Chapter 3Chapter 3Stoichiometry of Formulas & Equations Stoichiometry of Formulas & Equations

Mole - Mass Relationships in Chemical Systems

Determining the Formula of an Unknown Compound

Writing and Balancing Chemical Equations Calculating the Amounts of Reactant and

Product Calculating Limiting Reagent & Theoretical

Yield Fundamentals of Solution Stoichiometry

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StoichiometryStoichiometry Stoichiometry – The study of the quantitative relationships Stoichiometry – The study of the quantitative relationships

between elements, compounds, chemical formulas and between elements, compounds, chemical formulas and chemical reactionschemical reactions Mass of a substance relative to the chemical entities Mass of a substance relative to the chemical entities

(atoms, ions, molecules, formula units) comprising the (atoms, ions, molecules, formula units) comprising the mass.mass.

Different substances, having the same mass (weight) Different substances, having the same mass (weight) represent unique combinations of elements or groups of represent unique combinations of elements or groups of different elements (compounds).different elements (compounds).

Each element in a compound has a unique atomic mass; Each element in a compound has a unique atomic mass; thus the number of atoms (entities) in substances of equal thus the number of atoms (entities) in substances of equal mass is different.mass is different.

The concept of the “MOLE” was developed to relate the The concept of the “MOLE” was developed to relate the number of entities in a substance to the mass values we number of entities in a substance to the mass values we determine in the laboratory.determine in the laboratory.

From the relationship between the number of atoms and From the relationship between the number of atoms and the mass of a substance we can quantify the relationship the mass of a substance we can quantify the relationship between elements and compounds in chemical reactions.between elements and compounds in chemical reactions.

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Mass vs. AmountMass vs. Amount

The standard unit of mass in the metric system is the The standard unit of mass in the metric system is the gram (or kilogram)gram (or kilogram)

Each of 100 or so different elements has a unique Each of 100 or so different elements has a unique mass (atomic weight (expressed as either atomic mass (atomic weight (expressed as either atomic mass units (amu) or grams) determined by the mass units (amu) or grams) determined by the number of protons and neutrons in the nucleusnumber of protons and neutrons in the nucleus

The same mass (weight) of two different substances The same mass (weight) of two different substances will represent a different number of atoms will represent a different number of atoms

A chemical equation defines the relative number of A chemical equation defines the relative number of molecules of each component involved in the reactionmolecules of each component involved in the reaction

The “Mole” establishes the relationship between the The “Mole” establishes the relationship between the number of atoms of a given element and the mass of number of atoms of a given element and the mass of the substance used in a reactionthe substance used in a reaction

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Mass vs. AmountMass vs. Amount

Amounts in chemistry are expressed by the Amounts in chemistry are expressed by the molemole mole – mole – quantity of substance that contains the quantity of substance that contains the

same number of molecules or formula units as same number of molecules or formula units as exactly 12 g of carbon-12exactly 12 g of carbon-12

Number of atoms in 12 g of carbon-12 is Number of atoms in 12 g of carbon-12 is Avagadro’s number (NAvagadro’s number (NAA) which equals ) which equals 6.022 x 106.022 x 102323

The atomic mass of one atom expressed in atomic The atomic mass of one atom expressed in atomic mass units (amu) is numerically the same as the mass units (amu) is numerically the same as the mass of 1 mole of the element expressed in gramsmass of 1 mole of the element expressed in grams

Molar MassMolar Mass = mass of 1 mole of substance = mass of 1 mole of substance One molecule of Carbon (C) has an atomic mass of One molecule of Carbon (C) has an atomic mass of

12.0107 amu and a molar mass of 12.0107 g/mol12.0107 amu and a molar mass of 12.0107 g/mol 1 mole of Carbon contains 6.022 x 101 mole of Carbon contains 6.022 x 102323 atoms atoms 1 mole Sodium contains 6.022 x 101 mole Sodium contains 6.022 x 102323 atoms atoms

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Molecular & Formula WeightMolecular & Formula Weight

From Chapter 2From Chapter 2 Molecular Mass (also referred to Molecular Molecular Mass (also referred to Molecular

Weight (MW))Weight (MW)) is the sum of the atomic is the sum of the atomic weights of all atoms in a weights of all atoms in a covalentlycovalently bonded bonded molecule – organic compounds, oxides, etc.molecule – organic compounds, oxides, etc.

