Chapter 7: Stoichiometry - Mass Relations in Chemical Reactions

Transcription

1 Chapter 7: Stoichiometry - Mass Relations in Chemical Reactions How do we balance chemical equations? How can we used balanced chemical equations to relate the quantities of substances consumed and produced in chemical reactions? How can we determine a compound s elemental composition and chemical formula? Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 1

2 Law of Conservation of Mass The law of conservation of mass states that the sum of the masses of the reactants of a chemical equation is equal to the sum of the masses of the products. C(s) + O 2 (g) CO 2 (g) 12 g + 32 g = 44 g 1 mole 1 mole 1 mole Law of Conservation of Mass 2 C(s) + O 2 (g) 2 CO(g) 2 12 = 24 g + 32 g = 2 28 = 56 g 2 mole 1 mole 2 mole 2

3 Chemical Equations 2 C(s) + O 2 (g) 2 CO(g) The 2 s are called stoichiometric coefficients The symbol means reaction proceeds in this direction a symbol means the reaction is at equilibrium (s) = solid phase (l) = liquid phase (g) = gas phase (aq) = aqueous phase = heat Stoichiometric Coefficients: the ratios between reactants and/or products mol O 2 2 mol N 2 O 5 2 mol N 2 O 5 4 mol NO 2 3

4 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Guidelines for Balancing Chemical Equations 1. Write an epression using correct chemical formulas for the reactants and products, separated by an arrow ( ). Include symbols indicating physical states. 2. For each element, add up the numbers of atoms on each side. Check whether the epression is already balanced. If so, you re done! 3. Otherwise - if present - choose an element that appears in only one reactant and product to balance first. Insert the appropriate coefficient(s) to balance this element. 4. Choose the element that appears in the net fewest total reactants and products and balance it. Repeat the process for additional elements if necessary. 4

5 Eample 1: balance the reaction that occurs between nitrogen dioide and water to form nitric acid and nitrogen monoide 1. Write an epression using correct chemical formulas for the reactants and products, separated by an arrow ( ). Include symbols indicating physical states. NO 2 (g) + H 2 O(l) HNO 3 (aq) + NO(aq) 2. For each element, add up the numbers of atoms on each side. Check whether the epression is already balanced. If so, you re done! N = 1 N = 2 O = 3 O = 4 H = 2 H = 1 NO 2 (g) + H 2 O(l) HNO 3 (aq) + NO(aq) 3. Otherwise - if present - choose an element that appears in only one reactant and product to balance first. Insert the appropriate coefficient(s) to balance this element. = H, so try a 2 in front of HNO 3 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) N = 1 N = 3 O = 3 O = 7 H = 2 H = 2 5

6 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) 4. Choose the element that appears in the net fewest total reactants and products and balance it. Repeat the process for additional elements if necessary. = N, try a 3 in front of NO 2 3 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) N = 3 N = 3 O = 7 O = 7 H = 2 H = 2 Done! Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 6

7 Combustion Reactions The reaction of an organic compound with oygen to produce CO 2 + H 2 O, for eample, balance - CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) TIP FOR ALL COMBUSTION REACTIONS: Since oygen appears by itself, balance the other elements first, and then O 2 CH 4 (g) + O 2 (g) CO 2 (g) + 2 H 2 O(g) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) Done! Combustion Reactions Eample: C 2 H 6 (ethane) is combusted in oygen C 2 H 6 (g) + O 2 (g) CO 2 (g) + H 2 O(g) C 2 H 6 (g) + O 2 (g) 2 CO 2 (g) + H 2 O(g) C 2 H 6 (g) + O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) C 2 H 6 (g) + 7/2 O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(g) 7

8 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Stoichiometric Calculations and the Carbon Cycle Photosynthesis: 6 CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (aq) + 6 O 2 (g) 8

