + H 2 SO 4 3CaSO 4 + H 3 PO 4

25. Stoichiometry Balancing Equations Three sulfate groups are needed on the reactant side so place a coefficient of 3 in front of H 2 SO 4. 3H 2 SO 4 gives 6 H + and 3 SO 4 2-. Neither phosphate nor calcium is balanced. Ca 3 (PO 4 ) 2 + 3H 2 SO 4 3CaSO 4 + H 3 PO 4

26. Stoichiometry Balancing Equations A coefficient of 2 placed in front of H 3 PO 4 balances both hydrogen and phosphate. Ca 3 (PO 4 ) 2 + 3H 2 SO 4 3CaSO 4 + 2H 3 PO 4

27. Stoichiometry Balancing Equations The sulfate group breaks up. Each atom must be counted individually. Reactants: Cu – 1, H – 2, S – 1, O – 4 Products: Cu – 1, H – 2, S – 2, O – 7 Cu + H 2 SO 4 CuSO 4 + H 2 O + SO 2

28. Stoichiometry Balancing Equations Sulfur is not balanced. Place a two in front of sulfuric acid. Cu + 2H 2 SO 4 CuSO 4 + H 2 O + SO 2

29. Stoichiometry Balancing Equations Hydrogen needs to be balanced so place a 2 in front of the H 2 O. Count the number of atoms. Cu + H 2 SO 4 CuSO 4 + 2H 2 O + SO 2

30. Stoichiometry Balancing Equations Reactants: Cu – 1, H – 4, S – 2, O – 8 Products: Cu – 1, H – 4, S – 2, O – 8 It’s balanced! Cu + 2H 2 SO 4 CuSO 4 + 2H 2 O + SO 2

31. Stoichiometry Reaction Types

32. Stoichiometry Combination Reactions Examples: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) 2 Mg (s) + O 2 (g)  2 MgO (s) Two or more substances react to form one product

33. Stoichiometry 2 Mg (s) + O 2 (g)  2 MgO (s)

34. Stoichiometry Decomposition Reactions Examples: CaCO 3 (s)  CaO (s) + CO 2 (g) 2 KClO 3 (s)  2 KCl (s) + O 2 (g) 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g) One substance breaks down into two or more substances

35. Stoichiometry Combustion Reactions Examples: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air

36. Stoichiometry Formula Weights

37. Stoichiometry Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu These are generally reported for ionic compounds

38. Stoichiometry Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu

39. Stoichiometry Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

40. Stoichiometry Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

41. Stoichiometry Moles

42. Stoichiometry Avogadro’s Number 6.02 x 10 23 1 mole of 12 C has a mass of 12 g

43. Stoichiometry Molar Mass By definition, these are the mass of 1 mol of a substance (i.e., g/mol) –The molar mass of an element is the mass number for the element that we find on the periodic table –The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

44. Stoichiometry Using Moles Moles provide a bridge from the molecular scale to the real-world scale

45. Stoichiometry Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

46. Stoichiometry Finding Empirical Formulas

47. Stoichiometry Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

48. Stoichiometry Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

49. Stoichiometry Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

50. Stoichiometry Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

51. Stoichiometry Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

52. Stoichiometry Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

53. Stoichiometry Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

54. Stoichiometry Limiting Reactants

55. Stoichiometry How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

56. Stoichiometry How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

57. Stoichiometry Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount

58. Stoichiometry Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

59. Stoichiometry Limiting Reactants In the example below, the O 2 would be the excess reagent

60. Stoichiometry Theoretical Yield The theoretical yield is the amount of product that can be made –In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures

61. Stoichiometry Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100