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Chemical Equations and Stoichiometry

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

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Chemical Equations and Stoichiometry

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  1. Chemical Equations and Stoichiometry Chapter 4 Chapter 4

  2. Chemical Equations 2H2(g) + O2(g)2H2O(g) Chapter 4

  3. Chemical Equations 2H2(g) + O2(g)2H2O(g) • The materials you start with are called Reactants. Chapter 4

  4. Chemical Equations 2H2(g) + O2(g)2H2O(g) • The materials you start with are called Reactants. • The materials you make are called Products. Chapter 4

  5. Chemical Equations 2H2(g) + O2(g)2H2O(g) • The materials you start with are called Reactants. • The materials you make are called Products. • The numbers in front of the compounds (H2 and H2O) are called stoichiometriccoefficients. • Coefficients are multipliers, in this equation 2 in front of the H2 indicates that there are 2 molecules of H2 in the equation. Chapter 4

  6. Chemical Equations 2H2(g) + O2(g)2H2O(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. Chapter 4

  7. Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + ___H2O(g) Chapter 4

  8. Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + ___H2O(g) Chapter 4

  9. Chemical Equations Balancing Chemical Reactions ___NH4NO3(s)  ___N2O(g) + _2_H2O(g) Chapter 4

  10. Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  ___Mg(OH)2(s) + ___NH3(aq) Chapter 4

  11. Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  _3_Mg(OH)2(s) + ___NH3(aq) Chapter 4

  12. Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + ___H2O(l)  _3_Mg(OH)2(s) + _2_NH3(aq) Chapter 4

  13. Chemical Equations Balancing Chemical Reactions ___Mg3N2(s) + _6_H2O(l)  _3_Mg(OH)2(s) + _2_NH3(aq) Chapter 4

  14. Quantitative Information • 2 H2(g) + O2(g)  2 H2O(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 H2 : 1 O2 • 2 H2 : 2 H2O • 1 O2 : 2 H2O • The ratios can be used to predict • The amount of product formed • The amount of reactant needed Chapter 4

  15. Quantitative Information Chapter 4

  16. Quantitative Information 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? Chapter 4

  17. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Moles of C4H10 • F.W. 58.124g Chapter 4

  18. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Moles of C4H10 • F.W. 58.124g Chapter 4

  19. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Ratio of C4H10:CO2 • 2 C4H10 : 8 CO2 or Chapter 4

  20. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Set-up ratio and proportion between known and unknown quantities Chapter 4

  21. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Set-up ratio and proportion between known and unknown quantities Chapter 4

  22. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Set-up ratio and proportion between known and unknown quantities Chapter 4

  23. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Convert the moles of unknown substance into the desired units Chapter 4

  24. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Convert the moles of unknown substance into the desired units • FW of CO2: 44.011g/mol Chapter 4

  25. Quantitative Information • 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g) • How many grams of CO2 are formed if 1.00g of butane (C4H10) is allowed to react with excess oxygen? • Convert the moles of unknown substance into the desired units • FW of CO2: 44.011g/mol Chapter 4

  26. Limiting Reactants • “What runs out first” • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • If you have 2 moles of C8H18 and 20 moles of O2all the O2 will be used and the reaction will stop • O2 is call the limiting reagent (reactant) • Limiting Reactant – The reagent present in the smallest stoichiometric quantity in a mixture of reactants. Chapter 4

  27. Limiting Reactants • Example • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. • Convert grams to molesFW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol Chapter 4

  28. Limiting Reactants • Example • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. • Convert grams to molesFW(C8H18) 114.268g/mol FW(O2) = 32.00g/mol Chapter 4

  29. Limiting Reactants • Example • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. • Divide each reagent by its own coefficient Chapter 4

  30. Limiting Reactants • Example • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. • Divide each reagent by its own coefficient Chapter 4

  31. Limiting Reactants • Example • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Determine the limiting reagent of this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. • The substance with the smallest calculated value will be the limiting reagent. In this case, O2 is the limiting reagent. Chapter 4

  32. Limiting Reactants Theoretical Yield - The calculated amount of product based on the limiting reactant (Theoretical yield). Chapter 4

  33. Limiting Reactants Theoretical Yield 2 C8H18 + 25 O2 16 CO2 + 18 H2O Determine the theoretical yield of CO2 for this reaction if 10.0 grams of C8H18 and 25.0 grams of O2 are allowed to react. - already know that O2 is the limiting reactant. Chapter 4

  34. Limiting Reactants Theoretical Yield • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Calculate moles of oxygen • Calculate moles of CO2 Chapter 4

  35. Limiting Reactants Theoretical Yield • 2 C8H18 + 25 O2 16 CO2 + 18 H2O • Calculate moles of CO2 Chapter 4

  36. Limiting Reactants Percent Yield - Calculation which indicates how much of the theoretical yield was obtained. Chapter 4

  37. Empirical Formulas from Analyses Combustion Analysis Combustion Reaction: The “burning” of any substance in oxygen. • Typical example: • 2 C2H6(g) + 7 O2 4 CO2(g) + 6 H2O(g) • The combustion of any hydrocarbon produces CO2 and water. • This observation can be used to determine the empirical formula of the reactant. Chapter 4

  38. Empirical Formulas from Analyses Combustion Analysis Chapter 4

  39. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2 moles CO2  moles C  grams C Chapter 4

  40. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2 moles CO2  moles C  grams C Chapter 4

  41. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2 moles CO2  moles C  grams C Chapter 4

  42. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Carbon mass CO2 moles CO2  moles C  grams C Chapter 4

  43. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 4

  44. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 4

  45. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 4

  46. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Hydrogen mass H2O  moles H2O  moles H  grams H Chapter 4

  47. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 4

  48. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. What is the empirical formula for menthol? Mass of Oxygen mass O = mass of sample – (mass C +mass H) Chapter 4

  49. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. 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 Chapter 4

  50. Empirical Formulas from Analyses 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 CO2 and 0.1159g of H2O. 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 = Chapter 4

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