9 Mauna Loa Observatory CO 2 Concentrations since CO 2 emissions over the last 800,000 years - 9

10 Amounts of Reactants & Products = STOICHIOMETRY 1/MM ratio MM 1. Write the balanced chemical equation 2. Convert the mass of the reactant into moles 3. Use coefficients in the balanced equation to calculate the number of moles of product (stoichiometric ratio) 4. Convert moles of products into grams (or other desired quantities) Stoichiometry Eample, p. 282 If the combustion of fossil fuels adds kilograms of carbon to the atmosphere each year as CO 2, what is the mass of added carbon dioide? Step 1: Write the balanced chemical equation C(s) + O 2 (g) CO 2 (g) Step 2: Convert quantities of known substances (C) into moles kg C 1000 g C kg 1 mol C 12.0 g C = mol C 10

11 Stoichiometry Eample, p. 282 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) mol C 1 mol CO 2 = mol C 14 mol CO 2 Step 4: Convert moles of CO 2 into grams mol CO g CO 2 = mol CO 16 g CO 2 2 Stoichiometry Eample, p. 283 C 7 H 6 O 3 MW = C 4 H 6 O 3 MW = C 9 H 8 O 4 MW = Suppose we wish to prepare 1.00 kg of acetylsalicylic acid. How many grams of salicylic acid and how many grams of acetic anhydride are needed? 11

12 Stoichiometry Eample, p. 283 Step 1: Write the balanced chemical equation SA + AA ASA + Ac? g? g 1.00 kg Step 2: Convert quantities of known substances (ASA) into moles 1.00 kg ASA 1000 g ASA kg ASA 1 mol ASA g ASA = mol ASA Stoichiometry Eample, p. 283 SA + AA ASA + Ac Step 3: Use coefficients in balanced equation to calculate the number of moles of SA and AA (stoichiometric ratio) mol SA = mol ASA 1 mol SA 1 mol ASA = mol SA mol AA = mol ASA 1 mol AA 1 mol ASA = mol AA 12

13 Stoichiometry Eample, p. 283 Step 4: Convert moles of SA and AA into grams g SA = mol SA g SA 1 mol SA = 767 g SA g AA = mol AA g AA 1 mol AA = 567 g AA Sample Eercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Each year, power plants in the U.S. consume about kg of natural gas (CH 4 ). How many kg of CO 2 (MW = 44.01) are released into atmosphere from these power plants. Given that natural gas is mainly CH 4 (MW = 16.04), base the calculation on its combustion reaction Step 1: Write the balanced chemical equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) kg? g 13

14 Sample Eercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Step 2: Convert quantities of known substances (CH 4 ) into moles kg CH 1000 g CH 1 mol CH = mol kg CH g CH 4 CH 4 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) = mol CO mol CH 2 4 = mol CH 12 mol CO 2 4 Step 4: Convert moles of CO 2 into grams g CO mol CO 2 2 = g CO 2 1 mol CO 2 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 14

15 Percent Composition: the composition of a compound in terms of the percentage by mass of each element in the compound n molar mass of element molar mass of compound 100% n is the number of moles of the element in 1 mole of the compound 3 (1.008 g) %H = 100% = 3.086% g 1 (30.97 g) %P = 100% = 31.60% g 4 (16.00 g) %O = 100% = 65.31% g H 3 PO 4 MM = g/mol 3.086% % % = 99.96% = 100% Sample Eercise 7.4 Calculating Percent Composition from a Chemical Formula The mineral forsterite: Mg 2 SiO 4, MW = %Mg = %Si = %O = 2 (24.31 g Mg) g 1 (28.09 g Si) g 4 (16.00 g O) g 100% = 34.55% 100% = 19.96% 100% = 45.48% 34.55% % % = 99.99% = 100% 15

16 Empirical Formula from % Composition A formula showing the smallest whole number ratio of elements in a compound, e.g. Benzene:» Empirical = CH» Molecular = C 6 H 6 Glucose» Empirical = CH 2 O» Molecular = C 6 H 12 O 6 Empirical Formula from % Composition 1. Assume 100 g 2. Convert to moles 3. Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary 16

