Chapter 03- Introduction PDF

Chapter 3 (6 Lectures) Fifth Edition Stoichiometry: Ratios of Combination  Stoichiometry: Ratios of CHAPTER 3 Combination 3.1 Molecular and Formula Masses. 3.2 Percent Composition of Compounds. 3.3 Chemical...

Chapter 3 (6 Lectures) Fifth Edition Stoichiometry: Ratios of Combination  Stoichiometry: Ratios of CHAPTER 3 Combination 3.1 Molecular and Formula Masses. 3.2 Percent Composition of Compounds. 3.3 Chemical Equations. 3.4 The Mole and Molar Masses. 3.5 Combustion Analysis. 3.6 Calculations with Balanced Chemical Equations. 3.7 Limiting Reactants. 2 ©2020 McGraw-Hill Education  3.1 Molecular and Formula Masses Topics Molecular and Formula Masses Learning Objectives Define the terms molecular mass and formula mass of a substance. Calculate the molecular mass of the formula mass from a formula 3 ©2020 McGraw-Hill Education  3.1 Molecular and Formula Masses 1 Molecular and Formula Masses Using atomic masses from the periodic table and a molecular formula, we can determine the molecular mass, which is the mass in atomic mass units (amu) of an individual molecule. molecular mass of H2O = 2(atomic mass of H) + atomic mass of O = 2(1.008 amu ) + 16.00 amu= 18.02 amu 4 ©2020 McGraw-Hill Education  3.1 Molecular and Formula Masses 2 Molecular and Formula Masses Although an ionic compound does not have a molecular mass, we can use its empirical formula to determine its formula mass. 5 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.1 Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: (a) propane, (C3H8). (b) lithium hydroxide, (LiOH). (c) barium acetate, [Ba(C2H3O2)2]. 6 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.1 Solution Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: (a) propane, (C3H8). (b) lithium hydroxide, (LiOH). (c) barium acetate, [Ba(C2H3O2)2]. Solution (a) 3(12.01 amu ) + 8(1.008 amu ) = 44.09 amu. (b) 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu. (c) 137.3 amu + 4(12.01 amu + 6(1.008 amu ) + 4(16.00 amu ) = 255.4 amu. 7 ©2020 McGraw-Hill Education  3.1 Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: (a) propane, (C3H8) (b) lithium hydroxide, (LiOH) (c) barium acetate, [Ba(C2H3O2)2] Solution (a) 3(12.01 amu) + 8(1.008 amu) = 44.09 amu (b) 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu (c) 137.3 amu + 4(12.01 amu) + 6(1.008 amu) + 4(16.00 amu) = 255.4 amu 8 ©2020 McGraw-Hill Education 8  3.2 Percent Composition of Compounds Topics Percent Composition of Compounds Learning Objectives Define mass percentage. Calculate the percent composition of the elements in a compound. Calculate the mass of an element in a given mass of compound. 9 ©2020 McGraw-Hill Education  3.2 Percent Composition of Compounds 1 Percent Composition of Compounds A list of the percent by mass of each element in a compound is known as the compound’s percent composition by mass. n  atomic mass of element percent by mass of an element   100% molecular or formula mass of compound H2O2 2 ×1.008 amu H %H   100%  5.926% 34.02 amu H2O2 2 ×16.00 amu O %O   100%  94.06% 34.02 amu H2O2 10 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.2 Setup Calculate the percent composition by mass of lithium carbonate (Li2CO3). Setup The formula mass of Li2CO3 is 2(6.941 amu) + 12.01 amu + 3(16.00 amu) = 73.89 amu 11 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.2 Solution Calculate the percent composition by mass of lithium carbonate (Li2CO3). Solution 2  6.941 amu Li %Li   100%  18.79% 73.89 amu Li2CO3 12.01 amu C %C   100%  16.25% 73.89 amu Li2CO3 3  16.00 amu O %O   100%  64.96% 73.89 amu Li2CO3 12 ©2020 McGraw-Hill Education  3.3 Chemical Equations Topics Interpreting and Writing Chemical Equations Balancing Chemical Equations Learning Objectives Relate the coefficients in a balanced chemical equation to the number of molecules or atoms. Balance chemical equations. 