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1/9/2015 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 1/9/2015 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) + O2(g) → CO2(g) Law of Conservation of Mass 2 C(s) + O2(g) → 2 CO(g) 2 1/9/2015 Chemical Equations 2 C(s) + O2(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 1/9/2015 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 1/9/2015 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! NO2(g) + H2O(l) → HNO3(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 HNO3 5 1/9/2015 NO2(g) + H2O(l) → 2 HNO3(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 NO2 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 1/9/2015 Combustion Reactions The reaction of an organic compound with oxygen to produce CO2 + H2O, for example, balance CH4(g) + O2(g) → CO2(g) + H2O(g) TIP FOR ALL COMBUSTION REACTIONS: Since oxygen appears by itself, balance the other elements first, and then O 2 Combustion Reactions Example: C2H6 (ethane) is combusted in oxygen C2H6(g) + O2(g) → CO2(g) + H2O(g) 7 1/9/2015 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 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g) 8 1/9/2015 Mauna Loa Observatory CO2 Concentrations since 1960 - CO2 emissions over the last 800,000 years - 9 1/9/2015 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 10 12 kilograms of carbon to the atmosphere each year as CO2, 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 1/9/2015 Stoichiometry Example, p. 282 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO2 (stoichiometric ratio) Step 4: Convert moles of CO2 into grams Stoichiometry Example, p. 283 C7H6O3 MW = 138.12 C4H6O3 MW = 102.09 C9H8O4 MW = 180.16 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 1/9/2015 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 1/9/2015 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 10 11 kg of natural gas (CH4). How many kg of CO2 (MW = 44.01) are released into atmosphere from these power plants. Given that natural gas is mainly CH4 (MW = 16.04), base the calculation on its combustion reaction – Step 1: Write the balanced chemical equation 13 1/9/2015 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 CO2 (stoichiometric ratio) Step 4: Convert moles of CO2 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 1/9/2015 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 x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound 3 x (1.008 g) x 100% = 3.086% 97.99 g 1 x (30.97 g) %P = x 100% = 31.60% 97.99 g 4 x (16.00 g) %O = x 100% = 65.31% 97.99 g %H = H3PO4 MM = 97.99 g/mol 3.086% + 31.60% + 65.31% = 99.96% = 100% Sample Exercise 7.4 – Calculating Percent Composition from a Chemical Formula The mineral forsterite: Mg2SiO4, MW = 140.71 15 1/9/2015 Empirical Formula from % Composition A formula showing the smallest whole number ratio of elements in a compound, e.g. Benzene: » Empirical = CH » Molecular = C6H6 Glucose » Empirical = CH2O » Molecular = C6H12O6 Empirical Formula from % Composition 1. Assume 100 g 3. Divide by 4. Convert the mole fewest ratio from step 3 2. Convert to moles number of into small whole moles numbers if necessary 16 1/9/2015 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 = 100 – 21.73 – 13.18 – 13.03 = 52.06 % 1. Assume 100 g 2. Convert to moles Sample Exercise 7.6 3. Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary – NOT REQUIRED HERE 17 1/9/2015 Example Illustrating Step 4 Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is 63.15 %C and 5.30 %H; the rest is oxygen. What are the empirical and molecular formulas? % O = 100 – 63.15 – 5.30 = 31.55 % 1. Assume 100 g 2. Convert to moles 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 1/9/2015 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 = CH2O formula weight (FW) = 30 g/mol molecular = C2H4O2 Molecular weight (MW) = 60 g/mol Note that molecular formula = empirical x 2 glycolaldehyde C2H4O2 = (CH2O) x 2 = C2H4O2 Or more generally – molecular = (empirical) x n 19 1/9/2015 Empirical and Molecular Formulas Compared molecular = (empirical) x n MW = (FW) x n it follows that - and therefore - n = MW FW e.g. glucose – assume we know the MW = 180.0 Empirical = CH2O n= 180.0 30.0 FW = 30.0 =6 So molecular = (CH2O) x 6 = C6H12O6 Molecular Mass and Mass Spectrometry Acetylene C2H2 Benzene C6H6 20 1/9/2015 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 1/9/2015 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 1/9/2015 Experimental Determination of Empirical Formulas: Combustion Analysis CxHyOz + O2(g) x CO2(g) + y/2 H2O(g) mass sample excess Some of the oxygen comes from the sample, some from the excess O2 Calculation Outline: g CO2 mol CO2 mol C gC g H2O mol H2O mol H gH 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 1.000 grams of an organic compound known to contain only C, H and O produces 2.360 g CO2 and 0.640 g H2O. What is the empirical formula of the compound? 23 1/9/2015 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 1/9/2015 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 1/9/2015 Limiting Reagents 2 Br + 1 Ch + 1 Sal 1 sandwich Given: 8 4 3 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 1/9/2015 Sample Exercise 7.9: Identifying the Limiting Reagent in a Reaction Mixture The flame in an acetylene torch reaches temperatures as high as 3500 oC as a result of the combustion of a mixture of acetylene (C2H2) and pure oxygen. If these two gases flow from high-pressure tanks at the rates of 52.0 g C2H2 and 188 g O2 per minute, which reactant is the limiting reagent, or is the mixture stoichiometric? 2 C2H2 + 5 O2 4 CO2 + 2 H2O 188 g 52.0 g Given ratio: Sample Exercise 7.9: Identifying the Limiting Reagent in a Reaction Mixture 2 C2H2 + 5 O2 4 CO2 + 2 H2O 𝑖𝑠 𝑚𝑜𝑙𝑒𝑠 𝑂2 𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻2 > 𝑔𝑖𝑣𝑒𝑛 Given ratio: 5.88 mol O2 = 2.94 2.00 mol C2H2 5 𝑚𝑜𝑙 𝑂2 2 𝑚𝑜𝑙 𝐶2𝐻2 ??? 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 Stoichiometric ratio: 5 mol O2 = 2.5 2 mol C2H2 therefore O2 is in excess and so C2H2 is the LR 27 1/9/2015 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 x 100 Theoretical Yield 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 NH3 (MW = 17.03) is produced by a reaction mixture that initially contains 6.00 kg of H2 (MW = 2.016) and an excess of N2, what is the percent yield of the reaction? Actual Yield N2(g) + 3 H2(g) → 2 NH3(g) x 100 % Yield = Theoretical Yield Excess 6.00 kg Actual yield = (no LR) 18.2 kg Calculate the theoretical yield of NH3 based on H2, and then calculate the % yield - 28 1/9/2015 Sample Exercise 7.10 % Yield = Actual Yield x 100 Theoretical Yield 29