Formula Mass Formula Mass is sometimes used in a more is sometimes used in a more general sense to include Molecular Mass, general sense to include Molecular Mass, but its formal definition refers to the sum of but its formal definition refers to the sum of the atomic weights of the atoms in the atomic weights of the atoms in ionic ionic bonded bonded compoundscompounds

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Molecular & Formula WeightMolecular & Formula WeightThe computation of Molecular (covalent) orThe computation of Molecular (covalent) orFormula (ionic) molar masses is mathematically the sameFormula (ionic) molar masses is mathematically the same

Ex.Ex.

Molecular Molar Mass of Methane (CHMolecular Molar Mass of Methane (CH44) (covalent bonds)) (covalent bonds)

1 mol CH1 mol CH44 = (1mol C/mol CH= (1mol C/mol CH44 x 12.0107 g/mol C) + x 12.0107 g/mol C) +

(4mol H/mol CH(4mol H/mol CH44 x 1.00794 g/mol H) x 1.00794 g/mol H)

= 16.0425 g/mol CH= 16.0425 g/mol CH44

Formula Molar Mass of Aluminum Phosphate (ionic bonds)Formula Molar Mass of Aluminum Phosphate (ionic bonds)

1 mol AlPO1 mol AlPO44 = (1mol Al/mol AlPO= (1mol Al/mol AlPO44 x 26.981538 g/mol Al) + x 26.981538 g/mol Al) +

(1mol P/mol AlPO (1mol P/mol AlPO44 x 30.973761 g/mol P)x 30.973761 g/mol P)

++

(4mol O/mol AlPO(4mol O/mol AlPO4 4 x 15.9994 g/mol O)x 15.9994 g/mol O)

= 121.953 g/mol AlPO= 121.953 g/mol AlPO44

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The Concept of AmountThe Concept of Amount

Isotopic MassIsotopic Mass Mass of an isotope of an element atomic mass units Mass of an isotope of an element atomic mass units (amu)(amu)

Atomic MassAtomic Mass

(atomic (atomic weight)weight)

Average of the masses of the naturally occurring isotopes of an element

1 amu = 1.66054 x 101 amu = 1.66054 x 10-24-24 g g

O = 15.9994 amu = 15.9994 x 1.66054 O = 15.9994 amu = 15.9994 x 1.66054 x 10x 10-24-24 gg/amu/amu

= 2.65676 x 10= 2.65676 x 10-23-23gg

Molecular MassMolecular Mass

Formula MassFormula MassSum of the atomic masses of the atoms or ions in a molecule or formula unit

Mole Quantity of substance that contains the same number of molecules or formula units as exactly 12 g of carbon-12

Molar Mass Mass of 1 mole of a chemical (gram-molecular weight entity - atom, ion, molecule, formula unit)

Summary of Mass Terminology

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Mole and Formula UnitMole and Formula Unit

The widely applied herbicide The widely applied herbicide atrazineatrazine has the has the following molecular formulafollowing molecular formula

CC88HH1414ClNClN55

1 mol atrazine = 6.022e23 molecules1 mol atrazine = 6.022e23 molecules

1 mol atrazine = 8 mol C atoms1 mol atrazine = 8 mol C atoms

1 mol atrazine = 14 mol H atoms1 mol atrazine = 14 mol H atoms

1 mol atrazine = 1 mol Cl atoms1 mol atrazine = 1 mol Cl atoms

1 mol atrazine = 5 mol N atoms1 mol atrazine = 5 mol N atoms

1 mol atrazine = 8+14+1+5 = 28 mol total atoms1 mol atrazine = 8+14+1+5 = 28 mol total atoms

