Chapter1 2 Dian

8/12/2019 Chapter1 2 Dian

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Stoichiometry (2)


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Empirical & MolecularEmpirical & Molecular

FormulasFormulasA pure compound always consists of theA pure compound always consists of the

same elements combined in the samesame elements combined in the same

proportions by weight.proportions by weight.

Therefore, we can express the molecularTherefore, we can express the molecular

composition ascomposition as PERCENT BY WEIGHTPERCENT BY WEIGHT

Ethanol, CEthanol, C22HH66OO

52.13% C52.13% C13.15% H13.15% H

34.72% O34.72% O


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Percent CompositionPercent CompositionPercent CompositionPercent Composition

Consider 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 gx 100 % 69 .6%

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

46 .0 gx 100 % 69 .6%

Wt. % N =

14.0 g N

46.0 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?


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Determining FormulasDetermining FormulasDetermining FormulasDetermining Formulas

InIn chemical analysischemical analysis we determinewe determinethe % by weight of each element in athe % by weight of each element in a

given amount of pure compound andgiven amount of pure compound and

derive thederive the EMPIRICALEMPIRICAL oror SIMPLESTSIMPLESTformula.formula.

PROBLEMPROBLEM: A compound of B and H is: A compound of B and H is81.10% B. What is its empirical81.10% B. What is its empirical

formula?formula?


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Because it contains only B and H, it mustBecause it contains only B and H, it must

contain 18.90% H.contain 18.90% H.

In 100.0 g of the compound there areIn 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 theCalculate the number of molesnumber of moles of eachof each

constitutent.constitutent.

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


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Calculate theCalculate the number of molesnumber of moles of eachof 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 B

18.90 g H 1 mol

1.008 g = 18.75 mol H



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Now, recognize thatNow, recognize that atoms combine in theatoms combine in theratio of small whole numbersratio of small whole numbers

Find theFind the ratio of molesratio of moles of elements in theof elements in the

compound.compound.



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But we need aBut 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 H

7.502 mol B =

2.499 mol H

1.000 mol B =

2.5 mol H

1.0 mol B



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A compound of B and H is 81.10% B.A compound of B and H is 81.10% B. ItsItsempirical formulaempirical formula is Bis B22HH55. What is its. What is its

molecular formulamolecular formula ??Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, B, B66HH1515,,

BB88HH2020, etc.?, etc.?

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

B2H6


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ow to etermine t eow to etermine t emolar mass?molar mass?

Mass spectrometerMass spectrometer


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

BB22HH55. What is its molecular formula. What is its molecular formula??A compound of B and H is 81.10% B. Its empirical formula isA compound of B and H is 81.10% B. Its empirical formula is

BB22HH55. What is its molecular formula. What is its molecular formula??

We need to do anWe need to do an EXPERIMENTEXPERIMENTto findto findthe MOLAR MASS.the MOLAR MASS.

Here experiment givesHere 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/unitFind the ratio of these masses.Find the ratio of these masses.

53.3 g/mol

26.66 g/unit of B 2H5

=2 units of B 2H5

1 mol

Molecular formula = BMolecular formula = B44HH1010


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

CHEMICAL COMPOUNDSCHEMICAL COMPOUNDS are formed when atoms of different elementsare formed when atoms of different elementscombine with each other: A given compound always has the samecombine with each other: A given compound always has the samerelative numbers and types of atomsrelative numbers and types of atoms

Law of definite proportion [Joseph Proust, 1754Law of definite proportion [Joseph Proust, 1754--1826]1826]A given compound always contains the sameA given compound always contains the same

proportion of elements by massproportion of elements by mass

Law of multiple proportions [John Dalton, 1766Law of multiple proportions [John Dalton, 1766--1844]1844]When 2 elements form multiple compounds, the mass of the secondWhen 2 elements form multiple compounds, the mass of the second

element per gram of the first one can always be reduced to small wholeelement per gram of the first one can always be reduced to small whole

numbersnumbers

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

Mass is neither created nor destroyedMass is neither created nor destroyed

in a chemical reactionin a chemical reaction


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Chemical EquationsChemical EquationsChemical EquationsChemical Equations

Because the sameBecause the same

atoms are present in aatoms are present in a

reaction at the beginningreaction at the beginning

and at the end, theand at the end, theamount of matter in aamount of matter in a

system does not change.system does not change.

TheThe Law of theLaw of the

Conservation of MatterConservation of Matter

Demo of conservation ofmatter, See Screen 4.3.


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Because of the principle of theBecause of the principle of theconservation of matterconservation of matter,,

anan equation must beequation must bebalancedbalanced..

It must have the sameIt must have the same

number of atoms of thenumber of atoms of the

same kind onsame kind on

both sides.both sides.

Chemical EquationsChemical Equations

Lavoisier, 1788Lavoisier, 1788


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BalancingBalancingEquationsEquationsBalancingBalancingEquationsEquations

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


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Chemical EquationsChemical EquationsDepict the kind ofDepict the kind of reactantsreactants andand productsproducts and theirand their

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

2 Al(s) + 3 Br2 Al(s) + 3 Br22(g)(g) ------> Al> 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 ofThe letters (s), (g), and (l) are the physical states ofcompounds.compounds.


