1 © 2006 brooks/cole - thomson chapter 3 stoichiometry of formulas and equations chemical reactions

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© 2006 Brooks/Cole - Thomson

Chapter 3 Chapter 3 Stoichiometry of Stoichiometry of

Formulas and EquationsFormulas and Equations

Chemical Reactions

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3.1 The Mole 3.1 The Mole An atom of carbon weighs 12 amu = 1.99 x 10An atom of carbon weighs 12 amu = 1.99 x 10-23-23 g g

1 mol of 1 mol of 1212C = 6.022 x 10C = 6.022 x 102323 atoms of C atoms of C = 12.00 g of C = 12.00 g of C

• A mole is the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope.

• Just like a dozen = 12 items a pair = 2 items

a mole = 6.022 x 10a mole = 6.022 x 102323 itemsitems (also known as Avogadro’s number).

The point—it takes a lot of atoms to have something we can weigh.

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Avogadro’s Number Avogadro’s Number used as conversion factorsused as conversion factors

Avogadro’s number 6.02 x 1023 can be written as an equality and two conversion factors.

Equality:

1 mole = 6.02 x 1023 particles

Conversion Factors:

6.02 x 1023 particles and 1 mole ______

1 mole 6.02 x 1023 particles

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Molar Mass Molar Mass

1 mol of 1 mol of 1212C = 6.022 x 10C = 6.022 x 102323 atoms of C atoms of C = 12.00 g of C = 12.00 g of C

12.00 g of 12.00 g of 1212C is its C is its MOLAR MASSMOLAR MASS

Taking into account all of the Taking into account all of the

isotopes of C, the molar mass isotopes of C, the molar mass

of Carbon is of Carbon is 12.011 g/mol12.011 g/mol

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Where do we find molar mass of Where do we find molar mass of an element?an element?

- It’s on your periodic tableIt’s on your periodic table! The number that represents the weighted average atomic mass is the same amount a mole of those atoms would weigh in grams!

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One-mole One-mole AmountsAmounts

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PROBLEM: PROBLEM: How many moles How many moles of Mg are in 0.200 g? How many of Mg are in 0.200 g? How many

atoms?atoms?

PROBLEM: PROBLEM: How many moles How many moles of Mg are in 0.200 g? How many of Mg are in 0.200 g? How many

atoms?atoms?

Mg has a molar mass of 24.3050 g/mol.

0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol

8.23 x 10-3 mol • 6.022 x 1023 atoms

1 mol

= 4.95 x 10= 4.95 x 102121 atoms Mg atoms Mg

How many atoms in this piece of Mg?

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Converting between moles, mass, Converting between moles, mass, & number of chemical entities& number of chemical entities

Avogadro’s number is used to convert moles of a substance to # of entities of that substance

How many Cu atoms are in 0.50 mole Cu?

0.50 mole Cu x 6.02 x 1023 Cu atoms 1 mole Cu

= 3.0 x 1023 Cu atoms

How many moles of CO2 are in 2.50 x 1024 molecules CO2?

2.50 x 1024 molecules CO2 x 1 mole CO2

6.02 x 1023 molecules CO2

= 4.15 moles CO2

Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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MolecuLes, MolecuLes, Compounds, & Compounds, & MOLAR MASSMOLAR MASS

How many How many atoms of Catoms of C are there in a can of are there in a can of beer if it contains 21.3 g of Cbeer if it contains 21.3 g of C22HH66O?O?

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MolecuLes, MolecuLes, Compounds, & Compounds, & MOLAR MASSMOLAR MASS

What is the molar mass of ethanol, What is the molar mass of ethanol, CC22HH66O?O?

1 mol 1 mol CC22HH66OO contains contains

2 mol C (12.01 g C/1 mol) = 24.02 g C2 mol C (12.01 g C/1 mol) = 24.02 g C

6 mol H (1.01 g H/1 mol) = 6.06 g H6 mol H (1.01 g H/1 mol) = 6.06 g H

1 mol O (16.00 g O/1 mol) = 16.00 g O1 mol O (16.00 g O/1 mol) = 16.00 g O

TOTAL TOTAL = = molar mass = 46.08 g/molmolar mass = 46.08 g/mol

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How many How many molesmoles of ethanol are of ethanol are there in a can of beer if it there in a can of beer if it contains 21.3 g of Ccontains 21.3 g of C22HH66O?O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calculate moles of alcohol

21.3 g • 1 mol

46.08 g = 0.462 mol

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How many How many moleculesmolecules of alcohol of alcohol are there in a can of beer if it are there in a can of beer if it

contains 21.3 g of Ccontains 21.3 g of C22HH66O?O?

