Presentation on theme: "Chapter 11 – Stoichiometry"— Presentation transcript:
1 Chapter 11 – Stoichiometry 11.1 What is Stoichiometry?11.2 Stoichiometric Calculations11.3 Limiting Reactants11.4 Percent Yield
2 Section 11.1 Defining Stoichiometry The amount of each reactant present at the start of a chemical reaction determines how much product can form.Describe the types of relationships indicated by a balanced chemical equation.Explain the relationship between stoichiometry and the principle of the conservation of mass.State the mole ratios from a balanced chemical equation.
3 Section 11.1 Defining Stoichiometry Key ConceptsBalanced chemical equations can be interpreted in terms of moles, mass, and representative particles (atoms, molecules, formula units).The law of conservation of mass applies to all chemical reactions.Mole ratios are derived from the coefficients of a balanced chemical equation. Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction.
4 StoichiometryStudy of quantitative relationships between amounts of reactants used and amounts of products formed by a chemical reactionCombo of 2 Greek words – “to measure the elements”Really a tool (system) for accounting for the conservation of matterWhat goes in must come out
5 Rests on principle of conservation of matter (mass) StoichiometryRests on principle of conservation of matter (mass)2 Al(s) + 3 Br2(l) Al2Br6(s)
6 Stoichiometry - Questions If I have this much reactant, how much product can I form?If I have this much of one reactant, how much of a second reactant do I need so that after the reaction is completed I won’t have any of either reactant left over?
7 Conversion Map for Stoichiometry Calculations Requires balanced equation!MassreactantMassproductStoichiometricfactorMolesreactantMolesproduct
8 Stoichiometry & Balanced Equation 4Fe(s) + 3O2(g) 2Fe2O3(s)Can interpret as4 atoms Fe + 3 molecules O2 2 formula units Fe2O34 mol Fe + 3 mol O2 2 mol Fe2O3No direct information on massesCan use molar mass to get mass info
9 Stoichiometry & Balanced Equation 4Fe(s) + 3O2(g) 2Fe2O3(s)4 mol Fe g Fe/mol Fe = g Fe3 mol O2 g O2/mol O2 = g O22 mol Fe2O3 g Fe2O3/mol Fe2O3 = g Fe2O3Total mass reactants = gTotal mass products = g
10 Stoichiometry & Balanced Equation Example problem 11.1, p. 370C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)Interpret as molecules, moles & mass1 mol C3H8 g C3H8 /mol C3H8 = g C3H85 mol O2 g O2/mol O2 = g O23 mol CO2 g CO2/mol CO2 = g CO24 mol H2O g H2O /mol H2O = g H2OMass reactants = = gMass products = = g1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O
12 Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2) Mole RatioRatio between the numbers of moles of any two substances in a balanced chemical equation2Al(s) + 3Br2(l) 2AlBr3(s)Possible to write 3 unique mole ratios:Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2)Get 3 more ratios from inverses of above for a total of 6 for this reaction
13 3Fe(s) + 4H2O(l) Fe3O4(s) + 4H2(g) Mole RatioIn general, for n species (reactant, product) in chemical equation, have n(n-1) mole ratios3Fe(s) + 4H2O(l) Fe3O4(s) + 4H2(g)4 species, 4 3 = 12 possible mole ratios (6 + inverse of each)Fe:H2O Fe:Fe3O Fe:H2H2O:Fe3O4 H2O:H2 Fe3O4: H2
14 Mole RatioHow many mole ratios can be written for the following reaction?4H2(g) + O2(g) → 2H2O(l)A B C D. 2?
16 Chapter 11 – Stoichiometry 11.1 What is Stoichiometry?11.2 Stoichiometric Calculations11.3 Limiting Reactants11.4 Percent Yield
17 Section 11.2 Stoichiometric Calculations The solution to every stoichiometric problem requires a balanced chemical equation.List the sequence of steps used in solving stoichiometric problems.Solve stoichiometric problems (mole-to-mole, mole-to-mass, mass-to-mass and variations including number of particles) using the factor-label method.
18 Section 11.2 Stoichiometric Calculations Key ConceptsChemists use stoichiometric calculations to predict the amounts of reactants used and products formed in specific reactions.The first step in solving stoichiometric problems is writing the balanced chemical equation.Mole ratios derived from the balanced chemical equation are used in stoichiometric calculations.Stoichiometric problems make use of mole ratios to convert between mass and moles.