13 ©2020 McGraw-Hill Education  3.3 Chemical Equations 6 Balancing Chemical Equations Access the text alternative for these images 14 ©2020 McGraw-Hill Education  3.3 Chemical Equations 7 Balancing Chemical Equations Balancing a chemical equation requires something of a trial-and-error approach. In general, it will facilitate the balancing process if you do the following: 1. Change the coefficients of compounds (for example, CO2) before changing the coefficients of elements (for example, O2). 15 ©2020 McGraw-Hill Education  3.3 Chemical Equations 8 Balancing Chemical Equations 2. Treat polyatomic ions that appear on both sides of the equation (for example, CO32− as units, rather than counting their constituent atoms individually. 3. Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient. 16 ©2020 McGraw-Hill Education  3.3 Chemical Equations 9 Balancing Chemical Equations Combustion refers to burning in the presence of oxygen. C4H10  g   O2  g  CO2  g   H2O  l  4  C 1 10  H  2 2O3 C4H10  g   O2  g  4CO2  g   H2O  l  4 C4 10  H  2 2O9 17 ©2020 McGraw-Hill Education  3.3 Chemical Equations 10 Balancing Chemical Equations C4H10  g   O2  g  4CO2  g   5H2O  l  4 C4 10  H  10 2  O  13 13 C4H10  g   O2  g  4CO2  g   5H2O  l  2 4 C4 10  H  10 13  O  13 18 ©2020 McGraw-Hill Education  3.3 Chemical Equations 11 Balancing Chemical Equations 2C4H10  g   13O2  g  8CO2  g   10H2O  l  8C8 20  H  20 26  O  26 19 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.3 Setup Write and balance the chemical equation for the aqueous reaction of barium hydroxide and perchloric acid to produce aqueous barium perchlorate and water. Setup Ba  OH2  aq   HClO4  aq  Ba  ClO4 2  aq   H2O  l  20 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.3 Solution 1 Write and balance the chemical equation for the aqueous reaction of barium hydroxide and perchloric acid to produce aqueous barium perchlorate and water. Solution Ba  OH2  aq   HClO4  aq  Ba  ClO4 2  aq   H2O  l  1  Ba  1 2  O 1 3H2 (not including O atoms 1  ClO4  2 in ClO4− ions) 21 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.3 Solution 2 Solution Ba  OH2  aq   2HClO4  aq  Ba  ClO4 2  aq   H2O  l  1  Ba  1 2  O 1 4 H2 2  ClO4  2 22 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.3 Solution 3 Solution Ba  OH2  aq   2HClO4  aq  Ba  ClO4 2  aq   2H2O  l  1  Ba  1 2O2 4 H 4 2  ClO4  2 23 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.4 Setup Assuming that the only products are CO2 and H2O, write and balance the equation for the metabolism of butyric acid, C4H8O2. Setup C4H8O2  aq   O2  g  CO2  g   H2O  l  24 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.4 Solution Solution C4H8O2  aq   O2  g  CO2  g   H2O  l  C4H8O2  aq   O2  g  4CO2  g   H2O  l  C4H8O2  aq   O2  g  4CO2  g   4H2O  l  C4H8O2  aq   5O2  g  4CO2  g   4H2O  l  25 ©2020 McGraw-Hill Education  3.4 For Practice Balance the following reaction. ___C6H6(l) + ___O2(g) → ___H2O(g) + __CO2(g) 26 ©2020 McGraw-Hill Education 26  3.4 The Mole and Molar Masses Topics The Mole Determining Molar Mass Interconverting Mass, Moles, and Numbers of Particles Empirical Formula from Percent Composition Learning Objectives Define the mole and its relation to Avogadro’s number. Calculate the molar mass of a substance in units of g/mol. Calculate the mass of atoms and molecules in grams. Perform calculations using the mole. Interconvert from moles of substance to grams of substance to number of particles. 