N

N

N

NN CH(CH3)2

H

CH3CH2

H

Cl

Atrazine

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Quantities in Chemical Quantities in Chemical Reactions: ReviewReactions: Review

Molecular wtMolecular wt..:: sum of atomic weights sum of atomic weights in molecule of substance (units of amu)in molecule of substance (units of amu)

Formula wtFormula wt..:: sum of atomic weights in sum of atomic weights in formula unit of compound (units of formula unit of compound (units of amu)amu)

Mole (mol)Mole (mol):: quantity of substance that quantity of substance that contains equal numbers of molecules or contains equal numbers of molecules or formula units as in the number of atoms formula units as in the number of atoms in 12.00 g of C-12in 12.00 g of C-12

Avagadro’s number (NAvagadro’s number (NAA)):: 6.022e23 6.022e23 Molar MassMolar Mass:: mass of one mole of mass of one mole of

substance (units of g/mol)substance (units of g/mol)

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Mole Relationships: Example CalculationsMole Relationships: Example Calculations

How many How many moleculesmolecules of H of H22O are in 251 kg of O are in 251 kg of water?water?

251 kg x (1000 g/kg)251 kg x (1000 g/kg) = 2.51 x 10= 2.51 x 1055 g H g H22OO

2.51 x 102.51 x 1055 g H g H22O x (1 mol C/18.0153 g)O x (1 mol C/18.0153 g) = 1.39 x 10= 1.39 x 1044 mol mol

HH22OO

1.39 x 101.39 x 1044 mol x 6.022 x 10 mol x 6.022 x 102323/mol/mol = 8.39 x 10= 8.39 x 102727 molecules molecules

How many total How many total atomsatoms are in 251 kg of water? are in 251 kg of water?

8.39 x 108.39 x 102727 molecules x (3 atoms/1 molecule) = molecules x (3 atoms/1 molecule) =

2.52 x 102.52 x 102828 atoms atoms

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Practice ProblemsPractice Problems

What is the molar mass of caffeine, CWhat is the molar mass of caffeine, C88HH1010NN44OO22??

CC == 12.0107 g/mol12.0107 g/mol H = 1.00794 g/mol H = 1.00794 g/mol

NN == 14.0067 g/mol14.0067 g/mol O = 15.9994 g/mol O = 15.9994 g/mol

12.0107g/mol C 12.0107g/mol C x 8 mol C/mol Cx 8 mol C/mol C88HH1010NN44OO22 = 96.0856= 96.0856 g/mol g/mol

C8H10N4O2C8H10N4O2

1.00794g/mol H1.00794g/mol H x 10 mol H/mol Cx 10 mol H/mol C88HH1010NN44OO22 = 10.0794= 10.0794 g/mol g/mol

CC88HH1010NN44OO22

14.0067g/mol N14.0067g/mol N x 4 mol N/mol Cx 4 mol N/mol C88HH1010NN44OO22 = 56.0268= 56.0268 g/mol g/mol

CC88HH1010NN44OO22

15.9994g/mol O15.9994g/mol O x 2 mol O/mol Cx 2 mol O/mol C88HH1010NN44OO22 = 31.9988= 31.9988 g/mol g/mol

CC88HH1010NN44OO22

96.0856 + 10.0794 + 56.0268 + 31.9988 = 96.0856 + 10.0794 + 56.0268 + 31.9988 =

194.1906 g/mol C194.1906 g/mol C88HH1010NN44OO22

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Practice ProblemsPractice ProblemsHow many sulfur atoms are in 25 g of AlHow many sulfur atoms are in 25 g of Al22SS33??