19 Stoichiometric Calculations Start with balanced equationDetermine required mole ratiosUse mole-to-mass and mass-to-mole relations for 3 basic conversions:Mole to moleMole to massMass to massOnce have moles, can get # particles
20 Conversion Map for Stoichiometry Calculations Requires balanced equation!MassreactantMassproductStoichiometricfactorMolesreactantMolesproduct
21 nA + mB xC + yD Mass reactant Mass product Moles reactant Moles 41MolesreactantMolesproduct25# part.reactant# part.product36
22 Mole to Mole Conversions Given number of moles of a reactant (product), how many moles of a product (reactant) must be formed?Use initial moles and mole ratiomol known x [mol unknown/mol known] = mol unknownmole ratio from equation
23 Mole to Mole Conversions 2K(s) + 2H2O(l) 2KOH(aq) + H2(g)Mol H2 if start with 4.00x10-2 mol K?4.00x10-2 mol K [1 mol H2/2 mol K]mole ratio from equation= 2.00x10-2 mol H2
24 Mole to Mole Conversions How many moles of CO2 will be produced in following reaction if initial amount of reactants = 0.50 moles?2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)A B C D. 1.0?
26 Mole to Mass Conversions Step 1 – Do mole to mole conversion using mole ratioStep 2 – Do mole to mass conversion using molar mass
27 Conversion Map for Stoichiometry Calculations Requires balanced equation!MassreactantMassproductStoichiometricfactorMolesreactantMolesproduct
28 Mole to Mass Conversions 2Na(s) + Cl2(g) 2NaCl(s)Mass NaCl produced from 1.25 mol Cl2?1.25 mol Cl2 [2 mol NaCl/1 mol Cl2]mole ratio from equation= 2.50 mol NaCl2.50 mol NaCl [58.44 g NaCl/mol NaCl]molar mass of NaCl= 146 g NaCl
30 Mass to Mass Conversions Step 1 – Do mass to mole conversion using molar massStep 2 – Do mole to mole conversion using mole ratioStep 3 – Do mole to mass conversion using molar mass
31 Conversion Map for Stoichiometry Calculations Requires balanced equation!MassreactantMassproductStoichiometricfactorMolesreactantMolesproduct
32 Mass to Mass Conversions NH4NO3(s) N2O(g) + 2H2O(g)Mass of H2O from 25.0 g NH4NO3(s)?AN= NH4NO3(s)25.0 g AN [1 mol AN/80.04 g AN]1/molar mass of AN= mol AN0.312 mol AN [2 mol H2O/1 mol AN]mole ratio from equation= mol H2O
33 Mass to Mass Conversions NH4NO3(s) N2O(g) + 2H2O(g)Mass of H2O from 25.0 g NH4NO3(s)?0.624 mol H2O [18.02 g H2O/mol H2O]molar mass of H2O= 11.2 g H2O
35 Chapter 11 – Stoichiometry 11.1 What is Stoichiometry?11.2 Stoichiometric Calculations11.3 Limiting Reactants11.4 Percent Yield
36 Section 11.3 Limiting Reactants A chemical reaction stops when one of the reactants is used up.Identify the limiting reactant in a chemical equation.Identify the excess reactant, and calculate the amount remaining after the reaction is complete.Calculate the mass of a product when the amounts of more than one reactant are given.
37 Section 11.3 Limiting Reactants Key ConceptsThe limiting reactant is the reactant that is completely consumed during a chemical reaction. Reactants that remain after the reaction stops are called excess reactants.To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation.Stoichiometric calculations must be based on the limiting reactant.
38 Limiting ReactantsTypical case: reactants not present in exact stoichiometric ratiosMole ratio given by coefficients of reactants in balanced chemical equationN2 + 3H2 2NH3Stoichiometric ratio of reactants for ammonia synthesis is 1 N2: 3 H2If ratio 1:3 then one reactant is limiting
39 Limiting and Excess Reactants Ammonia Synthesis N2(g) + 3H2(g) 2NH3(g)Case of 1:1 N2:H2 mole ratioBefore ReactionAfter Reaction3N23H22NH32N2N2 is excess reactantH2 is limiting reactant
40 Limiting and Excess Reactants Limiting reactantLimits the extent of reactionDetermines the amount of productExcess reactantAlways have “left over”Often a reactant is deliberately used in excess to:Drive reaction to completion (equilibrium)Speed up reaction (kinetics)
41 Limiting and Excess Reactants Limiting and excess air (oxygen) in combustion reactionAir limiting – soot (carbon)Air excess – CO2 & H2O
42 Limiting Reactant - Identifying Step 1 – If masses given, convert to molesStep 2 – Compare actual mole ratio of reactants to mole ratio determined from balanced equationRatios same - no limiting reactantRatios different - compare to identify limiting reactant
43 Limiting Reactant S8(l) + 4Cl2(g) 4S2Cl2(l) 200.0 g sulfur reacts with g Cl2Part a: g of product produced?Step 1 – Convert masses to moles100.0 g Cl2 [1 mol Cl2 /70.91 g Cl2]1/molar mass of S8= mol Cl21/molar mass of Cl2200.0 g S8 [1 mol S8 /256.5 g S8]= mol S8
44 Limiting Reactant S8(l) + 4Cl2(g) 4S2Cl2(l) 200.0 g sulfur reacts with g Cl2Part a: g product ?Step 2 – Compare actual mole ratio to mole ratio from equationEquation: Cl2:S8 = 4:1Actual: mol Cl2: mol S8= 1.808:1Compare ratios Cl2 is limiting
