1. Moles

2. Like a recipe: ReactantsProducts 2H 2 (g) + O 2 (g)  2H 2 O(l) coefficients subscripts Balancing Eqns

3. Moles Symbols  Yields or Produces (s)solid (l)liquid (pure liquid) (aq)aqueous (dissolved in water) (g) or  gas Balancing Eqns

4. Moles Copper(II) Chloride reacts with Iron(III) Sulfate to form Copper(II)Sulfate and Iron(III) Chloride Aluminum nitrate reacts with Sodium hydroxide to form Aluminum hydroxide and Sodium nitrate Balancing

5. Moles Some Types of Reactions 1. Synthesis Al + Cl 2  AlCl 3 CaO(s) + CO 2 (g)  CaCO 3 (s) 2.Decomposition HgO(s)  Hg(l) + O 2 (g) CaCO 3 (s)  CaO(s) + CO 2 (g)

6. Moles 3.Combustion Fe + O 2  Fe 2 O 3 C 4 H 10 + O 2  CO 2 + H 2 O 4.Other Types Single Replacement Double Replacement Some Types of Reactions

7. Moles Know your reactions well!!!!

8. Moles Mole 1 dozen = 12 items 1 mole = 6.022 X 10 23 atoms/molecules 1 gram hydrogen = 6.022 X 10 23 atoms of hydrogen Stoichiometry

9. Moles

11. Stoichiometry GramsMolesAtoms 1 g H1 mole H6.02 X 10 23 atms 2 g H 12 g C 36 g C 48 g O

12. Moles Stoichiometry GramsMolesMoleculesAtoms 16 g CH 4 8 g CH 4 88 g CO 2 131 g Ba(NO 3 ) 2

13. Moles 1.Molar Mass = mass of one mole 2.Element= atomic mass 1 mole of O = 16.0 grams 3. Molecule or Compound – sum of all the atoms Molar Mass

14. Moles

15. Calculate the molar mass of Barium O 2 BaCl 2 ? Fe 2 (SO 4 ) 3? Molar Mass

16. Moles Grams Moles Atoms 1.How many C atoms are present in 18.0 g? (Ans: 9.03 X 10 23 C) 2.What is the mass of 1.20 X 10 24 atoms of Na? (Ans: 45.8 grams) GMA

17. Moles 3.What is the mass of 1.51 X 10 23 atoms of Be?(Ans: 2.26 g) 5.How many atoms and grams are in 0.400 mol of Radium? (Ans: 90.4 g, 2.41 X 10 23 atoms) GMA

18. Moles 1.Monoatomic Elements (C, Fe, Au) GMA 2.Molecules and Ionics (H 2 O, CaCl 2, O 2 ) GMMA

19. Moles 3.Molecules and formula units work the same when converting Molecules = Molecular Comps Formula Units = Ionic Compounds G M A GMMA

20. Moles 1.How many calcium and chlorine atoms are in 200.0 grams of Calcium Chloride? (Ans: 2.17 X 10 24 atoms Cl) GMMA

21. Moles 2. How many hydrogen and oxygen atoms are in 3.60 grams of H 2 O? (Ans: 2.41 X 10 23 atoms H) GMMA

22. Moles 3.Given 3.01X10 24 molecules of SO 3, find everything else. 4.Given 3.01 X 10 22 molecules of Iron(III)bromide, find everything else. GMMA

23. Moles 1.How many carbon atoms are in 36.0 grams of carbon (1.81 X 10 24 ) 2.How many carbon atoms are in 36.0 grams of C 2 H 6 ? (Ans:1.45 X 10 24 atoms of C) Mixed Examples

24. Moles Homework Problems (find everything else) a) 10.0 g C b) 10.0 g C 2 H 6 c) 4.0 X 10 23 atoms of S d) 4.0 X 10 23 molecules of SO 2 e) 0.44 moles of SO 2 Mixed Examples

25. Moles 1.Empirical formula - simplest ratio of the elements in a compound 2.FormulaEmpirical Form. C 2 H 2 Al 4 S 6 C 6 H 12 O 6 C 12 H 24 O 12 Empirical Formula

26. Moles 1.What is the EF of a compound that has 0.900 g Ca and 1.60 g Cl? Rules - Go to moles - Divide by the smaller Empirical Formula

27. Moles 2. What is the EF of a compound that is 66.0 % Ca and 34.0% P? 3. What is the EF of a compound that is 43.7 % P and 56.3 % O? Empirical Formula