Al = 26.9815 g/molAl = 26.9815 g/mol S = 32.065 g/molS = 32.065 g/mol

AlAl22SS33 = 26.98915g/mol Al x 2mol Al + 32.065g/mol S x 3mol S = 26.98915g/mol Al x 2mol Al + 32.065g/mol S x 3mol S

= 150.158 g/mol Al= 150.158 g/mol Al22SS33

25 g Al25 g Al22SS33 / 150.158 g/mol Al / 150.158 g/mol Al22SS33 = 0.166491 mol Al= 0.166491 mol Al22SS33

0.166491 mol Al0.166491 mol Al22SS33 x 3mol S/1mol Al x 3mol S/1mol Al22SS33 = 0.499474mol S atoms= 0.499474mol S atoms

0.499474 mol S x 6.022x100.499474 mol S x 6.022x102323 S atoms/1mol S atoms = S atoms/1mol S atoms = 3.008e23 atoms S3.008e23 atoms S

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Percent CompositionPercent Composition It is often necessary to determine the mass percentage of a It is often necessary to determine the mass percentage of a

component in a mixture or an element in a compoundcomponent in a mixture or an element in a compound

Example calculationExample calculation: what are the mass percentages of C, : what are the mass percentages of C, H and O in CH and O in C22HH44OO22 (acetic acid)? (acetic acid)?

1 mol acetic acid = 60.052 g1 mol acetic acid = 60.052 g

% C% C = [2 mol C= [2 mol C x (12.0107x (12.0107 g/mol C)]g/mol C)] 60.052 g/mol x 100 = 40.00% 60.052 g/mol x 100 = 40.00%% H% H = [4 mol H= [4 mol H x (1.00794x (1.00794 g/mol C)]g/mol C)] 60.052 g/mol x 100 = 6.71% 60.052 g/mol x 100 = 6.71%% O% O = [2 mol O= [2 mol O x (15.9994x (15.9994 g/mol C)]g/mol C)] 60.052 g/mol x 100 = 53.29% 60.052 g/mol x 100 = 53.29%

100xwholeofmasswholeinAofmass

Amass %

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Practice ProblemPractice ProblemWhat is the mass percentage of C in in l-carvone, CWhat is the mass percentage of C in in l-carvone, C1010HH1414O, O, which is the principal component of spearmint?which is the principal component of spearmint?

C = 12.0107 g/mol; H = 1.00794 g.mol;C = 12.0107 g/mol; H = 1.00794 g.mol; O = 15.9994 g/molO = 15.9994 g/mol

a. 30% b. 40% c. 60% d. 70% e. 80%a. 30% b. 40% c. 60% d. 70% e. 80%

Ans: e Molar Mass C= 12.0170 g/mol C x 10 mol C = 120.170 g C H= 1.00794 g/mol C x 14 mol H = 14.1112 g H O= 15.9994 g/mol C x 1 mol O = 15.9994 g O Molar Mass CC1010HH1414OO = 150.218 g/mol= 150.218 g/mol

Mass % C = 120.170 / 150.218 x 100Mass % C = 120.170 / 150.218 x 100 = 79.9971 = 79.9971 (80%)(80%)

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Empirical & Molecular FormulasEmpirical & Molecular Formulas

Empirical formulaEmpirical formula – formula of a substance – formula of a substance written with the smallest whole number written with the smallest whole number subscriptssubscripts

EF of Acetic acid = CEF of Acetic acid = C22HH55OO22

For small molecules, empirical formula is For small molecules, empirical formula is identical to the identical to the molecular formulamolecular formula: formula : formula for a single molecule of substancefor a single molecule of substance

For succinic acid, its molecular formula is For succinic acid, its molecular formula is CC44HH66OO44 Its empirical formula is C Its empirical formula is C22HH33OO22 (n = 2) (n = 2)

Molecular weight = n x empirical formula Molecular weight = n x empirical formula weightweight (n = number of empirical formula (n = number of empirical formula units in the molecule)units in the molecule)

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Practice ProblemPractice ProblemOf the following, the only empirical formula

is

a. C2H4 b. C5H12 c. N2O4 d. S8 e. N2H4

Ans: b subscript (5) cannot be further divided into whole numbers

Molecular Formula from Elemental Analysis: An Example Calculation

A moth repellant, para-dichlorobenzene, has the composition 49.1% A moth repellant, para-dichlorobenzene, has the composition 49.1%

C, 2.7% H and 48.2% Cl. Its molecular weight is determined from C, 2.7% H and 48.2% Cl. Its molecular weight is determined from

mass spectrometry (next slide). What is its molecular formula?mass spectrometry (next slide). What is its molecular formula?