27 ©2020 McGraw-Hill Education  3.4 The Mole and Molar Masses Learning Objectives (continued) Calculate the number of molecules in a given mass of substance. Determine the empirical formula from the percent composition. 28 ©2020 McGraw-Hill Education 28  3.4 Molecules and Molecular Compounds 1 The Mole People use a variety of counting groups to conveniently indicate the number of objects in some set: 1 “pair” objects 2 objects 1 “dozen” objects 12 objects 1 “gross” objects 144 objects 1 “million” objects 1,000,000 objects 1 “trillion” objects 1012 objects 1 “mole” objects 6.023 × 1023 objects 29 ©2020 McGraw-Hill Education  3.4 The Mole and Molar Masses 2 The Mole A “mole” is a counting group , defined as the number of atoms in exactly 12 g of carbon-12. This number of atoms in 12 g of carbon-12 is known as Avogadro’s Number (NA): NA = 6.0221418 × 1023 objects If you order one dozen doughnut, you are asking for 12 doughnuts. If you order one mole of doughnuts, you are asking for 6.022 × 1023 doughnuts! 30 ©2020 McGraw-Hill Education  3.5 A typical human body contains 30.00 moles of calcium. Determine (a) the number of Ca atoms in 30.00 moles of calcium and (b) the number of moles of calcium in a sample containing 1.00 × 1020 Ca atoms. 1 mol Ca atoms 6.022  1023 Ca atoms 6.022  1023 Ca atoms 1 mol Ca atoms Solution 31 31 ©2020 McGraw-Hill Education 18  SAMPLE PROBLEM 3.5 Setup A typical human body contains roughly 30 moles of calcium. Determine (a) the number of Ca atoms in 30.00 moles of calcium and (b) the number of moles of calcium in a sample containing 1.00 × 1020 Ca atoms. 1 mol Ca atoms 6.022  1023 Ca atoms 6.022  1023 Ca atoms 1 mol Ca atoms Solution 6.022  1023 Ca atoms  a 30.00 mol Ca   1.807  1025 Ca atoms 1 mol Ca 1 mol Ca b 1.00  10 Ca atoms  20  1.66  10 4 mol Ca 6.022  10 Ca atoms 23 32 ©2020 McGraw-Hill Education  3.4 The Mole and Molar Masses Determining Molar Mass The molar mass (M) of a substance is defined as the mass in grams of one mole of the substance. [SI Unit: g/mol] By definition, the mass of a mole of carbon-12 is exactly 12 g. For any element, its atomic mass in amu is equal to the weight of 1 mole of the element in grams. In general, an element’s molar mass in grams is numerically equal to its atomic mass in atomic mass units. The atomic mass of calcium is 40.08 amu and its molar mass is 40.08 g, the atomic mass of sodium is 22.99 amu and its molar mass is 22.99 g, etc. 33 ©2020 McGraw-Hill Education 33  3.4 The Mole and Molar Masses 6 Determining Molar Mass 1 amu = 1.661×10−24 g ⟶ 1g = 6.022×1023 amu Recap: the molar mass (in grams) of any compound is numerically equal to its molecular or formula mass (in amu). 34 ©2020 McGraw-Hill Education  3.4 The Mole and Molar Masses 7 Interconverting Mass, Moles, and Numbers of Particles Access the text alternative for these images 35 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.6 Setup Determine (a) the number of moles of C in 10.00 g of naturally occurring carbon and (b) the mass of 0.905 mole of sodium chloride. Setup The molar mass of carbon is 12.01 g/mol. The molar mass of a compound is numerically equal to its formula mass. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. 36 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.6 Solution Solution 1 mol C  a 10.00 g C   0.8326 mol C 12.01 g C 58.44 g NaCl b 0.905 mol NaCl   52.9 g NaCl 1mol NaCl 37 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.7 (a) Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water. (b) Determine the mass of 7.92 × 1019 carbon dioxide molecules. 