Chapter Three Lecture- Stoichiometry
- 1. 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
- 2. Chemical Equations Concise representations of chemical reactions
- 3. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
- 4. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the left side of the equation.
- 5. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Products appear on the right side of the equation.
- 6. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) The states of the reactants and products are written in parentheses to the right of each compound.
- 7. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Coefficients are inserted to balance the equation.
- 8. Subscripts and Coefficients Give Different Information • Subscripts tell the number of atoms of each element in a molecule
- 9. Subscripts and Coefficients Give Different Information • Subscripts tell the number of atoms of each element in a molecule • Coefficients tell the number of molecules
- 10. Reaction Types
- 11. Combination Reactions • Two or more substances react to form one product
- 12. Combination Reactions • Two or more substances react to form one product • Examples:
- 13. Combination Reactions • Two or more substances react to form one product • Examples: N2 (g) + 3 H2 (g) → 2 NH3 (g)
- 14. Combination Reactions • Two or more substances react to form one product • Examples: N2 (g) + 3 H2 (g) → 2 NH3 (g) C3H6 (g) + Br2 (l) → C3H6Br2 (l)
- 15. Combination Reactions • Two or more substances react to form one product • Examples: N2 (g) + 3 H2 (g) → 2 NH3 (g) C3H6 (g) + Br2 (l) → C3H6Br2 (l) 2 Mg (s) + O2 (g) → 2 MgO (s)
- 16. 2 Mg (s) + O2 (g) → 2 MgO (s)
- 17. Decomposition Reactions • One substance breaks down into two or more substances
- 18. Decomposition Reactions • One substance breaks down into two or more substances • Examples:
- 19. Decomposition Reactions • One substance breaks down into two or more substances • Examples: CaCO3 (s) → CaO (s) + CO2 (g)
- 20. Decomposition Reactions • One substance breaks down into two or more substances • Examples: CaCO3 (s) → CaO (s) + CO2 (g) 2 KClO3 (s) → 2 KCl (s) + O2 (g)
- 21. Decomposition Reactions • One substance breaks down into two or more substances • Examples: CaCO3 (s) → CaO (s) + CO2 (g) 2 KClO3 (s) → 2 KCl (s) + O2 (g) 2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
- 23. Combustion Reactions • Rapid reactions that produce a flame
- 24. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air
- 25. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air • Examples:
- 26. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air • Examples: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
- 27. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air • Examples: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
- 28. Formula Weights
- 30. Formula Weight (FW) • Sum of the atomic weights for the atoms in a chemical formula • So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu
- 31. Formula Weight (FW) • Sum of the atomic weights for the atoms in a chemical formula • So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu • These are generally reported for ionic compounds
- 33. Molecular Weight (MW) • Sum of the atomic weights of the atoms in a molecule
- 34. Molecular Weight (MW) • Sum of the atomic weights of the atoms in a molecule • For the molecule ethane, C2H6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu
- 35. 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: (number of atoms)(atomic weight) % element = x 100 (FW of the compound)
- 36. Percent Composition So the percentage of carbon in ethane is… (2)(12.0 amu) %C = (30.0 amu) 24.0 amu = x 100 30.0 amu = 80.0%
- 37. Moles
- 39. Avogadro’s Number • 6.02 x 1023
- 40. Avogadro’s Number • 6.02 x 1023 • 1 mole of 12C has a mass of 12 g
- 41. Molar Mass
- 42. Molar Mass • By definition, the mass of 1 mol of a substance (i.e., g/mol)
- 43. Molar Mass • By definition, 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
- 44. Molar Mass • By definition, 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)
- 45. Using Moles Moles provide a bridge from the molecular scale to the real-world scale
- 47. Mole Relationships • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles
- 49. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition
- 50. 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.
- 51. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x 1 mol = 5.105 mol C 12.01 g H: 5.14 g x 1 mol = 5.09 mol H 1.01 g N: 10.21 g x 1 mol = 0.7288 mol N 14.01 g O: 23.33 g x 1 mol = 1.456 mol O 16.00 g
- 52. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: 5.105 mol C: = 7.005 ≈ 7 0.7288 mol 5.09 mol H: = 6.984 ≈ 7 0.7288 mol 0.7288 mol N: = 1.000 0.7288 mol O: 1.458 mol = 2.001 ≈ 2 0.7288 mol
- 53. Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2
- 55. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
- 56. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced
- 57. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced – H is determined from the mass of H2O produced
- 58. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced – H is determined from the mass of H2O produced – O is determined by difference after the C and H have been determined
- 59. Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O
- 60. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products
- 61. 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)
- 62. Stoichiometric Calculations C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
- 63. Stoichiometric Calculations C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6…
- 64. Stoichiometric Calculations C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…
- 65. Stoichiometric Calculations C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O…
- 66. Stoichiometric Calculations C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams
- 67. SAMPLE EXERCISE 3.16 continued PRACTICE EXERCISE The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory: How many grams of O2 can be prepared from 4.50 g of KClO3?
- 68. SAMPLE EXERCISE 3.16 continued PRACTICE EXERCISE The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory: How many grams of O2 can be prepared from 4.50 g of KClO3? Answer: 1.77 g
- 69. SAMPLE EXERCISE 3.17 continued PRACTICE EXERCISE Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane?
- 70. SAMPLE EXERCISE 3.17 continued PRACTICE EXERCISE Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane? Answers: 3.64 g
- 72. How Many Cookies Can I Make?
- 73. How Many Cookies Can I Make? • You can make cookies until you run out of one of the ingredients
- 74. 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)
- 75. 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
- 76. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount
- 77. 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 H2)
- 78. Limiting Reactants In the example below, the O2 would be the excess reagent
- 79. PRACTICE EXERCISE Consider the reaction A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?
- 80. PRACTICE EXERCISE Consider the reaction A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction? Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2
- 81. PRACTICE EXERCISE A strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction?
- 82. PRACTICE EXERCISE A strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction? Answers: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn
- 84. Theoretical Yield • The theoretical yield is the amount of product that can be made
- 85. 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
- 86. 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.
- 87. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make
- 88. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Percent Yield = Theoretical Yield x 100
- 89. PRACTICE EXERCISE Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale: (a) If you start with 150 g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield?
- 90. PRACTICE EXERCISE Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale: (a) If you start with 150 g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? Answers: (a) 105 g Fe, (b) 83.7%