= 2.78 x 1023 molecules

We know there are 0.462 mol of C2H6O.

0.462 mol • 6.022 x 1023 molecules

1 mol

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How many How many atoms of Catoms of C are there are there in a can of beer if it contains 21.3 in a can of beer if it contains 21.3

g of Cg of C22HH66O?O?

= 5.57 x 1023 C atoms

There are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023 molecules • 2 C atoms1 molecule

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Mass PercentMass PercentA pure compound always consists of the A pure compound always consists of the

same elements combined in the same same elements combined in the same proportions by weight.proportions by weight.

Therefore, we can express molecular Therefore, we can express molecular composition as composition as PERCENT PERCENT COMPOSITION BY WEIGHTCOMPOSITION BY WEIGHT

Ethanol, CEthanol, C22HH66OO52.13% C52.13% C13.15% H 13.15% H 34.72% O34.72% O

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Percent Percent CompositionComposition

Percent Percent CompositionComposition

Consider NOConsider NO22, molar mass = 46.0 g,, molar mass = 46.0 g,

what is the weight percent of N and of O?what is the weight percent of N and of O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N

46.0 g NO2 • 100% = 30.4 %Wt. % N =

14.0 g N46.0 g NO2

• 100% = 30.4 %

What are the weight percentages of N and O in What are the weight percentages of N and O in NO?NO?

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Percent Composition of a HydratePercent Composition of a HydrateHydrate: A solid compound that includes water molecules in its crystal structureAnhydrous Compound: An ionic compound that is without water or from which water has been removed

CuSO4•5H2O (s) CuSO4 (s) + 5H2O (g)copper(II) sulfate pentahydrate copper(II) sulfate

Cu 39.82%S 20.09%O 40.09%

H2O 36.05%

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3.3 Writing and Balancing 3.3 Writing and Balancing Chemical EquationsChemical Equations

Chemical equations Chemical equations depict the kind of depict the kind of reactantsreactants and and productsproducts and their relative and their relative amounts in a reaction.amounts in a reaction.

The numbers in the front are calledThe numbers in the front are calledstoichiometric coefficientsstoichiometric coefficientsThe letters (s), (g), and (l) are the physical states The letters (s), (g), and (l) are the physical states

of compounds.of compounds.

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Balancing Chemical EquationsBalancing Chemical Equations

Reactants Products 1 C atom = 1 C atom 4 H atoms = 4 H atoms 4 O atoms = 4 O atoms

- start with the most complex substance- end with the least complex substance

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Balancing Polyatomic Ion as a UnitBalancing Polyatomic Ion as a UnitNaNa33POPO44((aqaq) + MgCl) + MgCl22((aqaq) ) NaCl( NaCl(aqaq) + Mg) + Mg33(PO(PO44))22((ss))

Balance POBalance PO443- 3- as a unitas a unit

2 2 NaNa33POPO44((aqaq) ) MgMg33(PO(PO44))22((ss))

2 PO2 PO443-3- == 2 PO 2 PO44

3-3-

Balance MgBalance Mg

3 Mg3 MgClCl22((aqaq) ) MgMg33(PO(PO44))22((ss))

3 Mg3 Mg2+ 2+ == 3 Mg3 Mg2+2+

Balance Na and Cl to complete balanced equationBalance Na and Cl to complete balanced equation

3 3 MgMgClCl22((aqaq) + ) + 2 Na2 Na33POPO44((aqaq) ) 6 Na6 NaClCl((aqaq) + Mg) + Mg33(PO(PO44))22((ss))

6 Na6 Na+ + = = 6 Na6 Na++

6 Cl6 Cl-- == 6 Cl 6 Cl--

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Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

__C__C33HH88(g) + __ O(g) + __ O22(g) ----> __CO(g) ----> __CO22(g) + __ H(g) + __ H22O(g)O(g)

__B__B44HH1010(g) + __ O(g) + __ O22(g) ----> (g) ----> __ B__ B22OO33(g) + __H(g) + __H22O(g)O(g)

__Al(s) + __H2SO4(aq) ---- >

__Al2(SO4)3(aq) + __H2(g)

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3.4 Calculating Quantites of 3.4 Calculating Quantites of Reactant and ProductReactant and Product

- the study of mass the study of mass relationships in relationships in chemical chemical reactionsreactions

- It rests on the It rests on the principle of the principle of the conservation of conservation of mattermatter. .

