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Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten Chapter 3 Stoichiometry Definition: Mathematical calculations for chemical formulas & equations. Mrs. Deborah Amuso Camden High School Camden, NY 13316 Stoichiometry © 2009, Prentice-Hall, Inc. Chemical Equations Chemical equations are concise representations of chemical reactions. (example) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Stoichiometry © 2009, Prentice-Hall, Inc. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants: Products: appear on the left side of the equation appear on the right side of the equation The states are written in parentheses. Coefficients are inserted to balance the equation. Stoichiometry © 2009, Prentice-Hall, Inc. Subscripts vs Coefficients • Subscripts tell the number of atoms of each element in a molecule. H2O (The 2 means 2 atoms of H) • Coefficients tell the number of molecules. 3 H2O (The 3 means 3 molecules of water) Stoichiometry © 2009, Prentice-Hall, Inc. Counting Atoms Total # atoms = Coefficient x Subscript Practice: 3 H2SO4 6-H 3-S 12-O or 3 – SO4 2 Ca3(PO4)2 6-Ca 4-P 16-O or 4 – PO4 Stoichiometry © 2009, Prentice-Hall, Inc. Balancing Equations When balancing equations use coefficients to get the # of each type of atom on both sides of the equation the same. Consult your teacher for balancing tips. Stoichiometry © 2009, Prentice-Hall, Inc. ReactionTypes 1. 2. 3. 4. 5. Combination Decomposition Combustion Single Replacement Double Replacement Stoichiometry © 2009, Prentice-Hall, Inc. Combination Reactions • In a combination reaction two or more substances react to form one product. Examples: 2 Mg (s) + O2 (g) 2 MgO (s) N2 (g) + 3 H2 (g) 2 NH3 (g) C3H6 (g) + Br2 (l) C3H6Br2 (l) Stoichiometry © 2009, Prentice-Hall, Inc. Decomposition Reactions • In a decomposition reaction one reactant breaks down into two or more substances. Examples: CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s) 2 KCl (s) + O2 (g) 2 NaN3 (s) 2 Na (s) + 3 N2 (g) Stoichiometry © 2009, Prentice-Hall, Inc. Combustion Reactions • In a combustion reaction a carbon compound reacts with oxygen in the air to produce carbon dioxide & water. Examples: • These are generally rapid reactions that produce a flame. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Stoichiometry © 2009, Prentice-Hall, Inc. Single Replacement Reactions In a single replacement reaction, an element & compound react to form a different element & compound. Examples: Zn (s) + CuSO4 (aq) Cu(s) + H2SO4 (aq) 2 HCl (aq) + Mg(s) MgCl2 (aq) + H2(g) Stoichiometry © 2009, Prentice-Hall, Inc. Double Replacement Reactions In a double replacement reaction, two compounds in solution react by exchanging ions to form two different compounds. Examples: ZnCl2 (aq) + 2 AgNO3 (aq) 2AgCl(s) + Zn(NO3)2(aq) Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaNO3 (aq) Stoichiometry © 2009, Prentice-Hall, Inc. Formula Mass (FM) • A formula mass is the sum of the atomic masses for the atoms in a chemical formula. • So, the formula mass of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) = 40.1 + Cl: 2(35.5 amu) = 71.0 111.1 amu • Formula masses are generally reported for ionic compounds. Stoichiometry © 2009, Prentice-Hall, Inc. Molecular Mass (MM) • A molecular mass is the sum of the atomic masses for the atoms in a molecule. • For the molecule ethane, C2H6, the molecular mass would be C: 2(12.0 amu) = 24.0 + H: 6(1.0 amu) = 6.0 30.0 amu • FM & MM are found the same way! Stoichiometry © 2009, Prentice-Hall, Inc. Mass Percent or Percent Composition The mass percent of each element in a compound is found by using this equation: % element = (number of atoms)(atomic mass) (FM of the compound) Or simply: x 100 % = Part/Whole x 100 Stoichiometry © 2009, Prentice-Hall, Inc. Percent Composition So the percentage of carbon in ethane, C2H6, is… (2)(12.0 amu) %C = (30.0 amu) 24.0 amu x 100 = 30.0 amu = 80.0% Stoichiometry © 2009, Prentice-Hall, Inc. Moles & Avogadro’s Number • • • • 1 mole = 6.02 x 1023 The mass of 1 molecule of water is 18.0 amu. The mass of 1 mole of water is 18.0 grams. Stoichiometry Why? 1 gram = 6.02 x 1023 amu. © 2009, Prentice-Hall, Inc. Molar Mass or GFM • Molar mass is the mass of 1 mole of a substance in g/mol. – The molar mass of an element is the mass number that we find on the periodic table. example: Cu = 63.5 O2 = 2(16.0) = 32 – The molar mass