Chapter 8 Quantities In Reactions

Homework Assigned Problems (odd numbers only) “Problems” 17 to 73 “Cumulative Problems” “Highlight Problems” 97-99

Calculations Using Balanced Equations: Stoichiometry Stoichiometry is the study of the quantitative relationships among reactants and products in a chemical reaction These chemical calculations can be used to determine the amount of one reactant needed to completely react with another Or, to determine the amount of reactant needed to produce a desired amount of product To calculate chemical quantities in reactions involves knowing how to interpret a balanced chemical equation

Calculations Using Balanced Equations: Law of Conservation of Mass As a chemical reaction proceeds: –Reactants are consumed and new materials with new chemical properties are produced –Bonds are broken, formed, or atoms are rearranged which produces new substances –No material is lost or gained as original substances change to new substances

Calculations Using Balanced Equations: Law of Conservation of Mass Law of Conservation of Mass –Quantity of matter does not change during a chemical reaction –The sum of the masses of products is equal to the sum of masses of reactants –Atoms are neither created nor destroyed in chemical reactions –Only a balanced equation obeys this law

Mole Relationships in Chemical Equations: Conservation of Mass A balanced equation has the same number of atoms on each side of the arrow 2 2

Information Available from a Balanced Chemical Equation “1 mole of methane gas reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide and 2 moles of water vapor.” Multiplying each of the molar masses by the coefficient will give the total mass of reactants and products

Making Molecules: Mole-Mole Conversions A balanced chemical equations tell us: –The formulas and symbols of the reactants and products –The physical state of each substance –If special conditions such as heat are required –The number of molecules, formula units, or atoms of each type of molecule involved in the reaction Number can be in terms of single atoms, or moles of atoms –The relative number of moles of each reactant and product

Making Molecules: Mole-Mole Conversions In a balanced equation, conversion from moles of one substance to another will be determined by the values of the coefficients Balancing an equation will generate the coefficients that equal the number of moles of each reactant and product To determine how many moles of methanol would be produced if moles of hydrogen gas is consumed: Requires a mole to mole relationship between methanol and hydrogen gas 2

Making Molecules: Mole-Mole Factors Converting the given mole amount of hydrogen gas enables you to find the number of moles methanol produced Coefficients of the balanced equation can be used to make mole to mole relationships between the different reactants and products 2 Given: mol H 2 Find: mol CH 3 OH 2 mol H 2 = 1 mol CH 3 OH

Making Molecules: Mole-Mole Factors The unit path begins with moles of H 2 and ends with moles of CH 3 OH From any mole-mole relationship, two mole-mole conversion factors can be made: 2 mol H 2 mol CH 3 OH Mole-mole factor 2 mol H 2 = 1 mol CH 3 OH

Making Molecules: Mole-Mole Factors Write all of the possible mole-mole factors for the following chemical equation mol H 2 = 2 mol H 2 O 1 mol O 2 = 2 mol H 2 O 2 mol H 2 = 1 mol O 2

Using Mole-Mole Factors in Calculations: Calculating Moles of a Product Calculations based on balanced equations require the use of mole to mole (conversion) factors –Equation must be balanced –Identify the known (given) and needed (find) substances –Make the conversion factor based on:

Using Mole-Mole Factors in Calculations: Calculating Moles of a Product Using mole-mole factors from a BALANCED chemical equation –You can convert moles of one compound to moles of another compound using the correct mole-mole factor moles A moles B grams A grams B MM of AMM of B moles B moles A Stoichiometry Mole-mole factor

Using Mole-Mole Factors in Calculations: Calculating Moles of a Product Calculate the moles of CO 2 formed when 4.30 moles of C 3 H 8 reacts with (the required) 21.5 moles of O 2 Balance the equation Plan to convert the given amount of moles to the needed amount of moles Use coefficients to state the relationships and mole-mole factors Set up the problem using the mole-mole factor and canceling units 5 3 4

Using Mole-Mole Factors in Calculations: Calculating Moles of a Product Given: Find: 1 mol C 3 H 8 = 3 mol CO 2 Mole-mole factor Mole-mole factor Mole-mole relation MM of A MM of B

Using Mole-Mole Factors in Calculations: Calculating Moles of a Product Set up the problem using the mole-mole factor that cancels given moles and provides needed moles Mole-mole factor