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Molar Ratios from a Molar Ratios from a Balanced EquationBalanced Equation

• A molar ratio is used to relate the number of moles of one reactant or product to another.

4 Fe(4 Fe(ss) + 3 O) + 3 O22((gg) -------- > 2 Fe) -------- > 2 Fe22OO33((ss))

4 moles4 moles Fe + Fe + 3 moles3 moles O O22 ------- > ------- > 2 moles2 moles Fe Fe22OO33

Molar ratios:Molar ratios:

Fe and OFe and O22 4 moles Fe 4 moles Fe and and 3 moles O3 moles O22

3 moles O3 moles O22 4 moles Fe 4 moles Fe

Fe and FeFe and Fe22OO33 4 moles Fe 4 moles Fe and and 2 moles Fe2 moles Fe22OO33

2 moles Fe2 moles Fe22OO33 4 moles Fe 4 moles Fe

OO22 and Fe and Fe22OO33 3 moles O3 moles O22 and and 2 moles Fe2 moles Fe22OO33

2 moles Fe2 moles Fe22OO33 3 moles O 3 moles O22

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Learning Check

How many moles of Fe are needed for the How many moles of Fe are needed for the

reaction of 12.0 moles Oreaction of 12.0 moles O22??

4 Fe(4 Fe(ss) + 3 O) + 3 O22((gg) ------- > 2 Fe) ------- > 2 Fe22OO33((ss))

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GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass of A

Molar ratio

(molesB/molesA)

Molesof A

Moles of B

Mass Of B

Molar mass of A

Molar mass of B

A is known; B is unknownA is known; B is unknown

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PROBLEM:PROBLEM: If 454 g of NH If 454 g of NH44NONO33 decomposes, how many grams of decomposes, how many grams of NN22O are formed? What is the O are formed? What is the theoretical yield of products? theoretical yield of products?

STEP 1:STEP 1:Write the balanced chemical equationWrite the balanced chemical equation

NHNH44NONO33 ---> N ---> N22O + 2 HO + 2 H22OO

STEP 2: STEP 2: Convert mass A to moles AConvert mass A to moles A

STEP 3: STEP 3: Convert moles A to moles BConvert moles A to moles B

STEP 4: STEP 4: Convert moles B to mass BConvert moles B to mass B

ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY

PROBLEMS!PROBLEMS!

ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY

PROBLEMS!PROBLEMS!

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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O

STEP 4 STEP 4 Convert moles B (5.70 mol) Convert moles B (5.70 mol) to mass B: to mass B: 250. g N250. g N22OO

This is called the This is called the THEORETICAL THEORETICAL YIELDYIELD

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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O

% yield = actual yield

theoretical yield • 100%

% yield = 131 g250. g

• 100% = 52.4%

In the above reaction,In the above reaction, if only 131 g of Nif only 131 g of N22O is O is

collected, what is the percent yield?collected, what is the percent yield?

Reaction YieldReaction Yield

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LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________

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PROBLEMPROBLEM: : Mix 5.40 g of Al with 8.10 Mix 5.40 g of Al with 8.10 g of Clg of Cl22. What mass of AlCl. What mass of AlCl33 can be can be formed?formed?

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PROBLEMPROBLEM: Mix 5.40 g of Al with : Mix 5.40 g of Al with 8.10 g of Cl8.10 g of Cl22. What mass of AlCl. What mass of AlCl33 can be can be formed?formed?

2 Al (s) + 3 Cl2 (g) 2 AlCl3 (s)

Solution: 5.40 g Al produces 0.200 mol AlCl3

8.10 g Cl2 produces 0.072 mol AlCl3

Cl2 is limiting reactant

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• ClCl22 was the limiting reactant. was the limiting reactant.

Therefore, Al was present in Therefore, Al was present in

excess. But how much?excess. But how much?

• First, First, based on LR:based on LR: Find the moles (or Find the moles (or

mass) of Al was required.mass) of Al was required.

• Then find the moles (or mass) of Al Then find the moles (or mass) of Al leftover leftover

(in excess).(in excess).

How much of which reactant How much of which reactant will remain when reaction is will remain when reaction is

complete?complete?

How much of which reactant How much of which reactant will remain when reaction is will remain when reaction is

complete?complete?

