1. Chapter 41 Chemical Equations and Stoichiometry Chapter 4

2. 2 2H 2 (g) + O 2 (g)  2H 2 O(g) Chemical Equations

3. Chapter 43 2H 2 (g) + O 2 (g)  2H 2 O(g) The materials you start with are called Reactants. Chemical Equations

4. Chapter 44 2H 2 (g) + O 2 (g)  2H 2 O(g) The materials you start with are called Reactants. The materials you make are called Products. Chemical Equations

5. Chapter 45 2H 2 (g) + O 2 (g)  2H 2 O(g) The materials you start with are called Reactants. The materials you make are called Products. The numbers in front of the compounds (H 2 and H 2 O) are called stoichiometric coefficients. –Coefficients are multipliers, in this equation 2 in front of the H 2 indicates that there are 2 molecules of H 2 in the equation. Chemical Equations

6. Chapter 46 2H 2 (g) + O 2 (g)  2H 2 O(g) Notice that the number of hydrogen atoms and oxygen atoms on the reactant side and the product side is equal. Law of Conservation of Mass Matter cannot be created or lost in any chemical reaction. Chemical Equations

7. Chapter 47 Balancing Chemical Reactions ___NH 4 NO 3 (s)  ___N 2 O(g) + ___H 2 O(g) Chemical Equations

8. Chapter 48 Balancing Chemical Reactions ___NH 4 NO 3 (s)  ___N 2 O(g) + ___H 2 O(g) Chemical Equations ReactantsProducts N2N2 H4H2 O3O2

9. Chapter 49 Balancing Chemical Reactions ___NH 4 NO 3 (s)  ___N 2 O(g) + _2_H 2 O(g) Chemical Equations ReactantsProducts N2N2 H4H2 4 O3O2 32 3

10. Chapter 410 Balancing Chemical Reactions ___Mg 3 N 2 (s) + ___H 2 O(l)  ___Mg(OH) 2 (s) + ___NH 3 (aq) Chemical Equations ReactantsProducts Mg3 1 N2N1 H2H5 O1O2

11. Chapter 411 Balancing Chemical Reactions ___Mg 3 N 2 (s) + ___H 2 O(l)  _3_Mg(OH) 2 (s) + ___NH 3 (aq) Chemical Equations ReactantsProducts Mg3 1 3 N2N1 H2H5 9 O1O2 6

12. Chapter 412 Balancing Chemical Reactions ___Mg 3 N 2 (s) + ___H 2 O(l)  _3_Mg(OH) 2 (s) + _2_NH 3 (aq) Chemical Equations ReactantsProducts Mg3 1 3 N2N1 2 H2H5 9 12 O1O2 6

13. Chapter 413 Balancing Chemical Reactions ___Mg 3 N 2 (s) + _6_H 2 O(l)  _3_Mg(OH) 2 (s) + _2_NH 3 (aq) Chemical Equations ReactantsProducts Mg3 1 3 N2N1 2 H2 12H5 9 12 O1 6O2 6

14. Chapter 414 2 H 2 (g) + O 2 (g)  2 H 2 O(g) The coefficients in a balanced equation represent both the number of molecules and the number of moles in a reaction. The coefficients can also be used to derive ratios between any two substances in the chemical reaction. 2 H 2 : 1 O 2 2 H 2 : 2 H 2 O 1 O 2 : 2 H 2 O Quantitative Information The ratios can be used to predict The amount of product formed The amount of reactant needed

15. Chapter 415 Quantitative Information

16. Chapter 416 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen?

17. Chapter 417 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 1.Moles of C 4 H 10 F.W. 58.124g

18. Chapter 418 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 1.Moles of C 4 H 10 F.W. 58.124g

19. Chapter 419 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 2.Ratio of C 4 H 10 :CO 2 2 C 4 H 10 : 8 CO 2 or

20. Chapter 420 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 3.Set-up ratio and proportion between known and unknown quantities

21. Chapter 421 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 3.Set-up ratio and proportion between known and unknown quantities

22. Chapter 422 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 3.Set-up ratio and proportion between known and unknown quantities

23. Chapter 423 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 4.Convert the moles of unknown substance into the desired units

24. Chapter 424 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 4.Convert the moles of unknown substance into the desired units FW of CO 2 : 44.011g/mol

25. Chapter 425 Quantitative Information 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) How many grams of CO 2 are formed if 1.00g of butane (C 4 H 10 ) is allowed to react with excess oxygen? 4.Convert the moles of unknown substance into the desired units FW of CO 2 : 44.011g/mol