Making Molecules: Mass-to-Mass Conversions From the balanced equation –It is also possible to start with a known mass of one substance –Then convert to moles of another substance –Start with a given amount (of grams) of a substance –Then use mole-mole factor to find the sought-after mass of another substance

Making Molecules: Mass-to-Mass Conversions The most common type of stoichiometric calculation is the mass-to-mass calculation In this type of problem: The mass of one substance involved in a chemical reaction is given Find the mass of another substance involved in the reaction If a chemist only has so many grams of a certain chemical –Can calculate how many grams of another substance can be produced –Can calculate how many grams of another reactant are required to react with it

Mass-to-Mass Conversions To convert the grams of one substance to grams of another substance: Find the mole-mole factor using the coefficients in the balanced equation –You can only relate (moles-moles) of two compounds, not grams-to-grams –Ratios ONLY apply to moles, NOT grams –Must convert grams to moles, then use mole-mole factor MM of A MM of B moles A moles B Mole-mole factor

Mass-to-Mass Conversions Mass-to-mass conversions begin with a given mass of substance A By use of the balanced equation, find the mass of another (substance B) 1.Convert grams of A to moles of A 2.Convert the moles of A to moles of B by use of mole-mole ratio 3.Convert moles of B to mass of B moles A moles B grams A grams B MM of AMM of B moles B moles A Stoichiometry

Mass-to-Mass Conversions (Mass of Product from Mass of Reactant) Calculate the mass of carbon dioxide produced when 96.1 g of propane react with sufficient oxygen. Balance the equation Plan to convert the given mass to given moles Convert the given moles to sought-after moles by the use of mole- mole factor Convert the needed moles to needed mass moles C 3 H 8 moles CO 2 grams C 3 H 8 grams CO 2 MM of C 3 H 8 MM of CO 2 moles CO 2 moles C 3 H 8 Stoichiometry Mole-mole factor

Mass-to-mass Conversions: Example 1 Write the equalities 1 mol C 3 H 8 = g C 3 H 8 1 mol CO 2 = g CO 2 1 mol C 3 H 8 = 3 mol CO 2 to create mole-mole factor Find: g of CO 2 Given: 96.1 g C 3 H 8 moles C 3 H 8 moles CO 2 grams C 3 H 8 grams CO 2 MM of AMM of B moles B moles A Stoichiometry

Mass-to-Mass Conversions: Example X g CO 2

Mass-to-Mass Conversions: Example 2 What mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the following reaction: 2

Mass-to-Mass Conversions: Example 2 2

Mass Calculations Example 2

Limiting Reactant, Theoretical Yield, and Percent Yield The chemical reactants are usually not present in the exact mole-mole ratios as stated in the balanced chemical equation Often, one of the reactants is purposely added in an excess amount Reasons include: Increase the rate of reaction To ensure that one reactant is completely used up (reacted) Reactants are not completely converted to products as stated on paper (theory)

Limiting Reactant, Theoretical Yield, and Percent Yield Chemical reactions with two or more reactants will continue until one of the reactants is used up (consumed) If one of the reactants is used up, the reaction will stop because there is not enough of the other reactant to react with it The reactant used up is called the limiting reactant (reagent) This reactant limits the amount of product that can be made

Limiting Reactant When you make peanut butter sandwiches Required: 2 slices of bread and 1 tbsp. peanut butter per sandwich The reaction of nitrogen gas and hydrogen gas forming ammonia gas Required: 1 molec. N 2 gas and 3 molec. H 2 gas 2 slices of br. +1 tbsp p.b. = 1 sndw. N 2 (g) + 3H 2 (g) = 2 NH 3 (g) N2N2 NH 3

Limiting Reactant To determine the limiting reactant between two reactants: 1.Balance the equation 2.State the mole-mole relationships to make conversion factors 3.Convert the initial masses (reactants) to moles of each reactant 4.Calculate how many moles of product can be produced by each reactant

Limiting Reactant 5.Convert the moles of product to number of grams of product that each of the reactants would produce 6.Compare the numbers: The reactant producing the least amount of product (grams) is the limiting reactant

Limiting Reactant Problem Lithium nitride, an ionic compound containing Li + and N 3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium metal. 2 6 X