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2 Al + 3 2 Al + 3 ClCl22 productsproducts

0.200 mol0.200 mol0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR0.114 mol = LR0.114 mol = LR

Calculating Excess AlCalculating Excess AlCalculating Excess AlCalculating Excess Al

Excess Al = Al available - Al requiredExcess Al = Al available - Al required

0.114 mol Cl2 • 2 mol Al

3 mol Cl2 = 0.0760 mol Al req' d

= 0.200 mol - 0.0760 mol = 0.200 mol - 0.0760 mol

= = 0.124 mol Al in excess0.124 mol Al in excess

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In a solution: In a solution: SOLVENTSOLVENT the component whose the component whose

physical state is physical state is preserved when solution preserved when solution formsforms

Usually in much larger Usually in much larger quantity than the solutequantity than the solute

• SOLUTESOLUTEthe other solution componentthe other solution component

3.5 Fundamentals of 3.5 Fundamentals of Solution StoichiometrySolution Stoichiometry

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Molarity (M)Molarity (M)

• Is a concentration term for Is a concentration term for solutions.solutions.

• Gives the moles of solute in 1 L Gives the moles of solute in 1 L solution.solution.

Molar (M) = Molar (M) = moles of solutemoles of solute liter of solutionliter of solution

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Step 1: Step 1: Calculate moles of acid required.Calculate moles of acid required.

(0.0500 mol/L)(0.250 L) = 0.0125 mol(0.0500 mol/L)(0.250 L) = 0.0125 mol

Step 2: Step 2: Calculate mass of acid required.Calculate mass of acid required.

(0.0125 mol )(90.00 g/mol) = (0.0125 mol )(90.00 g/mol) = 1.13 g1.13 g

Amounts – Mass – Number Amounts – Mass – Number Conversions for SolutionsConversions for Solutions

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, ,

is required to make 250. mL of a is required to make 250. mL of a 0.0500 M solution?0.0500 M solution?

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Less than 1.0 L Less than 1.0 L of water was of water was

used to make 1.0 used to make 1.0 L of solution. L of solution.

Notice the water Notice the water left over.left over.

Concentration of Solute Concentration of Solute = = The amount of solute The amount of solute in a solutionin a solution

Prepare Solution of Known Concentration

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PROBLEM:PROBLEM: How many grams of How many grams of NiClNiCl22 should be dissolved in should be dissolved in enough water to make 250. mL of enough water to make 250. mL of 0.154 M solution?0.154 M solution?

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22

Step 2: Step 2: Calculate mass of Calculate mass of NiClNiCl22

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A shortcutA shortcut

CCinitialinitial • V • Vinitialinitial = C = Cfinalfinal • V • Vfinalfinal

Preparing Solutions by Preparing Solutions by DilutionDilution

Preparing Solutions by Preparing Solutions by DilutionDilution

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PROBLEMPROBLEM: You have 50.0 mL of : You have 50.0 mL of 3.00 M NaOH and you want 0.500 M 3.00 M NaOH and you want 0.500 M NaOH. What do you do?NaOH. What do you do?

Add water to the 3.00 M solution to Add water to the 3.00 M solution to lower its concentration to 0.500 M lower its concentration to 0.500 M

Dilute the solution!Dilute the solution!

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

Add 250. mL of waterAdd 250. mL of water to to 50.0 mL of 3.00 M NaOH to 50.0 mL of 3.00 M NaOH to make 300. mL of 0.500 M make 300. mL of 0.500 M NaOH.NaOH.

3.00 M NaOH3.00 M NaOHconcentratedconcentrated

0.500 M 0.500 M NaOHNaOHdiluteddiluted

50.0 50.0 mLmL

250. mL water

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Stoichiometry of Stoichiometry of Reactions in SolutionReactions in Solution

1. Write and balance the chemical equation2. Find the moles of the known substance A3. Find the moles of the unknown substance B,

using the molar ratio 4. Convert moles B to the desired units

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Problems

1) How many mL of a 0.150 M Na2S solution are needed to completely react 18.5 mL of 0.225 M NiCl2 solution?

NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)

2) If 22.8 mL of 0.100 M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution?

MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)

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3.2 Determining the Formula 3.2 Determining the Formula of an Unknown Compoundof an Unknown Compound

In In chemical analysischemical analysis we determine the % by weight of we determine the % by weight of each element in a given amount of pure compound and each element in a given amount of pure compound and derive the derive the EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula. Then formula. Then MOLECULARMOLECULAR formula is determined from empirical formula is determined from empirical formula and molar mass.formula and molar mass.