49.1 g C49.1 g C x 1 mol C x 1 mol C // 12.010712.0107 g Cg C = 4.0880 mol C= 4.0880 mol C

2.7 g H x 1 mol H2.7 g H x 1 mol H // 1.007941.00794 g Hg H = 2.6787 mol H= 2.6787 mol H

48.2 g Cl x 1 mol Cl48.2 g Cl x 1 mol Cl // 35.45335.453g Clg Cl = 1.3595 mol Cl= 1.3595 mol Cl

Convert Mole values to “Whole” numbers (divide each value by Convert Mole values to “Whole” numbers (divide each value by

smallest) smallest)

4.0880 / 1.3595 = 3.01 (3 mol C)4.0880 / 1.3595 = 3.01 (3 mol C)

2.6787 / 1.3595 = 1.97 (2 mol H) 2.6787 / 1.3595 = 1.97 (2 mol H)

1.3595 / 1.3595 = 1.00 (1 mol Cl)1.3595 / 1.3595 = 1.00 (1 mol Cl)

Empirical Formula is: CEmpirical Formula is: C33HH22ClClCon’t on next slide 17

Molecular Formula from Elemental Analysis: An Example Calculation (Con’t)

Empirical formula = CEmpirical formula = C33HH22ClCl

Compute Empirical Formula Weight (EFW)Compute Empirical Formula Weight (EFW)

EFW = 3 x 12.01 + 2 x 1.01 + 1 x 35.45 = 73.51 amuEFW = 3 x 12.01 + 2 x 1.01 + 1 x 35.45 = 73.51 amu

Molecular weight (MMolecular weight (M++ ion from mass spectrum) = 146 ion from mass spectrum) = 146 amuamu

n = 146/73.51 = 1.99 = 2n = 146/73.51 = 1.99 = 2

Molecular Formula = CMolecular Formula = C66HH44ClCl22

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M+ = MW = 146 amu

M+ = molecular ion peak

146

M+

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The Chemical EquationThe Chemical EquationA chemical equation is the representation of the reactants and A chemical equation is the representation of the reactants and products in a chemical reaction in terms of chemical symbols and products in a chemical reaction in terms of chemical symbols and formulas.formulas.

The The subscriptssubscripts represent the number of atoms of an element in the represent the number of atoms of an element in the compound.compound.

The The coefficientscoefficients in front of the compound represents the number of in front of the compound represents the number of moles of each compound required to balance the equation.moles of each compound required to balance the equation.

A balanced equation will have an equal number of atoms of each A balanced equation will have an equal number of atoms of each element on both sides of equation element on both sides of equation

NN22(g) + O(g) + O22(g) (g) 2 NO(g) 2 NO(g)

1 mole nitrogen + 1 mole oxygen yields 2 moles nitrogen 1 mole nitrogen + 1 mole oxygen yields 2 moles nitrogen monoxidemonoxide

Phase representations in Chemical EquationsPhase representations in Chemical Equations

= yields, or forms= yields, or forms (g) = gas phase(g) = gas phase

(l)(l) = liquid phase= liquid phase (s) = Solid phase(s) = Solid phase

Reactants Products

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Stoichiometry in Chemical Equations

StiochiometryStiochiometry – calculation of the quantities – calculation of the quantities of reactants and products in a chemical of reactants and products in a chemical reactionreaction

Example: air oxidation of methane to form Example: air oxidation of methane to form ozone (pollutant) and hydroxyl radical (OHozone (pollutant) and hydroxyl radical (OH))

CHCH44 + 10 O + 10 O22 →→ CO CO22 + H + H22O + 5 OO + 5 O33 + 2 OH + 2 OH

molecules 1 10 1 1 5 2molecules 1 10 1 1 5 2

moles 1 10 1 1 5 2 moles 1 10 1 1 5 2

Mass/mol 16.0g 320g 44.0g 18.0g 240g Mass/mol 16.0g 320g 44.0g 18.0g 240g 34.0g 34.0g

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StoicStoichiomehiometry in Chemical Equationstry in Chemical EquationsWhen dinitrogen pentoxide, N2O5, a while solid, is heated, it decomposes to nitrogen dioxide and oxygen

2 N2O5(s) → 4 NO2(g) + O2(g)

If a sample of N2O5 produces 1.315 g of O2, how many grams of NO2 are formed? How many grams of N2O5 are consumed?

