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) Law of Conservation of Mass 2 C(s) + O 2 (g) 2 CO(g) 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 4 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 expression 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 expression 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 next fewest total reactants and products and balance it. Repeat the process for additional elements if necessary. 4

5 Example 1: balance the reaction that occurs between nitrogen dioxide and water to form nitric acid and nitrogen monoxide 1. Write an expression 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 expression is already balanced. If so, you re done! 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 5

6 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) 4. Choose the element that appears in the next 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 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 oxygen to produce CO 2 + H 2 O, for example, balance - CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) TIP FOR ALL COMBUSTION REACTIONS: Since oxygen appears by itself, balance the other elements first, and then O 2 Combustion Reactions Example: C 2 H 6 (ethane) is combusted in oxygen C 2 H 6 (g) + O 2 (g) CO 2 (g) + 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 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 product into grams (or other desired quantities) Stoichiometry Example, p. 282 If the combustion of fossil fuels adds 8.2 x kilograms of carbon to the atmosphere each year as CO 2, what is the mass of added carbon dioxide? Step 1: Write the balanced chemical equation Step 2: Convert quantities of known substances (C) into moles 10

11 Stoichiometry Example, p. 282 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) Step 4: Convert moles of CO 2 into grams Stoichiometry Example, 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 Example, p. 283 Step 1: Write the balanced chemical equation Step 2: Convert quantities of known substances (ASA) into moles Stoichiometry Example, 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) 12

13 Stoichiometry Example, p. 283 Step 4: Convert moles of SA and AA into grams Sample Exercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Each year, power plants in the U.S. consume about 1.1 x 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 13

14 Sample Exercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Step 2: Convert quantities of known substances (C) into moles Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) Step 4: Convert moles of CO 2 into grams 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 x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound 3 x (1.008 g) %H = x 100% = 3.086% g 1 x (30.97 g) %P = x 100% = 31.60% g 4 x (16.00 g) %O = x 100% = 65.31% g H 3 PO 4 MM = g/mol 3.086% % % = 99.96% = 100% Sample Exercise 7.4 Calculating Percent Composition from a Chemical Formula The mineral forsterite: Mg 2 SiO 4, MW =

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 Exercise 7.6 A sample of the carbonate mineral dolomite is 21.73% Ca, 13.18% Mg, 13.03% C, and the rest is oxygen. What is its empirical formula? % O = = % 1. Assume 100 g 2. Convert to moles Sample Exercise Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary NOT REQUIRED HERE 17

18 Example 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 oxygen. What are the empirical and molecular formulas? 1. Assume 100 g 2. Convert to moles % O = = % Example Illustrating Step 4 3. Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary 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 x 2 glycolaldehyde C 2 H 4 O 2 = (CH 2 O) x 2 = C 2 H 4 O 2 Or more generally molecular = (empirical) x n 19

20 Empirical and Molecular Formulas Compared molecular = (empirical) x n it follows that - MW = (FW) x n and therefore - n = MW FW e.g. glucose assume we know the MW = Empirical = CH 2 O FW = 30.0 n = = 6 So molecular = (CH 2 O) x 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 Exercise 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 Sample Exercise 7.7 Using Percent Composition and Molecular weight to Derive a Molecular Fomula 3. Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary NOT NEEDED 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 Experimental Determination of Empirical Formulas: Combustion Analysis 22

23 Experimental Determination of Empirical Formulas: Combustion Analysis C x H y O z + O 2 (g) x CO 2 (g) + y/2 H 2 O(g) mass sample Calculation Outline: excess Some of the oxygen comes from the sample, some from the excess 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 Exercise 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? 23

24 Sample Exercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data g of O = g of sample g of C - g of H and so the moles of O = Sample Exercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data Now divide by the smallest number of moles - 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 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 excess and B is the limiting reagent 26

27 Sample Exercise 7.9: Identifying the Limiting Reagent in a Reaction Mixture The flame in an acetylene torch reaches temperatures as high as 3500 o C as a result of the combustion of a mixture of acetylene (C 2 H 2 ) and pure oxygen. 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 mixture stoichiometric? Given ratio: 2 C 2 H O 2 4 CO H 2 O 52.0 g 188 g Sample Exercise 7.9: Identifying the Limiting Reagent in a Reaction Mixture 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 excess and so C 2 H 2 is the LR 27

28 Percent Yield Theoretical Yield is the maximum 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 x 100 Sample Exercise 7.10 The industrial process for making the ammonia used in fertilizer, explosives, 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 mixture that initially contains 6.00 kg of H 2 (MW = 2.016) and an excess 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 = x 100 Excess (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 - 28

29 Sample Exercise 7.10 % Yield = Actual Yield Theoretical Yield x