45 Limiting Reactant S8(l) + 4Cl2(g) 4S2Cl2(l) 200.0 g sulfur reacts with g Cl2Part a: g product ?Mole ratioCl2 limiting reactant, have molCalculate mol product using mole ratio1.410 mol Cl2 [4 mol S2Cl2 /4 mol Cl2]molar mass of S2Cl2= mol S2Cl21.410 mol S2Cl2 [135.0 g S2Cl2/mol S2Cl2]= g S2Cl2
46 Limiting Reactant S8(l) + 4Cl2(g) 4S2Cl2(l) 200.0 g sulfur reacts with g Cl ?Part b: g of sulfur actually reacted?Mole ratioCl2 limiting reactant, have molCalculate mol sulfur using mole ratio1.410 mol Cl2 [1 mol S8 /4 mol Cl2]molar mass of S8= mol S8mol S8 [256.5 g S8 /mol S8]= g S8 reacts g excess
47 Limiting Reactant S8(l) + 4Cl2(g) 4S2Cl2(l) 200.0 g sulfur reacts with g Cl ?Part b: g of sulfur actually reacted?Know that all Cl2(g) reacted (100.0 g)Previously calculated g product S2Cl2(l) producedCons. mass: g = starting mass = final mass = m[S8(l)] + m[4S2Cl2(l)]m[S8(l)]= g -m[4S2Cl2(l)] =109.6 g
48 Limiting ReactantExcess reactant in following reaction if start with 50.0g of each reactant?P4(s) + 5 O2(g) → P4O10(s)A. O B. P4C. None – quantities are stoichiometric?50.0 g P4 [1 mol P4 / g P4]= mol P450.0 g O2 [1 mol O2 /32.00 g O2] = 1.56 mol O2Need 5 mol = 2.02 mol O2 O2 limiting
50 Chapter 11 – Stoichiometry 11.1 What is Stoichiometry?11.2 Stoichiometric Calculations11.3 Limiting Reactants11.4 Percent Yield
51 Section Percent YieldPercent yield is a measure of the efficiency of a chemical reaction.Calculate the theoretical yield of a chemical reaction from data.Explain the difference between a theoretical yield of a chemical reaction and an actual yield.Provide reasons why the theoretical yield and actual yield can be different from each other.Determine the percent yield for a chemical reaction.Explain why the percent yield can exeed 100%
52 Section 11.4 Percent Yield Key Concepts The theoretical yield of a chemical reaction is the maximum amount of product that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation.The actual yield is the amount of product produced. Actual yield must be obtained through experimentation.Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is important in reducing the cost of every product produced through chemical processes.
53 Percent YieldActual (real world) reactions don’t produce amount of product expected from masses of reactant and the balanced chemical equation (stoichiometry)
54 Types of Yields Theoretical Yield Actual Yield Maximum amount of product that can be produced from given amounts of reactantsCalculated using stoichiometryActual YieldAmount of product actually obtained when the reaction is carried out
55 Types of Yields Theoretical Yield and Actual Yield can differ because Reaction stops before limiting reactant all used upProduct is not all collected – sticks to reaction vessel, filter paper, etcReactions other than the intended reaction occur, producing other productsImpurities in reactants
56 Alternate Reaction Paths Combustion of natural gas (methane)CH4(g) + 2O2(g) CO2(g) + 2H2O(g)Know this reaction takes place if have enough O2 – get nice blue flameKnow if limit O2, get yellow flame and soot (carbon) productionSame reactants yield different products
57 Percent YieldPercent yield is a way of expressing the efficiency of a reactionPercent yield = 100 actual yieldtheoretical yieldTheoretical yield computed using same procedures used in previous sections (stoichiometry)
58 % Percent YieldTheoretical yield of chemical reaction = 25 g. % yield if actual yield = 22 g ?A. 88% B. 44% C. 50% D. 97%?% yield = 100 22 g / 25 g = 88%
59 Percent YieldIn commercial production of chemicals, attempt to produce as close to 100% yield as possibleOptimum use of materialsMinimizes or eliminates need to separate product from unused reactants and from unwanted products
60 Percent Yield for Multi-Stage Attempt to produce as close to 100% yield as possibleCritical for multi-stage reaction processes where product of one reaction used as one of the reactants of a subsequent reactionA three-stage reaction with 90% yield from each stage gives only 73% overall yield (0.9 x 0.9 x 0.9 = 0.73)
61 >100% Percent YieldActual Yield: Amount of product actually obtained when from reactionObtained from a laboratory measurement – most common is measuring mass on a balanceAssumption made is that the only “stuff” you’re measuring is the product of interest
62 >100% Percent YieldPossible to get more “stuff” you think is product than predicted if:Haven’t completely separated intended product from other products in reaction or from leftover reactants and/or solvent (insufficient drying)Product reacts with air or water vapor before it is weighed
63 Practice Problems 28 - 30 page 387 Problem 35, page 388