STOICHIOMETRY Chapter 3 - the study of the quantitative aspects of chemical reactions

STOICHIOMETRYSTOICHIOMETRYChapter 3Chapter 3

STOICHIOMETRYSTOICHIOMETRYChapter 3Chapter 3

- the study of the - the study of the quantitative quantitative aspects of aspects of chemical chemical reactions.reactions.

Counting AtomsCounting Atoms

Chemistry is a quantitative Chemistry is a quantitative science—we need a science—we need a “counting unit.”“counting unit.”

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as C atoms in molecules) as C atoms in 12.0 g of 12.0 g of 1212C.C.

1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as C atoms in molecules) as C atoms in 12.0 g of 12.0 g of 1212C.C.

MOLEMOLE

Particles in a Particles in a MoleMole

6.02214199 x 106.02214199 x 102323

Avogadro’s NumberAvogadro’s Number

There is Avogadro’s number of particles in a mole of any substance.There is Avogadro’s number of particles in a mole of any substance.

Amedeo AvogadroAmedeo Avogadro1776-18561776-1856

Molar MassMolar Mass1 mol of 1 mol of 1212C C

= 12.00 g of C= 12.00 g of C = 6.022 x 10 = 6.022 x 102323 atoms atoms

of Cof C

12.00 g of 12.00 g of 1212C is its C is its

MOLAR MASSMOLAR MASS

Taking into account all Taking into account all

of the isotopes of C, of the isotopes of C,

the molar mass of C is the molar mass of C is

12.011 g/mol12.011 g/mol

One-mole AmountsOne-mole Amounts

PROBLEM: What amount of Mg PROBLEM: What amount of Mg is represented by 0.200 g? How is represented by 0.200 g? How many atoms?many atoms?

PROBLEM: What amount of Mg PROBLEM: What amount of Mg is represented by 0.200 g? How is represented by 0.200 g? How many atoms?many 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?

MOLECULAR WEIGHT MOLECULAR WEIGHT AND MOLAR MASSAND MOLAR MASS

Molecular weightMolecular weight = sum of the = sum of the

atomic weights of all atoms in the atomic weights of all atoms in the

molecule.molecule.

Molar massMolar mass = molecular weight in = molecular weight in

grams per mol.grams per mol.

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

1 mol contains1 mol contains

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

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

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

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

How many How many molesmoles of alcohol of alcohol (C(C22HH66O) are there in a O) are there in a “standard” can of beer if there “standard” can of beer if there are 21.3 g of Care 21.3 g of C22HH66O?O?

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

(b) Calc. moles of alcohol

21.3 g • 1 mol

46.08 g = 0.462 mol

How many How many moleculesmolecules of alcohol of alcohol are there in a “standard” can of are there in a “standard” can of beer if there are 21.3 g of Cbeer if there are 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

How many How many atoms of Catoms of C are there are there in a “standard” can of beer if there in a “standard” can of beer if there are 21.3 g of Care 21.3 g 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

Empirical & Molecular Empirical & Molecular FormulasFormulas

A 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 the molecular Therefore, we can express the molecular composition as composition as PERCENT BY PERCENT BY WEIGHTWEIGHT

Ethanol, CEthanol, C22HH66OO

52.13% C52.13% C13.15% H 13.15% H 34.72% O34.72% O

Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider NOConsider NO22, Molar mass = ?, Molar mass = ?

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

Determining Determining FormulasFormulas

Determining Determining FormulasFormulas

In In chemical analysischemical analysis we determine we determine the % by weight of each element in a given the % by weight of each element in a given amount of pure compound and derive the amount of pure compound and derive the

EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.

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

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

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

• Calculate the Calculate the number of molesnumber of moles of each of each constitutent.constitutent.

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

Calculate the Calculate the number of molesnumber of moles of each of each element in 100.0 g of sample.element in 100.0 g of sample.

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H


Now, recognize that Now, recognize that atoms combine in atoms combine in the ratio of small whole numbersthe ratio of small whole numbers

Find the Find the ratio of molesratio of moles of elements of elements in the compound.in the compound.