Limiting Reactant Problem 6 2 Find: g of Li 3 N Given: 56.0 g Li Given: 56.0 g N 2 moles Li moles Li 3 N grams Li grams Li 3 N MM of LiMM of Li 3 N moles Li 3 N moles Li Stoichiometry 1 mol Li = 6.941g Li 1 mol N 2 = g N 2 1 mol Li 3 N= g Li 3 N Equalities and Conversion Factors- Solution Map: MM of N 2 moles N 2 grams N 2 6 mol Li = 2 mol Li 3 N 1 mol N 2 = 2 mol Li 3 N

Limiting Reactant Problem g Li 3 N g Li 3 N Limiting reactant 8.07 mol Li 2.00 mol N 2

Limiting Reactant Problem Lithium is the limiting reactant. We calculated the number of grams of lithium nitride which is formed in the reaction based on the limiting reactant This is the calculated amount of lithium nitride formed if the reaction proceeds completely as described by its balanced chemical equation 93.7 g Li 3 N Theoretical yield

Percent Yield The calculated amount of product that should be obtained is called the theoretical yield Assumes all reactants are converted to product based on the mole-mole ratios of reactant to product Rarely do you get the maximum amount of product –Side reactions –Loss during transfer –Accidental spills

Percent Yield Theoretical Yield –The calculated amount of product Actual Yield –The actual amount of product –Something less than the theoretical Percent Yield –The fraction of the theoretical yield actually obtained is expressed as a percent

Percent Yield Example Suppose, in the previous limiting reactant problem, you actually produced 90.8 g of Li 3 N. What is the percent yield of this reaction? 96.9 % yield

Enthalpy Chemical reactions are associated with an absorption or evolution of heat –A change in energy occurs as bonds are broken (reactants) and new ones form (products) –Nearly all chemical reactions absorb or produce heat –Measured by the heat of reaction or enthalpy Enthalpy change is the amount of heat produced or consumed in a process (∆H )

Sign of ∆H rxn Endothermic reactions absorb heat as they occur –If (∆H ) is positive, then heat is added to the reaction –If heat supply is removed, the reaction stops

Sign of ∆H rxn Exothermic reactions produce heat as they occur –If (∆H ) is negative, then heat is evolved by the reaction

Sign of ∆H rxn Enthalpy of Reaction Photosynthesis reaction –Carbon dioxide reacts with water to produce glucose and oxygen Cell metabolism –Glucose reacts with oxygen to produce carbon dioxide and water H = kJ ∆H = kJ H = kJ ∆H = kJ

Stoichiometry of ∆H rxn The coefficients in a given chemical reaction represent the number of moles of reactants and products that produce the given heat of reaction (enthalpy change) The combustion of methane gas: This information gives a quantitative relationship between the heat evolved per mole of methane and oxygen gas H rxn = -890 kJ ∆H rxn = -890 kJ 1 mol CH 4 = -890 kJ 2 mol O 2 = -890 kJ

Stoichiometry of ∆H rxn If the combustion of 1 mol of CH 4 with 2 mol O 2 releases 890 kJ of heat, the combustion of 2 mol of CH 4 with 4 mol of O 2 produces twice as much heat The equivalence statements can be used to make conversion factors between the amounts of reactants or products and the amount of heat absorbed or emitted in a given reaction Calculate the amount of heat emitted when a certain amount in grams undergoes combustion H rxn = kJ ∆H rxn = kJ

H Stoichiometry Involving ∆H Calculate the heat associated with the complete combustion of 4.50 g of methane gas H rxn = -890 kJ ∆H rxn = -890 kJ Given: 4.50 g CH 4 Find: kJ Conversion Factors: 1 mol CH 4 = -890 kJ 1 mol CH 4 = g Solution Map: g CH 4 mol CH 4 kJ CH 4 Solution: × 10 2 kJ

H Stoichiometry Involving ∆H The combustion of sulfur dioxide It reacts with oxygen to produce sulfur trioxide Calculate the heat produced when 75.2 g of sulfur trioxide is produced H rxn = kJ ∆H rxn = kJ Given 75.2 g SO 3 Heat in kJ produced Heat in kJ produced when SO 3 is formed Find:

H Stoichiometry Involving ∆H Solution Map: Relation between g of SO 3 and heat released Grams Grams of SO 3 of SO 3 Moles kj Write the necessary conversion factors: Set up the problem: 46.5 kJ Heat Heat of rxn Molarmass

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