Strategy:

1.Compute actual no. moles oxygen produced

1.315 g O2 x (1 mol O2/32.00 g O2) = 0.04109 moles Oxygen

2.Determine molar ratio of NO2 & N2O5 relative to O2 - (4:1 & 2:1)

3.Compute mass of NO2 produced from molar ratio and actual moles O2

0.04109 mol O2 x (4 mol NO2/1 mol O2) x (46.01g NO2/1 mol NO2) = 7.562g NO2

4.Compute mass of N2O5 from molar ratio and actual moles O2

0.04109 mol O2 x (2 mol N2O5 /1 mol O2) x (108.0g N2O5 /mol N2O5) = 8.834g N2O5

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Stoichiometry in Chemical Equations

How many grams of HCl are required to react How many grams of HCl are required to react with 5.00 grams manganese dioxide according with 5.00 grams manganese dioxide according to this equation?to this equation?

4 HCl(aq) + MnO4 HCl(aq) + MnO22(s) (s) 2 H 2 H22O(l) + MnClO(l) + MnCl22(aq) + Cl(aq) + Cl22(g)(g)Strategy: 1. Determine the Molar Ratio of HCL to MnO2

2. Compute the no. moles MnO2 actually used 3. Use actual moles MnO2 & Molar ratio to compute mass HCL

Molar Ratio HCl : MnO2 = 4 : 1

5.00 g MnO2 x (1 mol MnO2/86.9368 g MnO2) = 0.575 mol

MnO2

0.575 mol MnO2 x (4 mol HCl/1 mol MnO2) x (36.461 g

HCl/mol HCl)

= 8.39 g HCl

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Limiting Reactants and YieldsLimiting Reactants and YieldsLimiting Reagent & Theoretical Yield

The “Limiting Reagent” is that reactant whose mass (on a molar equivalent basis) actually consumed in the reaction is less than the amount of the other reactant, i.e., the reactant in excess.

From the Stoichiometric balanced reaction equation determine the molar ratio among the reactants and products, i.e., how many moles of reagent A react with how many moles of reagent B to yield how many moles of product C, D, etc.

The moles of product(s) will be the same as the “limiting Reagent” on a molar equivalent basis.

If the ratio of moles of A to moles of B actually used is greater than the Stoichiometric molar ratio of A to B, then the A reagent is in “Excess” and the B reagent is “Limiting.”

If, however, the actual molar ratio of A to B used is less than the Stoichiometric molar ratio, then B is in excess and A is “Limiting.”

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Limiting Reactants and YieldsLimiting Reactants and YieldsExample 1

A + B C Molar Ratio A:B = 1

Moles actually used: A = 0.345 B = 0.698

Ratio of moles actually used (A/B)

0.345/0.698 = 0.498

0.498 < 1.0

B is in excess) & A is Limiting

Since 1 mol A produces 1 mol C

Theoretical Yield of C = 0.345 moles

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Limiting Reactants and YieldsLimiting Reactants and YieldsExample 2

A + B C

Stoichiometric Molar ratio A:B = 1 : 1 = 1.0

Moles actually used: A = 0.20 B = 0.12

Ratio of Moles actually used (A/B):

0.20 / 0.12 = 1.67

The ratio of A:B is greater than 1.00

A is in excess and B is limiting.

Only 0.12 moles of the 0.2 moles of A would be required to react with the 0.12 moles of B.