But we need a But we need a whole number ratiowhole number ratio. .

2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol B

EMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55

Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlwaysdivide by the smaller number.divide by the smaller number.

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B


A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. ItsIts empirical formulaempirical formula is Bis B22HH55. .

What is itsWhat is its molecular molecular formula formula ??Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, ,

BB66HH1515, B, B88HH2020, etc.? , etc.?

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.

B2H6

How to Determine the molar How to Determine the molar mass?mass?

Mass spectrometerMass spectrometer

A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??

We need to do an We need to do an EXPERIMENTEXPERIMENT to to find the MOLAR MASS.find the MOLAR MASS.

Here experiment gives Here experiment gives 53.3 g/mol53.3 g/molCompare with the mass of BCompare with the mass of B22HH55

= = 26.66 g/unit26.66 g/unit

Find the ratio of these masses.Find the ratio of these masses.53.3 g/mol

26.66 g/unit of B2H5 =

2 units of B2H51 mol

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

Molecular formula = BMolecular formula = B44HH1010

ELEMENTSELEMENTS are composed of identical particles, atomsare composed of identical particles, atoms

CHEMICAL COMPOUNDSCHEMICAL COMPOUNDS are formed when atoms of different elements combine with each other: A given are formed when atoms of different elements combine with each other: A given compound always has the same relative numbers and types of atomscompound always has the same relative numbers and types of atoms

Law of definite proportion [Joseph Proust, 1754-1826]Law of definite proportion [Joseph Proust, 1754-1826]

A given compound always contains the same A given compound always contains the same proportion of elements by massproportion of elements by mass

Law of multiple proportions [John Dalton, 1766-1844]Law of multiple proportions [John Dalton, 1766-1844]

When 2 elements form multiple compounds, the mass When 2 elements form multiple compounds, the mass of the second element per gram of the first one can of the second element per gram of the first one can

always be reduced to small whole numbersalways be reduced to small whole numbers

Law of conservation of mass [Lavoisier, 1743-1794]Law of conservation of mass [Lavoisier, 1743-1794]

Mass is neither created nor destroyed Mass is neither created nor destroyed in a chemical reactionin a chemical reaction

Chemical EquationsChemical EquationsChemical EquationsChemical Equations

• Because the same atoms are Because the same atoms are present in a reaction at the present in a reaction at the beginning and at the end, the beginning and at the end, the amount of matter in a system amount of matter in a system does not change. does not change.

• The The Law of the Law of the Conservation of MatterConservation of Matter

Demo of conservation of matter, See Screen 4.3.

http://xbeams.chem.yale.edu/~batista/113/movies/04m03an1.mov

Because of the principle of the Because of the principle of the conservation of matterconservation of matter, ,

an an equation must be equation must be balancedbalanced..

It must have the same It must have the same number of atoms of the number of atoms of the

same kind on both same kind on both sides.sides.

Chemical EquationsChemical Equations

Lavoisier, 1788Lavoisier, 1788

Balancing Balancing EquationsEquationsBalancing Balancing EquationsEquations

___ Al(s) + ___ Br___ Al(s) + ___ Br22(liq) ---> ___ Al(liq) ---> ___ Al22BrBr66(s)(s)

http://xbeams.chem.yale.edu/~batista/113/movies/04m02vd1.mov

Chemical EquationsChemical EquationsDepict the kind of Depict the kind of reactantsreactants and and productsproducts and their and their

relative amounts in a reaction.relative amounts in a reaction.

2 Al(s) + 3 Br2 Al(s) + 3 Br22(g) ---> Al(g) ---> Al22BrBr66(s)(s)

The numbers in the front are calledThe numbers in the front are called

stoichiometric coefficientsstoichiometric coefficients

The letters (s), (g), and (l) are the physical states of The letters (s), (g), and (l) are the physical states of compounds.compounds.