38 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.7 Solution Solution 1mol H2O 6.022  1023 H2O molecules  a 3.26 g H2O   18.02 g H2O 1mol H2O  1.09  1023 H2O molecules 2 H atoms 1.09  10 H2O molecules  23  2.18  1023 H atoms 1H2O molecule 1O atom 1.09  10 H2O molecules  23  1.09  1023 O atoms 1H2O molecule 39 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.7 Solution 2 Solution (b) 1mol CO2 44.01 g CO2 7.92  10 CO2 molecules  19  6.022  10 CO2 molecules 1 mol CO2 23  5.79  103 g CO2 40 ©2020 McGraw-Hill Education  3.4 The Mole and Molar Masses 8 Empirical Formula from Percent Composition With the concepts of the mole and molar mass, we can now use the experimentally determined percent composition to determine the empirical formula of a compound. 41 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.8 Strategy Determine the empirical formula of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass. Strategy Assume a 100-g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles. 42 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.8 Setup Setup The empirical formula of a compound consisting of N and O is NxOy. The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One hundred grams of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass contains 30.45 g N and 69.55 g O. 43 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.8 Solution Solution 1 mol N 30.45 g N   2.173 mol N 14.01 g N 1 mol O 69.55 g O   4.347 mol O 16.00 g O N2.173O4.347 4.347/2.173  2 NO2 44 ©2020 McGraw-Hill Education  3.5 Combustion Analysis Topics Determination of Empirical Formula Determination of Molecular Formula Learning Objectives Use the mass of CO2 and H2O produced from combustion of a compound containing C,H, and O, to determine the empirical formula of that compound. Understand the relationship between the molar mass of a substance and its empirical formula mass. Determine the molecular formula from the empirical formula and molar mass. 45 ©2020 McGraw-Hill Education  3.5 Combustion Analysis Determination of Empirical Formula Suppose that in one such experiment the combustion of 18.8 g of glucose produced 27.6 g of CO2 and 11.3 g of H2O. We can calculate the mass of carbon and hydrogen in the original 18.8-g sample of glucose as follows: 46 ©2020 McGraw-Hill Education 46  3.5 Combustion Analysis Determination of Empirical Formula Thus, 18.8 g of glucose contains 7.53 g of carbon and 1.26 g of hydrogen. The remaining mass [18.8 g – (7.53 g + 1.26 g) = 10.0 g] is oxygen. 47 ©2020 McGraw-Hill Education 47  3.5 Combustion Analysis Determination of Empirical Formula 48 ©2020 McGraw-Hill Education 48  3.5 Combustion Analysis 6 Determination of Empirical Formula 1mol C moles of C  7.53 g C   0.627 mol C 12.01 g C 1 mol H moles of H  1.26 g H   1.25 mol H 1.008 g H 1 mol O moles of O  10.0 g O   0.626 mol O 16.00 g O C0.627H1.25O0.626 49 ©2020 McGraw-Hill Education  3.5 Combustion Analysis 7 Determination of Empirical Formula 0.627/0.626 ≈ 1, 1.25/0.626 ≈ 2, and 0.626/0.626 = 1 Answer: CH2O 50 ©2020 McGraw-Hill Education  3.5 Combustion Analysis 8 Determination of Molecular Formula The molar mass of glucose is about 180 g. The empirical-formula mass of CH2O is about 30 g [12.01 g + 2(1.008 g) + 16.00 g]. To determine the molecular formula, first divide the molar mass by the empirical-formula mass: 180 g/30 g = 6. This tells us that there are six empirical-formula units per molecule in glucose. 51 ©2020 McGraw-Hill Education  3.5 Combustion Analysis 9 Determination of Molecular Formula Multiplying each subscript by 6 (recall that when none is shown, the subscript is understood to be a 1) gives the molecular formula, C6H12O6. 