of a compound is the same number as its formula mass or molecular mass. Stoichiometry example: C2H6 = 30.0 © 2009, Prentice-Hall, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale. # moles = given mass GFM GFM is the same as Molar Mass (It’s the mass in grams of 1 mole of the substance)Stoichiometry © 2009, Prentice-Hall, Inc. Three Helpful Equations Use these 3 equations to convert between masses, moles, molecules (or units) , & atoms (or ions): #moles = mass/GFM #molecules = (#moles) (6.02 x 1023) (or # units) #atoms = (#molecules) (#atoms/molecule) (or # ions) Stoichiometry © 2009, Prentice-Hall, Inc. Mole Relationships: Check Your Understanding How many atoms of oxygen are in 10.0 grams of CH3COOH? Think: Mass Moles Molecules Atoms #moles = mass/GFM = 10.0g / 60.0 g/mol = 0.167 moles #molecules = (#moles)((6.02 x 1023) = (0.167 moles) (6.02 x 1023) = 1.00 x 1023 molecules #atoms = (#molecules) (#atoms/molecule) = (1.00 x 1023 molecules)(2 oxygen atoms/molecule) = 2.00 x 1023 oxygen atoms Stoichiometry © 2009, Prentice-Hall, Inc. Mole Relationships: Check Your Understanding How many hydroxide ions are in 10.0 grams of Ca(OH)2 Think: Mass Moles Units Ions #moles = mass/GFM = 10.0g / 74.1g/mol = 0.135 moles Ca(OH)2 #units = (#moles) (6.02 x 1023) = (0.135 mol) (6.02 x 1023) = 8.12 x 1022 #ions = (#units) (#ions/unit) = (8.12 x 1022) (2 OH-/unit) = 1.62 x 1023 OH- ions Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition of each element in the formula. Steps: 1) Assume percent of each element = its mass 2) Divide mass of each element by that element’s atomic mass to get # moles 3) Calculate the mole ratio by dividing by the smallest # of moles 4) These are the #’s in the empirical formula Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. C: 61.31 g / 12.00 g/mol = 5.105 mol C H: 5.14 g / 1.01 g/mol = 5.09 mol H N: 10.21 g / 14.01 g/mol = 0.7288 mol N O: 23.33 g / 16.00 g/mol = 1.456 mol O Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: 5.105 mol 0.7288 mol = 7.005 7 H: 5.09 mol 0.7288 mol = 6.984 7 N: 0.7288 mol 0.7288 mol = 1.000 O: 1.458 mol 0.7288 mol = 2.001 2 Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Find the empirical formula of a compound whose mass percentage is 50.0% sulfur and 50.0% oxygen. Step 1 S: 50.0 g / 32.1 g/mol O: 50.0 g / 16.0 g/mol = 1.56 mol S = 3.13 mol O Step 2 Step 3 S: 1.56 / 1.56 = 1 O: 3.13 / 1.56 = 2 Empirical Formula is SO2 Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Molecular Formulas From Empirical Formulas & Molar Mass What is the MF if the EF is C4H5N2O and the MM is 194 g/mol? Step 1: Determine molar mass of empirical formula C4H5N2O = 97 g/mol Step 2: Divide: Molar Mass given/Molar Mass of empirical formula 194 / 97 = 2 Step 3: Multiply: Answer from Step 2 * Empirical Formula (2)(C4H5N2O) = C8H10N4O2 Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Molecular Formulas From Empirical Formulas & Molar Mass What is the MF if the EF is CH2 and the molecular mass is 84.16 amu? Step 1: Determine molecular mass of empirical formula CH2 = 14.03 amu Step 2: Divide: Molecular mass given/molecular mass of emp formula 84.16 amu /14.03 amu = 6 Step 3: Multiply: Answer from Step 2 * Empirical Formula (6)(CH2) = C6H12 Stoichiometry © 2009, Prentice-Hall, Inc. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. – C is determined from the mass of CO2 produced. – H is determined from the mass of H2O produced. – O is determined by difference after the C and H have been determined. – Then you can determine the Empirical Formula of Stoichiometry the compound © 2009, Prentice-Hall, Inc. Combustion Analysis When 0.255 g of caproic acid is burned, you get 0.512 g CO2 and 0.209 g H2O. What is the emp formula of the acid? #mole CO2 = mass/GFM = 0.512g / 44.0g/mol = 0.0116 mol 0.0116 mol CO2 * (1 mole C/1 mol CO2) = 0.0116 mol C 0.0116 mol C * (12.0 g/mol) = 0.140 g C #mole H2O = mass/GFM = 0.209g / 18.0g/mol = 0.0116 mol 0.0116 mol H2O * (2 mole H/1 mol H2O)= 0.0232 mol H 0.0232 mol H * (1.01 g/mol) = 0.0235 g H Mass O = Mass caproic acid – (mass C + mass H) Mass O = 0.225 g – (0.140 g + 0.0235 g) = .0615 g Stoichiometry © 2009, Prentice-Hall, Inc. 2nd Part of Problem For carbon: 0.140 g C / 12.0 g/mol = 0.0117 mol For hydrogen: 0.0235 g H / 1.01 g/mol = 0.0233 mol For oxygen: 0.0615 g O / 16.0 g/mol = 0.00384 mol For carbon: 0.0117 / 0.00384 = 3.05 For hydrogen: 0.0233 / 0.00384 = 6.07 For oxygen: 0 0.00384 / 0.00384 = 1 Empirical Formula = C3H6O Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations The coefficients in the balanced equation represent the # of molecules or # of moles of reactants and products. The coefficients tell the ‘recipe’ for the rxn. Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams. Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculation Steps 1) Complete & balance equation using coefficients. 2) Identify Given & Unknown. 3) Convert Given mass into moles: #molesgiven = massgiven/GFMgiven 4) Use mole ratio to find moles of Unknown. #molesgiven = #molesunk Coeffgiven Coeffunk 5) Convert moles of ‘unknown’ to answer. massunknown = (#molesunk) (GFMunk) Stoichiometry © 2009, Prentice-Hall, Inc. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) How many grams of water are produced when 32.0 grams of methane completely react? 1) Given = 32.0 g CH4 Unk = ? g H2O 2) Find # moles CH4 #moles CH4 = mass/GFM = 32.0 g / 16.0 g/mol = 2.00 moles CH4 3) Find # moles H2O 2.00 moles CH4 1 mole = ? moles H2O 2 mole Ans. = 4.00 moles H2O 4) Find mass of H2O mass H2O = (#moles)(GFM) = (4.00 mol)(18.0 g/mol) = Stoichiometry Ans. = 72.0 g H2O © 2009, Prentice-Hall, Inc. 2 H2 (g) + O2 (g) 2 H2O (g) How many grams of water are produced when 6.0 grams of oxygen gas completely react? 1) Given = 6.0 g O2 Unk = ? g H2O 2) Find # moles O2 #moles O2 = mass/GFM = 6.0 g / 32.0 g/mol = 0.19 moles O2 3) Find # moles H2O 0.19 moles O2 1 mole = ? moles H2O 2 mole Ans. = 0.38 moles H2O 4) Find mass of H2O mass H2O = (#moles)(GFM) = (0.38 mol)(18.0 g/mol) = Stoichiometry Ans. = 6.8 g H2O © 2009, Prentice-Hall, Inc. Limiting Reactants: How Many Cookies Can I Make? • You can make cookies until you run out of one of the ingredients. • Once you run out of sugar, you can’t make any more cookies. • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make. Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants • The limiting reactant (or limiting reagent) is the reactant you’ll run out of first • What is the LR in this reaction? Stoichiometry H2 © 2009, Prentice-Hall, Inc. How to Determine Which Reactant is the Limiting Reactant Step 1: Convert the masses given to moles. Step 2: Use these equations: # moles reactant A / coefficient A = x # moles reactant B / coefficent B = y Step 3: The smaller # (x or y) identifies the Limiting Reagent. Step 4: Use the # of moles of the Limiting Reagent to solve the rest of the problem. Stoichiometry © 2009, Prentice-Hall, Inc. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Given 50.0 g of CH4 and 60.0 g of O2 Which is the Limiting Reagent? Step 1: #moles CH4 = mass/GFM = 50.0 g/16.0 g/mol = 3.13 moles #moles O2 = mass/GFM = 60.0 g/32.0 g/mol = 1.88 moles Step 2: Use these equations: # moles reactant CH4 / coefficient CH4= 3.13 / 1 = 3.12 # moles reactant O2 / coefficent O2 = 1.88 / 2 = 0.94 Step 3: The smaller # (x or y) identifies the Limiting Reagent 0.94 is smaller, so O2 is the L.R. Stoichiometry © 2009, Prentice-Hall, Inc. 2 CHCl3 (g) + 2 Cl2 (g) 2 CClO4 (g) + 2 HCl (g) Given 15.9 g of CHCl3 and 12.6 g of Cl2 Which is the Limiting Reagent? Step 1: #moles CHCl3 = mass/GFM = 15.9 g/119.5 g/mol = 0.133 mol #moles Cl2 = mass/GFM = 12.6 g/71.0 g/mol = 0.177 mol Step 2: Use these equations: # mol reactant CHCl3 / coefficient CHCl3= 0.133/ 2 = 0.0665 # mol reactant Cl2 / coefficent Cl2 = 0.177/ 2 = 0.0885 Step 3: The smaller # (x or y) identifies the Limiting Reagent 0.0665 is smaller, so CHCl3 is the L.R. Stoichiometry © 2009, Prentice-Hall, Inc. Theoretical Yield • The theoretical yield is the maximum amount of product that can be made. – In other words it’s the amount of product possible as calculated through the stoichiometry problem. • This is different from the actual yield, which is the amount one actually produces and measures. Stoichiometry © 2009, Prentice-Hall, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Percent Yield = Theoretical Yield x 100 Stoichiometry © 2009, Prentice-Hall, Inc. Percent Yield Problem CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Previously, you determined 72.0 g of water should be produced from 32.0 g of methane. Assume only 68.0 g of water are produced when this experiment is actually conducted. What is the percent yield? % yield = [actual yield/theoretical yield] x 100 % yield = [68.0g / 72.0 g] x 100 % yield = 94.4% Stoichiometry © 2009, Prentice-Hall, Inc.