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Basic Stoichiometry

Basic Stoichiometry. Pisgah High School M. Jones Revision history: 5/16/03, 02/04/12, 04/27/12. Part One. The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

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Basic Stoichiometry

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  1. Basic Stoichiometry Pisgah High School M. Jones Revision history: 5/16/03, 02/04/12, 04/27/12 Part One

  2. The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

  3. Stoichiometry deals with the amounts of reactants and products in a chemical reaction.

  4. Stoichiometry deals with moles.

  5. Recall that … 1 mole = 6.022 x1023 atoms or molecules 1 mole = the molar mass 1 mole = 22.4 L of any gas at STP

  6. The word mole is one that represents a very large number. Much like “dozen” means 12, … “mole” means 6.022 x 1023

  7. The key to doing stoichiometry is the balanced chemical equation. 2 H2 + O2 2 H2O 2 2

  8. The coefficients give the relative number of atoms or molecules of each reactant or product … as well as the number of moles. 2 H2 + O2 2 H2O

  9. 2 H2 + O2 2 H2O 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.

  10. 2 H2 + O2 2 H2O 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water The balanced chemical equation also gives the smallest integer number of moles.

  11. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water The balanced chemical equation also gives the smallest integer number of moles.

  12. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water Two moles of hydrogen combine with one mole of oxygen to make two moles of water.

  13. How can we show that this is really true? Consider the combustion of hydrogen in oxygen …

  14. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water What do each of the reactants and product weigh?

  15. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water 2 x 2.0 g 1 x 32.0 g 2 x 18.0 g + = 4.0 g 32.0 g 36.0 g

  16. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water The Law of Conservation of Matter + = 4.0 g 32.0 g 36.0 g

  17. 2 H2 + O2 2 H2O 2 moles of hydrogen 1 mole of oxygen 2 moles of water Matter can neither be created nor destroyed, only changed in form. + = 4.0 g 32.0 g 36.0 g

  18. The oxidation of iron

  19. Consider the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 4 moles 3 moles 2 moles The coefficients give the ratio of moles.

  20. Consider the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 If these react … then we have the following: 4 moles 3 moles 2 moles 8 moles 6 moles 4 moles 2 moles 1.5 moles 1 mole 0.50 mol 0.375 mol 0.25 mol

  21. Consider the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 The 0.375 moles was not as easy to predict. 0.375 mole O2

  22. Consider the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 The 0.375 moles was not as easy to predict. 0.375 mole O2 Use a conversion factor to determine the number of moles of an unknown.

  23. Consider the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 0.375 mole O2 The conversion factor comes from the coefficients in the balanced equation.

  24. Back to the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 4 moles 3 moles 2 moles Calculate the masses of these moles.

  25. Back to the oxidation of iron: 4 Fe(s) + O2(g)  Fe2O3(s) 3 2 4 moles 3 moles 2 moles 2 x 159.6 g 4 x 55.8 g 3 x 32.0 g + = 223.2 g 96.0 g 319.2 g Mass is conserved.

  26. The decomposition of ammonium carbonate

  27. Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O Suppose 96.0 grams of ammonium carbonate decomposes. How many grams of each of the gases will be produced?

  28. Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g The coefficients tell moles, not grams. Convert 96.0 g (NH4)2CO3 to moles.

  29. Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g Find the molar mass of ammonium carbonate. 2x14 + 8 +12 + 3x16 = 96.0 g/mol

  30. Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g Isn’t that convenient! We have one mole of ammonium carbonate.

  31. Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g 1 mol 2 mol 1 mol 1 mol 2 moles of ammonia are produced, along with 1 mole of carbon dioxide and 1 mole of water vapor.

  32. 2 x 17.0g 44.0 g 18.0 g Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g 1 mol 2 mol 1 mol 1 mol How many grams of each product are formed?

  33. 2 x 17.0g 44.0 g 18.0 g Now consider the decomposition of solid ammonium carbonate. (NH4)2CO3 2 NH3 + CO2 + H2O 96.0 grams ? g ? g ? g 1 mol 2 mol 1 mol 1 mol 34.0 g + 44.0 g + 18.0 g = 96.0 g

  34. The reaction between dinitrogen pentoxide and water

  35. Consider the reaction between dinitrogen pentoxide and water. What kind of reaction is it? Is it a … Double replacement reaction? Combustion reaction? Decomposition reaction? Single replacement reaction? So it must be a synthesis reaction

  36. Consider the reaction between dinitrogen pentoxide and water. Which kind of synthesis reaction is it? • Hydrogen + nonmetal = binary acid • Metal + nonmetal = salt • Metal + water = base • Nonmetal + water = ternary acid N2O5 + HOH  2 HNO3 HNO3 is a ternary acid; HNO3 is nitric acid

  37. N2O5 + HOH  2 HNO3 Suppose we needed to make 100.0 grams of nitric acid. How many grams of dinitrogen pentoxide would we need to react with excess water?

  38. N2O5 + HOH  2 HNO3 ??? g 100.0 g 1.59 mol/2 0.794 mole 1.59 mole = 1.59 mole HNO3 = 85.7 g N2O5

  39. N2O5 + HOH  2 HNO3 85.7 g 100.0 g 1.59 mol/2 0.794 mole 1.59 mole But. Suppose that we find out that there are only 60.0 grams of dinitrogen pentoxide. How many grams of nitric acid could we actually make?

  40. N2O5 + HOH  2 HNO3 60.0 g ??? g 0.556 mol x 2 0.556 mole 1.11 mole = 0.556 mol N2O5 = 70.0 g HNO3

  41. Description of stoichiometry problems

  42. Stoichiometry problems will usually take one of the following forms: • Mole-mole problem where you might be given moles and asked to find moles of another substance. • Mole-mass problem where you might be given moles and asked find the mass of another substance.

  43. Mass-mass problem where you might be given a mass and asked to find the mass of another substance. • Mass-volume problem where you might be given a mass and asked to find the volume of a gas. • Volume-volume problem where you might be given a volume and asked to find another volume.

  44. Volume-volume stoichiometry problems are easiest when you use Gay-Lussac’s Law. “The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.”

  45. Simply put, Gay-Lussac’s Law says this: The volumes of the gases are in the same ratio as the coefficients in the balanced equation. 2 H2 + O2 2 H2O 2 moles 1 mole 2 moles 2 L 1 L 2 L

  46. Applications of Gay-Lussac’s Law

  47. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) 5 mols 3 mols 4 mols 1 mol 5 L 3 L 4 L 1 L Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O2

  48. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) 5 mols 3 mols 4 mols 1 mol 5 L 3 L 4 L 1 L Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O2 How many liters of carbon dioxide gas and water vapor at STP would be produced? 10.5 L CO2 and 14.0 L H2O

  49. CH4(g) + 4 Cl2(g)  CCl4(g) + 4 HCl(g) When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced. How many L of methane will react with 0.800 L of chlorine gas at STP? 0.200 L Cl2

  50. Stoichiometry problems involving gases

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