Chemical EquationsChemical EquationsChemical EquationsChemical Equations4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g)

---> 2 Al---> 2 Al22OO33(s)(s)

This equation meansThis equation means

4 Al atoms + 3 O4 Al atoms + 3 O22 molecules molecules ---give--->---give--->

2 molecules of Al2 molecules of Al22OO33

4 moles of Al + 3 moles of O4 moles of Al + 3 moles of O22 ---give--->---give--->

2 moles of Al2 moles of Al22OO33

STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYIt rests on the principle of the It rests on the principle of the conservation of matterconservation of matter..

2 Al(s) + 3 Br2(liq) ------> Al2Br6(s)

PROBLEM: PROBLEM: If 454 g of NHIf 454 g of NH44NONO33 decomposes, how much N decomposes, how much N22O O and Hand H22O are formed? What is the theoretical O are formed? What is the theoretical yield of products?yield of products?

PROBLEM: PROBLEM: If 454 g of NHIf 454 g of NH44NONO33 decomposes, how much N decomposes, how much N22O O and Hand H22O are formed? What is the theoretical O are formed? What is the theoretical yield of products?yield of products?

STEP 1STEP 1

Write the balanced chemical Write the balanced chemical equationequation

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

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

STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) --> moles(454 g) --> moles

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3

STEP 3 STEP 3 Convert moles reactant Convert moles reactant (5.68 mol) --> moles product(5.68 mol) --> moles product


STEP 3 STEP 3 Convert moles reactant --> moles Convert moles reactant --> moles productproduct

Relate moles NHRelate moles NH44NONO33 to moles product to moles product expected. expected.

1 mol NH1 mol NH44NONO33 --> 2 mol H --> 2 mol H22OO

Express this relation as the Express this relation as the

STOICHIOMETRIC FACTORSTOICHIOMETRIC FACTOR..

2 mol H2O produced1 mol NH4NO3 used

2 mol H2O produced1 mol NH4NO3 used


= 11.4 mol H= 11.4 mol H22O producedO produced

5.68 mol NH4NO3 • 2 mol H2O produced1 mol NH4NO3 used

STEP 3 STEP 3 Convert moles reactant (5.68 Convert moles reactant (5.68 mol) --> moles productmol) --> moles product


11.4 mol H2O • 18.02 g1 mol

= 204 g H2O

STEP 4 STEP 4 Convert moles product Convert moles product (11.4 mol) --> mass product(11.4 mol) --> mass product

Called the Called the THEORETICAL THEORETICAL YIELDYIELD

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!

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product


STEP 5 STEP 5 How much NHow much N22O is formed?O is formed?

Total mass of reactants = total mass of productsTotal mass of reactants = total mass of products

454 g NH454 g NH44NONO33 = ___ g N = ___ g N22O + 204 g HO + 204 g H22OO

mass of Nmass of N22O = 250. gO = 250. g


STEP 6 STEP 6 Calculate the Calculate the percent yieldpercent yield

If you isolated only 131 g of NIf you isolated only 131 g of N22O, what is the O, what is the

percent yield?percent yield?

This compares the This compares the theoreticaltheoretical (250. g) and (250. g) and

actualactual (131 g) yields. (131 g) yields.


% yield = actual yield

theoretical yield • 100%

STEP 6 STEP 6 Calculate the percent yieldCalculate the percent yield

% yield = 131 g250. g

• 100% = 52.4%

PROBLEM: Using 5.00 g of PROBLEM: Using 5.00 g of HH22OO22, what mass of O, what mass of O22 and and of Hof H22O can be obtained?O can be obtained?

PROBLEM: Using 5.00 g of PROBLEM: Using 5.00 g of HH22OO22, what mass of O, what mass of O22 and and of Hof H22O can be obtained?O can be obtained?

2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)

Reaction is catalyzed by MnOReaction is catalyzed by MnO22

Step 1: moles of HStep 1: moles of H22OO22

Step 2: use STOICHIOMETRIC FACTOR to Step 2: use STOICHIOMETRIC FACTOR to calculate moles of Ocalculate moles of O22

Step 3: mass of OStep 3: mass of O22

Reactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANTReactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANT

• In a given reaction, there is not enough of In a given reaction, there is not enough of one reagent to use up the other reagent one reagent to use up the other reagent completely.completely.