52 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.9 Setup Combustion of a 5.50-g sample of benzene produces 18.59 g CO2 and 3.81 g H2O. Determine the empirical formula and the molecular formula of benzene, given that its molar mass is approximately 78 g/mol. Setup The necessary molar masses are CO2, 44.01 g/mol; H2O, 18.02 g/mol; C, 12.01 g/mol; H, 1.008 g/mol; and O, 16.00 g/mol. 53 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.9 Solution 1 Solution 1 mol CO2 1 mol C 12.01 g C moles of C  18.59 g CO2     5.073 g C 44.01 g CO2 1 mol CO2 1 mol C 1 mol H2O 2 mol H 1.008 g H moles of H  3.81 g H2O     0.426 g H 18.02 g H2O 1 mol H2O 1 mol H The total mass of products is 5.073 g + 0.426 g = 5.499 g. Because the combined masses of C and H account for the entire mass of the original sample (5.499 g ≈ 5.50 g), this compound must not contain O. 54 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.9 Solution 2 Solution 1 mol C moles of C  5.073 g C   0.4224 mol C 12.01 g C 1 mol H moles of H  0.426 gH   0.423 mol H 1.008 g H C0.4224H0.423 CH 12.01g/mol  1.008 g/mol  13.02 g/mol 1278/13.02  6 C6H6 55 ©2020 McGraw-Hill Education  3.6 Calculations with Balanced Chemical Equations Topics Moles of Reactants and Products Mass of Reactants and Products Learning Objectives Use the coefficients in a chemical reaction to perform calculations. Relate the moles of a reactant to the moles of another reactant or to the moles of a product. Relate the mass of a reactant to the mass of another reactant or to the mass of a product. 56 ©2020 McGraw-Hill Education  3.6 Calculations with Balanced Chemical Equations1 Moles of Reactants and Products In stoichiometric calculations, we say that 2 moles of CO are equivalent to 2 moles of CO2, which can be represented as 2mol CO≏2 mol CO2 57 ©2020 McGraw-Hill Education  3.6 Calculations with Balanced Chemical Equations2 Moles of Reactants and Products 2 mol CO ≏ 2 mol CO2 Stoichiometric conversion factors: 2 mol CO 1 mol CO or 2 mol CO2 1 mol CO2 2 mol CO2 1 mol CO2 or 2 mol CO 1 mol CO 58 ©2020 McGraw-Hill Education  3.6 Calculations with Balanced Chemical Equations3 Moles of Reactants and Products Consider the complete reaction of 3.82 moles of CO to form CO2. 1mol CO2 moles CO2 produced  3.82 mol CO   3.82 mol CO2 1mol CO We can determine the stoichiometric amount of O2 (how many moles of O2 are needed to react with 3.82 moles of CO): 1 mol O2 moles O2 needed  3.82 mol CO   1.91 mol O2 2 mol CO 59 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.10 Urea can be synthesized in the laboratory by the combination of ammonia and carbon dioxide according to the equation (a) Calculate the amount of urea that will be produced by the complete reaction of 5.25 moles of ammonia. (b) Determine the stoichiometric amount of carbon dioxide required to react with 5.25 moles of ammonia. 60 ©2020 McGraw-Hill Education  3.10 Solution (a) (b) 61 ©2020 McGraw-Hill Education 61  SAMPLE PROBLEM 3.10 Solution Solution  a moles NH2 2 CO produced  1 mol (NH2 )2CO 5.25 mol NH3   2.63 mol NH2 2 CO 2 mol NH3 b moles CO2 required  1 mol CO2 5.25 mol NH3   2.63 mol CO2 2 mol NH 3 62 ©2020 McGraw-Hill Education  3.6 Calculations with Balanced Chemical Equations4 Moles of Reactants and Products Balanced chemical equations give us the relative amounts of reactants and products in terms of moles. However, because we measure reactants and products in the laboratory by weighing them, most often such calculations start with mass rather than the number of moles. 63 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.11 Nitrous oxide (N2O) is commonly used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is NH4NO3  s   Δ N2O  g   2H2O  g  (a) Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water produced in the reaction. 