28. Moles 1.Empirical –ratios of the elements 2.Molecular –true number of each element Molecular Formula

29. Moles EFMF CH 2 OCH 2 O (30 g/mol) C 2 H 4 O 2 (60 g/mol) C 3 H 6 O 3 (90 g/mol) C 4 H 8 O 4 (120 g/mol) Molecular Formula

30. Moles 1.What is the MF of benzene if it has an EF of CH and a molar mass of 78.0 g? 2. What is the MF of a compound that is 40.9% C, 4.58 % H and 54.5 % O? It has a molar mass between 350 and 360 g/mol. Molecular Formula

31. Moles What coefficients mean: 2 Na+Cl 2  2NaCl 2 Na1 Cl 2 2NaCl 4 Na 6 Na Reaction Stoich.

32. Moles 2 Na+Cl 2  2NaCl 4 Cl 2 2 moles Na 10 moles Na ONLY WORKS FOR MOLES AND MOLECULES Reaction Stoich.

33. Moles 1.How many moles of H 2 and O 2 must react to form 6 moles of H 2 O? 2. How many moles of KCl and O 2 are formed from the decomposition of 6 moles of KClO 3 ? Reaction Stoich.

34. Moles 3.How many grams of oxygen are needed to react with 14.6 g of Na to form Na 2 O? (Ans: 5.08 g) 4.How many grams of P 4 and O 2 are needed to make 3.62 g of P 2 O 5 ? (Ans: 1.58 g, 2.04 g) Reaction Stoich.

35. Moles 5. What mass of oxygen is needed to react with 16.7 g of iron to form Iron(III)oxide? (Ans: 7.18 g) 4Fe + 3O 2  2Fe 2 O 3 Reaction Stoich.

36. Moles Calculate the mass of sodium bromide and oxygen that are formed from the decomposition of 50.0 grams of sodium bromate (NaBrO 3 ). (34.1 g NaBr, 15.9 g O 2 )

37. Moles 1.Sandwich analogy: 13 slices of bread 4 pieces of turkey Maximum # of sandwiches? 2. Limiting Reactant – Totally consumed in a reaction. No leftovers Limiting Reactant

38. Moles 1.How many grams of H 2 SO 4 can be formed from the rxn of 5.00 moles of SO 3 and 2.00 moles of H 2 O? SO 3 + H 2 O  H 2 SO 4 (Ans: 196 g) Limiting Reactant

39. Moles 1.How many grams of H 2 O can be formed from the rxn of 6.00 moles of H 2 and 4.00 moles of O 2 ? O 2 + H 2  H 2 O Limiting Reactant

40. Moles 2. How many grams of NaCl can be formed from the reaction of 0.300 mol of Na and 0.100 mol of Cl 2 ? 2Na + Cl 2  2NaCl (Ans: 11.7 g) Limiting Reactant

41. Moles 3.How many grams of Ag can be formed from the rxn of 2.00 g of Zn and 2.50 g of silver nitrate? How much excess reactant remains? Zn + AgNO 3  Ag + Zn(NO 3 ) 2 (Ans: 1.59 g Ag, 1.52 g xs zinc) Limiting Reactant

42. Moles 4.How many grams of Ba 3 (PO 4 ) 2 can be formed from the rxn of 3.50 g of Na 3 PO 4 and 6.40 g of Ba(NO 3 ) 2 ? Na 3 PO 4 + Ba(NO 3 ) 2  Ba 3 (PO 4 ) 2 + NaNO 3 (Ans: 4.92 g) Limiting Reactant

43. Moles 6.How many grams of Ag 2 S can be formed from the rxn of 15.6 g of Ag and 2.97 g of H 2 S? (Assume O 2 is in excess) 4Ag + 2H 2 S + O 2  2Ag 2 S + 2H 2 O (Ans: 18.1 g) Limiting Reactant

44. Moles A.Formula: Actual Yield X 100 = % Yield Theoretical Yield Percent Yield

45. Moles 1.What is the % yield if you start with 10.00 grams of C and obtain 1.49 g of H 2 gas? C + H 2 O  CO + H 2 (Ans: 89.4%) Percent Yield

46. Moles 2.Carbon was heated strongly in sulfur(S 8 ) to form carbon disulfide. What is the percent yield if you start with 13.51 g of sulfur and collect 12.5 g of CS 2 ? 4C + S 8  4CS 2 (Ans: 78.0%) Percent Yield

