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Chapter 3 Chemical Equations Stoichiometry: Calculations with Chemical Formulas and Equations Reactants Products C(s) + O2(g) CO2(g) Combination Reaction (Combustion) The rusting of iron Fe + O2 Ways of presenting molecular composition Fe2O3 •The mass of each element Need to balance 4 Fe + 3O2 •A molecular formula 2 Fe2O3 •Percent composition The first thing you always do is examine to see whether the equation is balanced. Stoichiometry Stoichiometry Quantitative relationship between amounts of reactants and products. aA + bB cC + dD Answers questions: •How much product is formed from some given amount of reactant? (Theoretical Yield) •How much of one reactant is required to bring a reaction to completion given some amount of another reactant? (Limiting Reagent) The Concept of the mole: We talk here about the chemist’s way for counting the amount of “stuff” as compared to some standard. One mole of “stuff” contains the same number of “particles” as there are atoms in one mole of carbon-12 # moles of an element = mass of element/ molar mass You have 53.0 grams of Na metal. How many moles will you have? 53.0 g/(22.99 g/mol) = 2.31 mol MOLE: Important Concepts # moles of an element = mass of element/ molar mass 1 mol contains Avogadro’s Number •One mole of carbon-12 will weigh 12.0000 grams and contain 6.02*1023 atoms (Avogadro’s number) of carbon-12. •One mole of oxygen-16 will weigh 16.0000 grams and contain 6.02*1023 atoms (Avogadro’s number) of oxygen-16. •One mole of nitrogen-14 will weigh 14.0000 grams and contain 6.02*1023 atoms (Avogadro’s number) of nitrogen-14. # moles of an element = mass of element/ molar mass How many grams of Fe are found in 0.25 mole? (0.25 mol) X (55.87 g/mol) = 14 g Problem How many atoms of Hg (mercury) will be contained in 15 mL (cm3) of Hg? The density of Hg is 13.534 g/mL. Molecules, Compounds and the Mole Avogadro’s number: (N) 6.02 x 1023 “things” per mol of “things” # Things = # mol x N Stoichiometry: concerned with mass/mass relationships in chemical reactions 2 A + B C #mol = mass/Molar mass (Molar mass (Molecular Weight )) = Sum Individual atomic masses Stoichiometry: concerned with mass/mass relationships in chemical reactions Mass B Mass A 2 A + Mass C B C MM C MM A MM B Moles A Moles B REAL! In theory 2 Molecules + 1 Molecule 1 Molecule From any balanced equation: If you know masses and molar masses you can calculate any missing quantity. # mol = mass / molar mass mass = # mol x molar mass Moles C Structural Formula Percent Composition H2 C H 2C EXAMPLE: What is the Percent composition of the elements in the sex attractant of the Asian Civet cat? CH2 CH H 3C C8H14O3 CH O CO2H C H2 Molar mass = 8 C *12.01 g/mol C + 14 H *1.008 g/mol H + 3 O *15.994 g/mol C = 158.174 g/mol C8H14O3 What percent of this mass is due to oxygen? Empirical Formulas %C = mass C/Molar mass (*100) = (96.08 g/mol) / (158.2 g/mol) * 100 = 60.73% The simplest formula of a compound in small whole numbers. %H = mass of H/Molar mass(*100) = (14.11 g/mol) / (158.2 g/mol) * 100 = 8.92% %O = mass of O/ Molar Mass (*100) = (47.98 g/mol) / (158.2 g/mol) * 100 = 30.33% You have a sample of the venom of the fire ant. You send it for analysis and the company sends back a printout telling you that your sample contains: Method: 1. ASSUME 100 grams of sample Thus your sample contains 80.4 grams C, 13.8 grams H and 5.8 grams N 2. What is mole ratio of the constituents? # mol = mass/molar mass of atoms 80.4 %C, 13.8 %H, and 5.8 %N #mol C = 80.4 g / (12.01 g/mol) = 6.69 mol #mol H = 13.8 g / (1.001 g/mol) = 13.8 mol #mol N = 5.8 g / (14.00 g/mol) = .414 mol C6.69H13.8N.414 3. Find a common denominator (divide by smallest) H2 C H 2C 6.69 mol /.414 mol = 16.17 13.8 mol /.414 mol = 33.3 .414 mol /.414mol = 1.0 CH2 CH H 3C C H2 C H2 C16H33N H2 C H2 C H2 C H2 C CH N H C H2 C H2 CH3 C H2 Determining an Empirical Formula C,H,O Burn in