64 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.11 Setup Setup The molar masses are as follows: 80.05 g/mol for NH4NO3, 44.02 g/mol for N2O, and 18.02 g/mol for H2O. 1 mol NH4NO3 2 mol H2O and 1 mol N2O 1 mol N2O 65 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.11 Solution 1 Solution (a) 1 mol N2O 10.0 g N2O   0.227 mol N2O 44.02 g N2O 1 mol NH4NO3 0.227 mol N2O   0.227 mol NH4NO3 1 mol N2O 80.05 g NH4NO3 0.227 mol NH4NO3   18.2 g NH4NO3 1 mol NH4NO3 Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide. 66 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.11 Solution 2 Solution (b) 2 mol H2O 0.227 mol N2O   0.454 mol H2O 1 mol N2O 18.02 g H2O 0.454 mol H2O   8.18 g H2O 1 mol H2O Therefore, 8.18 g of water will also be produced in the reaction. 67 ©2020 McGraw-Hill Education  3.7 Limiting Reactants Topics Determining the Limiting Reactant Reaction Yield Types of Chemical Reactions Learning Objectives Understand how a limiting reactant or limiting reagent determines the moles of product formed during a chemical reaction and calculate how much excess reactant remains. Perform limiting reactant calculations involving moles. Perform limiting reactant calculations involving masses. Define and calculate the theoretical yield of chemical reactions. Determine the percentage yield of a chemical reaction. 68 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 1 Determining the Limiting Reactant The reactant used up first in a reaction is called the limiting reactant, because the amount of this reactant limits the amount of product that can form. When all the limiting reactant has been consumed, no more product can be formed. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. 69 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 2 Determining the Limiting Reactant CO(g) + 2H2(g)⟶CH3OH(l) Suppose that initially we have 5 moles of CO and 8 moles of H2: 70 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 3 Determining the Limiting Reactant 2 mol H2 moles of H2  5mol CO   10 mol H2 1mol CO Because there are only 8 moles of H2 available, there is insufficient H2 to react with all the CO. Therefore, H2 is the limiting reactant and CO is the excess reactant. 71 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 4 Determining the Limiting Reactant To determine how much CO will be left over when the reaction is complete, we must first calculate the amount of CO that will react with all 8 moles of H2: 1 mol CO moles of CO  8 mol H2   4 mol CO 2 mol H2 Thus, there will be 4 moles of CO consumed and 1 mole (5 mol − 4 mol) left over. 72 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 (1) Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) react to form carbon dioxide gas, among other products. 3NaHCO3(aq) + H3C6H5O7(aq) ⟶ 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq) 73 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 (2) An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 g of citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO2 forms. 74 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 Setup Setup 3NaHCO3(aq) + H3C6H5O7(aq) ⟶ 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq) 3 mol NaHCO3 1 mol H3C6H5O7 1 mol H3C6H5O7 1 mol NaHCO3 3mol CO2 3 mol CO2 3 mol NaHCO3 1 mol H3C6H5O7 75 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 Solution 1 Solution 1 mol NaHCO3  a 1.700 g NaHCO3   0.02024 mol NaHCO3 84.01 g NaHCO3 1 mol H3C6H5O7 1.000 gH3C6H5O7   0.005205 mol H3C6H5O7 192.12 g H3C6H5O7 1 mol H3C6H5O7 0.02024 mol NaHCO3   0.006745 mol H3C6H5O7 3 mol NaHCO3 The amount of H3C6H5O7 required to react with 0.02024 mol of NaHCO3 is more than a tablet contains. Therefore, citric acid is the limiting reactant. 