The reaction would have a theoretical yield of:

0.12 moles of C (Molar Ratio of B:C = 1)

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27

Limiting Reactants and Yields

Example 3

A + 2B C

Stoichiometric Molar ratio A:B = 1 : 2 = 0.5

Moles actually used: A = 0.0069; B = 0.023

Ratio of Moles actually used (A/B):

0.0069 / 0.023 = 0.30 < 0.5 A is limiting

Only 0.0069 2 = 0.0138 moles of the 0.023 moles of B are required to react with 0.0069 moles of A.

Since 0.0138 < 0.023 B is in excess, A is limiting.

The reaction would have a theoretical yield of:

0.0069 moles of C (Molar Ratio of A:C = 1)

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Limiting Reactants and Yields

Theoretical Yield & Percent Yield The Theoretical Yield, in grams, is computed from the number of moles of the “Limiting Reagent”, the Stoichiometric Molar Ratio, and the Molecular Weight of the product.

Yield = mol (Lim) x Mol Ratio Prod/Lim x Mol Wgt Product

The Percent Yield of a product obtained in a “Synthesis” experiment is computed from the amount of product actually obtained in the experiment and the Theoretical Yield.

% Yield = Actual Yield / Theoretical Yield x 100

Note: The yield values can be expressed in either grams or moles

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Example Yield Calculation

Methyl Salicylate (MSA) is prepared by heating salicylic acid (SA), Methyl Salicylate (MSA) is prepared by heating salicylic acid (SA),

CC77HH66OO33, with methanol (ME), CH, with methanol (ME), CH33OHOH

CC77HH66OO33 + CH + CH33OH OH C C88HH88OO33 + H + H22OO

1.50 g of salicylic acid (SA) is reacted with 11.20 g of methanol 1.50 g of salicylic acid (SA) is reacted with 11.20 g of methanol

(ME). The yield of methyl salicylate is 1.27 g. What is the limiting (ME). The yield of methyl salicylate is 1.27 g. What is the limiting

reactant? What is the percent yield of Methyl Salicylate (MSA)?reactant? What is the percent yield of Methyl Salicylate (MSA)?

Molar Ratio:Molar Ratio: 1 mole SA reacts with 1 mole ME to produce 1 mole MSA1 mole SA reacts with 1 mole ME to produce 1 mole MSA

Moles SA:Moles SA: 1.50 g SA x (1 mol SA/138.12 g SA )1.50 g SA x (1 mol SA/138.12 g SA ) = 0.0109 mol SA= 0.0109 mol SA

Moles ME:Moles ME: 11.20 g ME x (1 mol ME/32.04 g ME)11.20 g ME x (1 mol ME/32.04 g ME) = 0.350 mol ME= 0.350 mol ME

0.0109 mol SA x (1mol SA/1 mol ME) < 0.350 mol ME0.0109 mol SA x (1mol SA/1 mol ME) < 0.350 mol ME

Salicylic acid (SA) is limiting; Methanol is in “Excess”Salicylic acid (SA) is limiting; Methanol is in “Excess”

Theoretical yieldTheoretical yield = 0.0109 mol SA x (1 mol MSA/1 mol SA) x = 0.0109 mol SA x (1 mol MSA/1 mol SA) x (152.131 g MSA/1 mol MSA) = 1.66 g MSA(152.131 g MSA/1 mol MSA) = 1.66 g MSA

% Yield% Yield = actual/theoretical x 100 = 1.27 g/1.66 g x 100 = 76.5% = actual/theoretical x 100 = 1.27 g/1.66 g x 100 = 76.5%

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Example Yield CalculationExample Yield Calculation

Hydrogen is a possible clean fuel because it reacts with Hydrogen is a possible clean fuel because it reacts with Oxygen to form non-polluting waterOxygen to form non-polluting water

2 H2(g) + O2(g) → 2 H2O(g)

If the yield of this reaction is 87% what mass of Oxygen is required to produce 105 kg of water?