O2 Grams All the carbon in carbon dioxide comes from here CO2 H2O Grams Grams Moles Moles #moles H2O = 0.7425g/(2.011g/mol + 15.994g/mol) = 0.04024 mol But, 1 mole of H2O has 2 moles of H Grams 0.523 Moles = CO2 H2O Grams Grams 1.612 0.03664 0.7425 Moles(C) = 0.03664 Moles(H) = Burn in O2 0.04024 0.08248 CO2 H2O Grams Grams All the hydrogen in water comes from here Moles #moles CO2 = 1.612g/(12.011g/mol + 31.988g/mol) = 0.03664 mol C,H Grams 0.523 Burn in O2 Moles = Moles(C) = What is the formula? Burn in O2 C,H,O Grams Exercise 0.523 g of CxHy was burned to give 1.612 g of carbon dioxide and 0.7425 g of water. A separate mass spectrum showed the molar mass = 114 g/mole. C,H Determining an Empirical Formula CO2 H2O Grams Grams 1.612 0.03664 0.7425 0.03664 Find a common denominator (divide by smallest) .08248 mol /.03664 mol = 2.251 .03664 mol /.03664 mol = 1.000 Another example C4H9 Empirical Formula Molar Mass = (4 x 12g/mol) + (9 x 1g/mol) = 57g/mol Analysis gives the following data: %C = 74.0 %H = 8.64 %N = 17.3 What is the empirical formula? Given: Molar Mass = 114g/mol Thus, Molecular Formula is Tobacco contains a number of alkaloids . . the most important is nicotine C8H18 If you know the molar mass = 162, what is the molecular formula? Using formulas(Balancing Reactions) CH4 + O2 CO2 + H2O Methane + oxygen burn to give carbon dioxide and water We have to conserve atoms: Have to have same number of each type of atom on both sides of arrow CH4 + O2 C =1 H=4 CO2 + 2H2O C=1 O=2 We need 2 more O’s in reactant H=4 O=4 Put a 2 as coefficient to oxygen. CH4 + O2 CO2 + H2O C =1 H=4 C=1 H=2 O=3 O=2 We need 2 more H’s in product CH4 +2 O 2 C =1 H=4 Put a 2 as coefficient to water. CO2 + 2H2O C=1 O=4 H=4 O=4 We have a balanced equation. It says “ 1 mole of methane + 2 moles of oxygen will produce 1 mole of carbon dioxide + 2 moles of water.” 24 grams CH4 +2 O 2 CO2 + 2H2O If we have 24 grams of methane (and plenty of air!) how much water will be formed? (i.e. methane is the Limiting Reagent) CH4 +2 1 mole of methane ? grams CO2 O2 + 2H2O 2 moles of water gives Molar mass of methane = 16.0 g/mol Molar mass of water=18 g/mol Moles Methane = 24 g /16 g/mol = 1.5 mol The process: We need to work in moles. How many moles of methane do we REALLY have? Thus, REALLY will have: 1.5 mol CH4 * (2 mol H2O mol CH4) = 3 moles of H2O 3 mol H2O *18 g/mol = 54 grams H 2O Combustion of Carbon Compounds A sugar: Glucose: C6H12O6 Our metabolism is a combustion process: Carbon compounds in . . . carbon dioxide and water out! A fat: Glyceryl stearate: C37H110O6 Lets compare fat with sugar. C6H12O6 +O2 CO2 + Balance: C6H12O6 + 6O2 6 CO2 + 6H2O H2O Molar Mass =72.07g C +12.096g H + 95.96g O =180.13 g/mol Thus, 1.00 g = 1.00 g / (180.13 g/mol) = 5.53 x 10-3 mol Lets examine the amount of oxygen needed to “burn” 1.00 gram of each. We’ll also see how much water and carbon dioxide we produce from each. C6H12O6 + 6O2 6 CO2 +6H2O 5.53 x 10 -3 mol C6H12O6 We would expect to use 5.53 x 10-3 mol C6H12O6 * (6 mol of O2 / mol C6H12O6) = 3.33 x 10 -2 mol O2 This is equal to 3.33 x 10 -2 mol O2 * 31.98 g/mol = 1.07 grams O2 C37H110O6 + O2 CO2 + H2O Balance: C37H110O6 + 123/2 O2 37 CO2 +55 H2O Molar Mass =444.4g C +110.9g H + 95.96g O =651.3 g/mol Thus, 1.00 g = 1.00 g / (651.3 g/mol) = 1.54 x 10-3 mol Limiting Reagent Problems: AgNO3 + NaCl -----> AgCl +NaNO 3 How much silver nitrate? How much sodium chloride? How much silver chloride will be formed? Stoichiometry Involves • Balancing the equations – This gives theory mole ratios • Assigning molar masses • Calculating “real” moles – This gives real mole ratios • Converting moles to mass C37 H110O6 +(123/2)O2 37 CO2 +55 H2O 1.54 x 10 -3 mol C37H110O6 We would expect to use 1.54 x 10-3 mol C37H110O6 * (123 mol of O 2 / 2 mol C6H12O6) = 9.47 x 10 -2 mol O2 This is equal to 9.47 x 10 -2 mol O2 * 31.98 g/mol = 3.03 grams O2 Determining Percent Yield % Yield = Actual/Theory