17 Sample Eercise 7.6 A sample of the carbonate mineral dolomite is 21.73% Ca, 13.18% Mg, 13.03% C, and the rest is oygen. What is its empirical formula? % O = = % 1. Assume 100 g 2. Convert to moles Ca = g Mg = g 1 mol Ca g 1 mol Mg g = mol Ca = mol Mg C = g 1 mol C g = mol C O = g 1 mol O g = mol O Sample Eercise Divide by fewest number of moles Ca = = 1.0 Mg = C = = 1.0 = 2.0 Empirical formula = CaMgC 2 O 6 O = = Convert the mole ratio from step 3 into small whole numbers if necessary NOT REQUIRED HERE 17

18 Eample Illustrating Step 4 Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is %C and 5.30 %H; the rest is oygen. What are the empirical and molecular formulas? 1. Assume 100 g 2. Convert to moles % O = = % C = g 1 mol C g = mol C H = 5.30 g 1 mol H g = mol H O = g 1 mol O g = mol O Eample Illustrating Step 4 3. Divide by fewest number of moles C = H = O = = 2.67 = 2.67 = 1.0 Note that 2.67 = So multiply by 3 which = 8 4. Convert the mole ratio from step 3 into small whole numbers if necessary C = H = O = = = = 3 Empirical formula = C 8 H 8 O 3 18

19 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Empirical and Molecular Formulas Compared The molecular formula can be determined from the empirical formula if the molecular weight of the compound is known. empirical = CH 2 O formula weight (FW) = 30 g/mol molecular = C 2 H 4 O 2 Molecular weight (MW) = 60 g/mol Note that molecular formula = empirical 2 glycolaldehyde C 2 H 4 O 2 = (CH 2 O) 2 = C 2 H 4 O 2 Or more generally molecular = (empirical) R 19

20 Empirical and Molecular Formulas Compared molecular = (empirical) R it follows that - MW = (FW) R and therefore - R = MW FW e.g. glucose assume we know the MW = Empirical = CH 2 O FW = 30.0 R = = 6 So molecular = (CH 2 O) 6 = C 6 H 12 O 6 Molecular Mass and Mass Spectrometry Acetylene C 2 H 2 Benzene C 6 H 6 20

21 Sample Eercise 7.7 Using Percent Composition and Molecular weight to Derive a Molecular Fomula Pheromones are chemical substances secreted by members of a species to stimulate a response in other individuals of the same species. The percent composition of eicosene, a compound similar to the Japanese beetle mating pheromone, is 85.63% C and 14.37% H. Its molecular mass, as determined by mass spectrometry, is 280 amu. What is the molecular formula of eicosene? 1. Assume 100 g 2. Convert to moles C = g 1 mol C g = mol C H = g 1 mol H g = mol H Sample Eercise 7.7 Using Percent Composition and Molecular weight to Derive a Molecular Fomula 3. Divide by fewest number of moles C = H = = 1.00 = 2 Empirical formula = CH 2 FW = Convert the mole ratio from step 3 into small whole numbers if necessary NOT NEEDED R = MW FW = = 20 So molecular = (CH 2 ) 20 = C 20 H 40 21

22 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Eperimental Determination of Empirical Formulas: Combustion Analysis 22

23 Eperimental Determination of Empirical Formulas: Combustion Analysis C H y O z + O 2 (g) CO 2 (g) + y/2 H 2 O(g) mass sample Calculation Outline: ecess Some of the oygen comes from the sample, some from the ecess O 2 g CO 2 mol CO 2 mol C g C g H 2 O mol H 2 O mol H g H g of O = g of sample g of C - g of H Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data Combustion of grams of an organic compound known to contain only C, H and O produces g CO 2 and g H 2 O. What is the empirical formula of the compound? mol C = g CO 2 1 mol CO g CO 2 1 mol C mol CO 2 = mol C g C = mol C g C 1 mol C = g C mol H = g H 2 O 1 mol H 2 O 18.0 g H 2 O 2 mol H mol H 2 O = mol H g H = = mol H g H 1 mol H = g H 23