Molar Ratio: 2 mol H2 reacts with 1 mol O2 to form 2 mol water (H2O)

Moles H2O: 105 kg H2O x (1000g/1 kg) x (1 mol H2O/18.01 g/mol H2O)

= 5,830 mol H2O

Moles O2 = 1 mol O2/2 mol H2O x 5,830 mol H2O = 11,660 mol O2

Mass O2 = 11,660 mol O2 x (32.0 g O2 / 1 mol O2) x (1 kg/1000g)

= 373 kg O2

Required O2 = 373 kg x 100%/87% = 429 kg O2 required

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Solution StoichiometrySolution Stoichiometry Solute – A substance dissolved in another Solute – A substance dissolved in another

substancesubstance Solvent – The substance in which the “Solute” is Solvent – The substance in which the “Solute” is

dissolveddissolved Concentration – The amount of solute dissolved in a Concentration – The amount of solute dissolved in a

given amount of solventgiven amount of solvent MolarityMolarity – Expresses the concentration of a – Expresses the concentration of a

solution in units of moles solute per liter of solutionsolution in units of moles solute per liter of solution

MolalityMolality is another concentration term which is another concentration term which expresses the number of moles dissolved in 1000g expresses the number of moles dissolved in 1000g (1KG) of solvent.(1KG) of solvent.

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soln L

solute mol M

solution of liters

solute of moles Molarity

solvent Kg

solute mol m

Solvent Kilogram

Solute Moles Molality

Solution Volume vs. Solvent VolumeSolution Volume vs. Solvent Volume

The The VolumeVolume term in the denominator of term in the denominator of the the molaritymolarity expression is the expression is the solutionsolution volume volume notnot the volume of the solvent the volume of the solvent

1 mole of solute dissolved in 1 Liter of a 1 mole of solute dissolved in 1 Liter of a solvent solvent does not does not product a 1 M solution.product a 1 M solution.

The The MassMass term in the denominator of the term in the denominator of the molalitymolality expression expression isis the the MassMass of solvent of solvent

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Solution StoichiometrySolution Stoichiometry Mole – Mass Conversions involving Solutions

Calculating the Mass of a substance given theVolume and Molarity

Ex. How many grams of sodium hydrogen phosphate (Na2HPO4) are in 1.75 L of a 0.460 M solution?

Moles Na2HPO4 = 1.75 L x 0.460 mol Na2HPO4/1 L soln

= 0.805 mol Na2HPO4

Mass Na2HPO4 = 0.805 mol x 141.96 g Na2HPO4/mol Na2HPO4

= 114. g

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Solution StoichiometrySolution StoichiometryPractice ProblemPractice Problem

Calculate the volume of a 3.30 M Sucrose (FW – 342.30 g/mol) solution containing 135 g of solute.

Ans: moles solute

135 g sucrose x 1 mol sucrose / 342.3 g sucrose = 0.3944 mol

Vol soln

0.3944 mol sucrose x 1.00 L solution/3.30 mol sucrose = 0.120 L

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DilutionDilution

The amount of solute in a solution is the The amount of solute in a solution is the same after the solution is diluted with same after the solution is diluted with additional solventadditional solvent

Dilution problems utilize the following Dilution problems utilize the following relationship between the molarity (M) and relationship between the molarity (M) and volume (V)volume (V)

MMdildil x V x Vdildil = number moles = M = number moles = Mconcconc x x VVconcconc

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Practice ProblemsPractice Problems

Calculate the Molarity of the solution prepared Calculate the Molarity of the solution prepared by diluting 37.00 mL of 0.250 M Potassium by diluting 37.00 mL of 0.250 M Potassium Chloride (KCl) to 150.00 mL.Chloride (KCl) to 150.00 mL.

Ans: Dilution problem (MAns: Dilution problem (M11VV11 = M = M22VV22))

MM11 = 0.250 M KCl V= 0.250 M KCl V11 = 37.00 mL= 37.00 mL

MM22 = ? V= ? V22 = 150.00 mL= 150.00 mL

MM11VV11 = M= M22VV22 M M22 = M= M11VV11 / V / V22

MM22 == (0.250 M) x (37.00 mL / 150.0 mL)(0.250 M) x (37.00 mL / 150.0 mL)

= 0.0617 M = 0.0617 M

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