26. Chapter 426 “What runs out first” 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O If you have 2 moles of C 8 H 18 and 20 moles of O 2 all the O 2 will be used and the reaction will stop O 2 is call the limiting reagent (reactant) Limiting Reactant – The reagent present in the smallest stoichiometric quantity in a mixture of reactants. Limiting Reactants

27. Chapter 427 Example 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. 1.Convert grams to moles FW(C 8 H 18 ) 114.268g/mol FW(O 2 ) = 32.00g/mol Limiting Reactants

28. Chapter 428 Example 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. 1.Convert grams to moles FW(C 8 H 18 ) 114.268g/mol FW(O 2 ) = 32.00g/mol Limiting Reactants

29. Chapter 429 Example 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. 2.Divide each reagent by its own coefficient Limiting Reactants

30. Chapter 430 Example 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. 2.Divide each reagent by its own coefficient Limiting Reactants

31. Chapter 431 Example 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the limiting reagent of this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. 3.The substance with the smallest calculated value will be the limiting reagent. In this case, O 2 is the limiting reagent. Limiting Reactants

32. Chapter 432 Theoretical Yield - The calculated amount of product based on the limiting reactant (Theoretical yield). Limiting Reactants

33. Chapter 433 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Determine the theoretical yield of CO 2 for this reaction if 10.0 grams of C 8 H 18 and 25.0 grams of O 2 are allowed to react. - already know that O 2 is the limiting reactant. Limiting Reactants Theoretical Yield

34. Chapter 434 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Calculate moles of oxygen Limiting Reactants Theoretical Yield Calculate moles of CO 2

35. Chapter 435 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O Calculate moles of CO 2 Limiting Reactants Theoretical Yield

36. Chapter 436 Percent Yield - Calculation which indicates how much of the theoretical yield was obtained. Limiting Reactants

37. Chapter 437 Combustion Analysis Empirical Formulas from Analyses Typical example: 2 C 2 H 6 (g) + 7 O 2  4 CO 2 (g) + 6 H 2 O(g) -The combustion of any hydrocarbon produces CO 2 and water. -This observation can be used to determine the empirical formula of the reactant. Combustion Reaction: The “burning” of any substance in oxygen.

38. Chapter 438 Combustion Analysis Empirical Formulas from Analyses

39. Chapter 439 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2  moles CO 2  moles C  grams C Empirical Formulas from Analyses

40. Chapter 440 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2  moles CO 2  moles C  grams C Empirical Formulas from Analyses

41. Chapter 441 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2  moles CO 2  moles C  grams C Empirical Formulas from Analyses

42. Chapter 442 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Carbon mass CO 2  moles CO 2  moles C  grams C Empirical Formulas from Analyses

43. Chapter 443 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O  moles H 2 O  moles H  grams H Empirical Formulas from Analyses

44. Chapter 444 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O  moles H 2 O  moles H  grams H Empirical Formulas from Analyses

45. Chapter 445 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O  moles H 2 O  moles H  grams H Empirical Formulas from Analyses

46. Chapter 446 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Hydrogen mass H 2 O  moles H 2 O  moles H  grams H Empirical Formulas from Analyses

47. Chapter 447 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Empirical Formulas from Analyses

48. Chapter 448 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Empirical Formulas from Analyses

49. Chapter 449 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Mass of elements: C  0.07721g H  0.01299g O  0.01030g Empirical Formulas from Analyses

50. Chapter 450 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = H  0.01299g/1.01g/mol = O  0.01030g/16.00g/mol = Empirical Formulas from Analyses

51. Chapter 451 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.012861 mol O  0.01030g/16.00g/mol = 0.0006428mol Empirical Formulas from Analyses

52. Chapter 452 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.012861 mol O  0.01030g/16.00g/mol = 0.0006428 mol Empirical Formulas from Analyses

53. Chapter 453 Combustion Analysis Menthol, the substance we can smell in mentholated cough drops, is composed of C, H and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO 2 and 0.1159g of H 2 O. What is the empirical formula for menthol? Now we can determine the empirical formula Moles of elements: C  0.07721g/12.011g/mol = 0.006428 mol H  0.01299g/1.01g/mol = 0.012861 mol O  0.01030g/16.00g/mol = 0.0006428 mol Empirical Formulas from Analyses

54. Chapter 454 Practice Problems 4, 10, 14, 22, 26, 28, 38, 46, 50, 54