24 Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data g of O = g of sample g of C - g of H g of O = g g C g H g of O = g and so the moles of O = g = mol O 1 mol O g O Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data Now divide by the smallest number of moles - C = mol C = 3 H = mol H = 4 Empirical formula = C 3 H 4 O O = mol O = 1 24

25 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Limiting Reagents Limiting Reagents - a reactant that is consumed completely in a chemical reaction before the other reactant(s) run out. The amount of product formed depends on the amount of the limiting reagent available. Each sandwich consists of 2 slices bread, 1 slice of cheese, and 1 slice of salami 8 slices bread 4 slices cheese 3 slices salami 25

26 Limiting Reagents 2 Br + 1 Ch + 1 Sal 1 sandwich Given: Here s the logic compare the given number of moles to the stoichiometric ratio Given ratio: 8 Br = 2.7 tip: use a ratio > 1 3 Sal Stoichiometric ratio: 2 Br = Sal Br in ecess, so need more salami! = LR Strategy for determining which reactant is the Limiting Reagent: aa + bb C 1. Convert grams of each into moles 2. Calculate the stoichiometric ratio that s the largest, e.g. a/b or b/a 3. Calculate the given mole ratio in the same way 4. Compare to identify the LR if moles A moles B given > a b stoichiometric then A is in ecess and B is the limiting reagent 26

27 Sample Eercise 7.9: Identifying the Limiting Reagent in a Reaction Miture The flame in an acetylene torch reaches temperatures as high as 3500 o C as a result of the combustion of a miture of acetylene (C 2 H 2 ) and pure oygen. If these two gases flow from high-pressure tanks at the rates of 52.0 g C 2 H 2 and 188 g O 2 per minute, which reactant is the limiting reagent, or is the miture stoichiometric? Given ratio: mol C 2 H 2 2 C 2 H O 2 4 CO H 2 O 52.0 g 188 g 1 mol C = 52.0 g C 2 H 2 H 2 2 = 2.00 mol C 26.0 g C 2 H 2 H 2 2 tip: use a ratio > mol O mol C H 1 mol O 2 2 mol O 2 = 188 g O g O = 5.88 mol O 2 2 = 2.94 Sample Eercise 7.9: Identifying the Limiting Reagent in a Reaction Miture 2 C 2 H O 2 4 CO H 2 O is moles O 2 moles C 2 H 2 given > 5 mol O 2 2 mol C 2 H 2 stoichiometric??? Given ratio: Stoichiometric ratio: 5.88 mol O mol C 2 H 2 = mol O 2 2 mol C 2 H 2 = 2.5 therefore O 2 is in ecess and so C 2 H 2 is the LR 27

28 Percent Yield Theoretical Yield is the maimum amount of product formed from given quantities of reactants (check if a limiting reagent is present). Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield 100 Sample Eercise 7.10 The industrial process for making the ammonia used in fertilizer, eplosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure. If 18.2 kg of NH 3 (MW = 17.03) is produced by a reaction miture that initially contains 6.00 kg of H 2 (MW = 2.016) and an ecess of N 2, what is the percent yield of the reaction? Actual Yield N 2 (g) + 3 H 2 (g) 2 NH 3 (g) % Yield = 100 Ecess (no LR) 6.00 kg Actual yield = 18.2 kg Theoretical Yield Calculate the theoretical yield of NH 3 based on H 2, and then calculate the % yield g g NH 3 = 6.00 kg H 2 kg mol H g H 2 2 mol NH g NH 3 3 mol H 2 mol NH 3 = g NH 3 = kg NH 3 = 33.8 kg NH 3 28

29 Sample Eercise 7.10 % Yield = Actual Yield Theoretical Yield 100 Actual Yield = 18.2 kg Theoretical Yield = kg % Yield = 18.2 kg kg 100 = 53.8 % 29