76 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 Solution 2 Solution (b) 3 mol NaHCO3 0.005205 mol H3C6H5O7   0.01562 mol NaHCO3 1 mol H3C6H5O7 Thus, 0.01562 mol of NaHCO3 will be consumed, leaving 0.00462 mol unreacted. 84.01 g NaHCO3 0.00462 mol NaHCO3   0.388 g NaHCO3 1mol NaHCO3 77 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.12 Solution 3 Solution (c) 3 mol CO2 0.005205 mol H3C6H5O7   0.01562 mol CO2 1 mol H3C6H5O7 44.01 g CO2 0.01562 mol CO2   0.6874 g CO2 1mol CO2 To summarize the results: (a) citric acid is the limiting reactant, (b) 0.388 g sodium bicarbonate remains unreacted, and (c) 0.6874 g carbon dioxide is produced. 78 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 5 Determining the Limiting Reactant When you use stoichiometry to calculate the amount of product formed in a reaction, you are calculating the theoretical yield of the reaction. The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. It is the maximum obtainable yield, predicted by the balanced equation. 79 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 6 Reaction Yield In practice, the actual yield—the amount of product actually obtained from a reaction—is almost always less than the theoretical yield. The percent yield tells what percentage the actual yield is of the theoretical yield. actual yield % yield   100% theoretical yield 80 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.13 Aspirin, acetylsalicylic acid (C9H8O4), is produced by the reaction of salicylic acid (C7H6O3) and acetic anhydride (C4H6O3) according to the following equation: In a certain aspirin synthesis, 104.8 g of salicylic acid and 110.9 g of acetic anhydride are combined. Calculate the percent yield of the reaction if 105.6 g of aspirin are produced. 81 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.13 Setup Setup The necessary molar masses are 138.12 g/mol for salicylic acid, 102.09 g/mol for acetic anhydride, and 180.15 g/mol for aspirin. Solution 1mol C7H6O3 104.8 g C7H6O3   0.7588 mol C7H6O3 138.12 g C7H6O3 1mol C4H6O3 110.9 g C4H6O3   1.086 mol C4H6O3 102.09 g C4H6O3 Salicylic acid is the limiting reactant. 82 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.13 Solution Solution 1 mol salicylic acid (C7H6O3) ≏ 1 mol aspirin (C9H8O4) Therefore, the theoretical yield of aspirin is 0.7588 mol. 180.15 g C9H8O4 0.7588 mol C9H8O4   136.7 g C9H8O4 1 mol C9H8O4 Thus, the theoretical yield is 136.7 g. 105.6 g % yield   100%  77.25% yield 136.7 g 83 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 7 Types of Chemical Reactions Combination. A reaction in which two or more reactants combine to form a single product is known as a combination reaction. NH3(g) + HCl(g) ⟶ NH4Cl(s) 84 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 8 Types of Chemical Reactions Decomposition. A reaction in which two or more products form from a single reactant is known as a decomposition reaction. CaCO3  s   Δ CaO  s  + CO2  g  85 ©2020 McGraw-Hill Education  3.7 Limiting Reactants 9 Types of Chemical Reactions Combustion. A combustion reaction is one in which a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and water. By convention, we will consider the water produced in a combustion reaction to be liquid water. CH2O(l) + O2(g)⟶CO2(g) + H2O(l) 86 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.14 Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: a) H2(g) + Br2(g) ⟶ 2HBr(g). b) 2HCO2H(l) + O2(g) ⟶ 2CO2(g) + 2H2O(l). c) 2KClO3(s) ⟶ 2KCl(s) + 3O2(g). 87 ©2020 McGraw-Hill Education  SAMPLE PROBLEM 3.14 Solution Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: a) H2(g) + Br2(g) ⟶ 2HBr(g). b) 2HCO2H(l) + O2(g) ⟶ 2CO2(g) + 2H2O(l). c) 2KClO3(s) ⟶ 2KCl(s) + 3O2(g). Solution (a) Combination. (b) Combustion. (c) Decomposition. 88 ©2020 McGraw-Hill Education