Molecular Formula

Molar massAssume 100 g compound

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• Because it contains only B and H, it must contain Because it contains only B and H, it must contain 18.90% H.18.90% H.

• In 100.0 g of the compound there are 81.10 g of B and In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.18.90 g of H.

Solution: Solution:

- the number of moles of each constituent the number of moles of each constituent (7.502 mol B (7.502 mol B and 18.75 mol H)and 18.75 mol H)

- The mole ratio of these 2 elements The mole ratio of these 2 elements

- (2.500 mol H / 1 mol B) (2.500 mol H / 1 mol B) NEVER ROUND OFF!!NEVER ROUND OFF!!

- The empirical formula of the compoundThe empirical formula of the compound: : BB22HH55

A compound of B and H is 81.10% A compound of B and H is 81.10% B. What is its empirical formula?B. What is its empirical formula?

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A compound of B and H is 81.10% B. Its A compound of B and H is 81.10% B. Its empirical formula is Bempirical formula is B22HH55, and its molar mass , and its molar mass

is 53.3 g/mol. What is its molecular formulais 53.3 g/mol. What is its molecular formula??

A compound of B and H is 81.10% B. Its A compound of B and H is 81.10% B. Its empirical formula is Bempirical formula is B22HH55, and its molar mass , and its molar mass

is 53.3 g/mol. What is its molecular formulais 53.3 g/mol. What is its molecular formula??

The molecular formula of a compound is found by determination of the number of empirical formula units in the molecule

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

Molecular formula = BMolecular formula = B44HH1010

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Determine the Formula of Determine the Formula of a compound by a compound by

CombustionCombustion

Burn 0.115 g of a hydrocarbon, CxHy, and produce

0.379 g of CO2 and 0.1035 g of H2O.

CxHy + some oxygen --->

0.379 g CO2 + 0.1035 g H2O

What is the empirical formula of CxHy?

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Determine the Formula Determine the Formula of a Compound by of a Compound by


Recognize that all C in CO2 and all H in H2O is from CxHy.

CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O

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Determine the Formula Determine the Formula of a Compound by of a Compound by


Solution:

8.61 mol CO2 and 0.00575 mole H2O

8.61mol C and 0.01149 mol H

Mole ratio: 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C

= 1.33 mol H / 1.00 mol C

= 4 mol H / 3 mol C

Empirical formula: CEmpirical formula: C33HH44

CCxxHHy y + oxygen ---> 0.379 g CO + oxygen ---> 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

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Determine a FormulaDetermine a Formula

If this hydrocarbon has molar mass 80.0 g/mol. What is its molecular formula?

Empirical formula: CEmpirical formula: C33HH44

• Molecular formula: (CMolecular formula: (C33HH44))nn

• Mass of CMass of C33HH44 group: 40.0 g/C group: 40.0 g/C33HH44 group group• n = (80.0 g/mol)(1 group/40.0 g) = 2n = (80.0 g/mol)(1 group/40.0 g) = 2• Molecular formula: CMolecular formula: C66HH88

CCxxHHy y + oxygen ---> 0.379 g CO + oxygen ---> 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

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Determine Molar MassDetermine Molar Mass

If 1.44 x 101.44 x 10-3-3 mol mol of this hydrocarbon was used in the experiment. What is its molar mass?

0.1035 g H0.1035 g H22O = 1.149 x 10 O = 1.149 x 10 -2 -2 mol Hmol H

0.379 g CO0.379 g CO22 = 8.61 x 10 = 8.61 x 10-3 -3 mol C mol C

Total mass of the 1.44 x 10Total mass of the 1.44 x 10-3-3 mol of compound: mol of compound:

(1.149 x 10(1.149 x 10-2-2 mol H x 1.008 g/mol H) mol H x 1.008 g/mol H)

+ (8.61 x10+ (8.61 x10-3-3 mol C x 12.00 g/mol C) = 0.115 g compound mol C x 12.00 g/mol C) = 0.115 g compound

Molar mass of hydrocarbon = 0.115 g / 1.44 x 10Molar mass of hydrocarbon = 0.115 g / 1.44 x 10-3-3 mol mol

Molar mass: 79.9 g/molMolar mass: 79.9 g/mol

1.44 x 101.44 x 10-3-3 mol C mol CxxHHy y + O + O22 ---> 0.379 g CO ---> 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

1 © 2006 brooks/cole - thomson chapter 3 stoichiometry of formulas and equations chemical reactions

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