• The reagent in short supply The reagent in short supply LIMITSLIMITS the the quantity of product that can be formed.quantity of product that can be formed.

LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

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

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

LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTS

Demo of limiting reactants on Screen 4.7Demo of limiting reactants on Screen 4.7


Rxn 1: Balloon inflates fully, some Zn leftRxn 1: Balloon inflates fully, some Zn left* * More than enough Zn to use up the 0.100 mol HClMore than enough Zn to use up the 0.100 mol HCl

Rxn 2: Balloon inflates fully, no Zn leftRxn 2: Balloon inflates fully, no Zn left* Right amount of each (HCl and Zn)* Right amount of each (HCl and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.Rxn 3: Balloon does not inflate fully, no Zn left.* Not enough Zn to use up 0.100 mol HCl* Not enough Zn to use up 0.100 mol HCl


React solid Zn with 0.100 React solid Zn with 0.100 mol HCl (aq)mol HCl (aq)Zn + 2 HCl ---> ZnClZn + 2 HCl ---> ZnCl22 + H + H22

1 2 3

(See CD Screen 4.8)(See CD Screen 4.8)


Rxn 1Rxn 1 Rxn 2Rxn 2 Rxn 3Rxn 3

mass Zn (g)mass Zn (g) 7.007.00 3.273.27 1.311.31

mol Znmol Zn 0.1070.107 0.0500.050 0.0200.020

mol HClmol HCl 0.1000.100 0.1000.100 0.1000.100

mol HCl/mol Znmol HCl/mol Zn 0.93/10.93/1 2.00/12.00/1 5.00/15.00/1

Lim ReactantLim Reactant LR = HClLR = HCl no LRno LR LR = ZnLR = Zn


React solid Zn with 0.100 React solid Zn with 0.100 mol HCl (aq)mol HCl (aq)

Zn + 2 HCl ---> ZnClZn + 2 HCl ---> ZnCl22 + H + H2 2

Reaction to be StudiedReaction to be StudiedReaction to be StudiedReaction to be Studied

2 Al + 3 Cl2 Al + 3 Cl22 ---> Al ---> Al22ClCl66

PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 g of Mix 5.40 g of Al with 8.10 g of ClCl22. What mass of Al. What mass of Al22ClCl66 can form? can form?PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 g of Mix 5.40 g of Al with 8.10 g of ClCl22. What mass of Al. What mass of Al22ClCl66 can form? can form?

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product

Determine the formula of a Determine the formula of a compound of Sn and I using compound of Sn and I using the following data.the following data.

Determine the formula of a Determine the formula of a compound of Sn and I using compound of Sn and I using the following data.the following data.

• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.

• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g

• Mass of iodine (IMass of iodine (I22) used = 1.947 g) used = 1.947 g

• Mass of Sn remaining Mass of Sn remaining = 0.601 g= 0.601 g

Find the Find the mass of Sn that combinedmass of Sn that combined with with 1.947 g I1.947 g I22..

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 gMass of Sn used = 0.455 g

Find Find moles of Sn usedmoles of Sn used::

0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •

1 mol118.7 g

= 3.83 x 10-3 mol Sn

Tin and Iodine Tin and Iodine CompoundCompound


Now find the Now find the number of moles of Inumber of moles of I22 that that combined with 3.83 x 10combined with 3.83 x 10-3-3 mol Sn. Mass mol Sn. Mass of Iof I22 used was 1.947 g. used was 1.947 g.

1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2

How many moles of How many moles of iodine atomsiodine atoms? ?

= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2


Now find the ratio of number of moles of moles Now find the ratio of number of moles of moles of I and Sn that combined.of I and Sn that combined.

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

Empirical formula is Empirical formula is

SnISnI44

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STOICHIOMETRY Chapter 3 - the study of the quantitative aspects of chemical reactions