47. Moles 3. 36.7 grams of CO 2 were formed from the rxn of 40.0 g of CH 3 OH and 46.0 g of O 2. What is the % yield? 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (ANS: 87%) Percent Yield

48. Moles

49. In this experiment, magnesium chloride was prepared and its empirical formula was compared to the accepted formula of MgCl 2. To prepare magnesium chloride, 0.40 grams of magnesium powder was combined with 10 mL of 0.10 M HCl. The mixture was allowed to react, and heated to dryness. The mass of the resulting crystals was used to calculate the empirical formula. The average calculated formula of MgCl 1.8 had a 10% error and a range was 0.40 chlorine atoms. This procedure was not effective because while it was accurate, it was not precise.

50. Moles 8a) SO 3 + H 2 O  H 2 SO 4 b)B 2 S 3 + 6H 2 O  2H 3 BO 3 + 3H 2 S c)4PH 3 + 8O 2  6H 2 O + P 4 O 10 d)2Hg(NO 3 ) 2  2HgO + 4NO 2 + O 2 e)Cu + 2H 2 SO 4  CuSO 4 + SO 2 + 2H 2 O

51. Moles 12.a) 6 1 214.a) 1 1 1 b) 1 3 2b) 1 6 2 3 c) 2 2 1 4c) 1 2 2 1 d) 1 6 3 2d) 2 2 4 1 e) 3 2 1 6e) 1 2 1 1 2 f) 2 1 1 2 g) 4 9 4 10 2

52. Moles 18.a) 4Al + 3O 2  2Al 2 O 3 b) Cu(OH) 2  CuO + H 2 O c) C 7 H 16 + 11O 2  7CO 2 + 8H 2 O d) 2C 5 H 12 O + 15O 2  10CO 2 + 12H 2 O 20. 2 9 6 6 1 1 2 1 6 5 3 1 3 2 1 1 2 22.a) 44.0 g/molb) 122.0 g/mol c) 58.3 g/mold) 60.0 g/mol e) 130.0 amu

53. Moles 24. 26.0 g/mole92.3% C 176.0 g/mole4.5% H 132.1 g/mole6.1% H 300.1 g/mole65.01% Pt 272.0 g/mole11.8% O 305.0 g/mole70.8 % C

54. Moles 46a)K 3 PO 4 b) Na 2 SiF 6 c) C 12 H 12 N 2 O 3 48a) H 2 C 2 O 4 b) C 4 H 8 O 2 50a) C 13 H 18 O 2 b) C 5 H 14 N 2 c) C 9 H 13 O 3 N

55. Moles 58.a) 0.800 mol CO 2 b) 14.7 g C 6 H 12 O 6 c) 7.16 g CO 2 60.a) 0.939 mol Fe 2 O3b) 78.9 g CO c) 105 g Fed) 229 g= 229 g

56. Moles 48.a) H 2 C 2 O 4 b) C 4 H 8 O 2 50. a) C 13 H 8 O 2 b)C 5 H 14 N 2 c) C 9 H 13 O 3 N 58.a) 0.800 mol CO 2 b) 14.7 g C 6 H 12 O 6 c) 7.18 g CO 2 60.a) Fe 2 O 3 + 3CO  2Fe + 3CO 2 b) 78.9 g CO 2 c) 124 g CO 2 d) 229 g = 229 g

57. Moles 62.a) CaH 2 + 2H 2 O  Ca(OH) 2 + 2H 2 b) 88.75 g CaH2 64. a) 15.6 mol O 2 b) 35.0 g O 2 c) 9175.1 g 72.0.167 mol Al 2 (SO 4 ) 3 form 0.333 mol Al(OH) 3 react 0.167 mol AL(OH) 3 remain 74.a) O 2 is limiting reactant b) 1.86 g H 2 O produced c) 0.329 g NH 3 remain d) 4.25 g = 4.25 g

58. Moles 76. 5.24 g H 2 SO 4 6.99 g PbSO 4 2.77 g HC 2 H 3 O 2 78. C 2 H 6 + Cl 2  C 2 H 5 Cl + HCl 232 g C 2 H 5 Cl (theoretical yield) 88.8% yield 80. Actual yield of Na 2 S = 1.80 g (1.95 g is the theoretical yield)

59. Moles The atmosphere of Jupiter is composed almost entirely of hydrogen (H 2 ) and helium (He). If the average molar mass of Jupiter’s atmosphere is 2.254 g/mole, calculate the percent composition.

60. Moles