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Chemical EquationsChemical EquationsChemical EquationsChemical Equations

4 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 moleculesmolecules------givegive------>>

2 molecules of Al2 molecules of Al22OO33

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

2 moles of Al2 moles of Al22OO33


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STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRY

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

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


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PROBLEM:PROBLEM:If 454 g of NHIf 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O andO and

HH22O are formed? What is the theoretical yield ofO are formed? What is the theoretical yield ofproducts?products?

PROBLEM:PROBLEM:If 454 g of NHIf 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O andO and

HH22O are formed? What is the theoretical yield ofO are formed? What is the theoretical yield ofproducts?products?

STEP 1STEP 1

Write the balanced chemicalWrite the balanced chemicalequationequation

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


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454 g of NH454 g of NH44NONO33 ----> N> N22O + 2 HO + 2 H22OO454 g of NH454 g of NH44NONO33 ----> N> N22O + 2 HO + 2 H22OO

STEP 2STEP 2 Convert mass reactantConvert mass reactant

(454 g)(454 g) ----> moles> moles

454 g

1 mol

80.04 g = 5.68 mol NH4NO3

STEP 3STEP 3 Convert moles reactant (5.68Convert moles reactant (5.68

mol)mol) ----> moles product> moles product


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STEP 3STEP 3 Convert moles reactantConvert moles reactant ----> moles> moles

productproduct

Relate moles NHRelate moles NH44NONO33 to moles productto moles product

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

Express this relation as theExpress this relation as the

STOICHIOMETRIC FACTORSTOICHIOMETRIC FACTOR..

2 mol H2O produced

1 mol NH4NO3 used


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= 11.4 mol H= 11.4 mol H22O producedO produced

5.68 mol NH4NO

3

2 mol H2O produced

1 mol NH4NO3used

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


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11.4 mol H2O 18.02 g

1 mol= 204 g H2O

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

Called theCalled the

THEORETICAL YIELDTHEORETICAL YIELD

ALWAYS FOLLOW THESE STEPS IN SOLVINGALWAYS FOLLOW THESE STEPS IN SOLVINGSTOICHIOMETRY PROBLEMS!STOICHIOMETRY PROBLEMS!


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GENERAL PLAN FORGENERAL PLAN FORSTOICHIOMETRYSTOICHIOMETRY

CALCULATIONSCALCULATIONS

GENERAL PLAN FORGENERAL PLAN FORSTOICHIOMETRYSTOICHIOMETRY

CALCULATIONSCALCULATIONS

Mass

reactant

Stoichiometric

factorMoles

reactant

Moles

product

Mass

product


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

Total mass of reactants = total mass ofTotal mass of reactants = total mass of

productsproducts

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

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


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STEP 6STEP 6 Calculate theCalculate the percent yieldpercent yield

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

percent yield?percent yield?

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

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


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% yield =actual yield

theoretical yield

100%

STEP 6STEP 6 Calculate the percent yieldCalculate the percent yield

% yield =131 g

250. g 100% = 52.4%


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PROBLEM: Using 5.00PROBLEM: Using 5.00g of Hg of H22OO22, what mass, what mass

of Oof O22 and of Hand of H22O canO canbe obtained?be obtained?

PROBLEM: Using 5.00PROBLEM: Using 5.00g of Hg of H22OO22, what mass, what mass

of Oof O22 and of Hand of H22O canO canbe obtained?be obtained?2 H2 H22OO22(liq)(liq) ------> 2 H> 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 FACTORStep 2: use STOICHIOMETRIC FACTOR

to calculate moles of Oto calculate moles of O22

Step 3: mass of OStep 3: mass of O22


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

In a given reaction, there is not enough ofIn a given reaction, there is not enough of

one reagent to use up the other reagentone reagent to use up the other reagent

completely.completely.

The reagent in short supplyThe reagent in short supply LIMITSLIMITS thethequantity of product that can be formed.quantity of product that can be formed.


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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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LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTS

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


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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 mol HClReact solid Zn with 0.100 mol HCl

(aq)(aq)

Zn + 2 HClZn + 2 HCl ------> ZnCl> ZnCl22 + H+ H22

1 2 3

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


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

Zn + 2 HClZn + 2 HCl ------> ZnCl> ZnCl22 + H+ H22


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

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


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PROBLEM:PROBLEM: Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22..What mass of AlWhat mass of Al22ClCl66 can form?can form?PROBLEM:PROBLEM: Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22..What mass of AlWhat mass of Al22ClCl66 can form?can form?

Mass

reactant

Stoichiometric

factorMoles

reactant

Moles

product

Mass

product


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Determine the formula of aDetermine the formula of acompound of Sn and I using thecompound of Sn and I using the

following data.following data.

Determine the formula of aDetermine the formula of acompound of Sn and I using thecompound of Sn and I using the

following data.following data.

Reaction of Sn and IReaction of Sn and I22 is done using excessis done using excess

Sn.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 remainingMass of Sn remaining = 0.601 g= 0.601 g


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Find theFind the mass of Sn that combinedmass of Sn that combined withwith

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

FindFind moles of Sn usedmoles of Sn used::

0.455 g Sn

1 mol

118.7 g = 3.83 x 10-3

mol Sn

Tin and Iodine CompoundTin and Iodine Compound


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Now find theNow find the number of moles of Inumber of moles of I22 thatthat

combined with 3.83 x 10combined with 3.83 x 10--33 mol Sn. Massmol 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 I2

How many moles ofHow 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 I 2


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Now find the ratio of number of moles ofNow find the ratio of number of moles of

moles of I and Sn that combined.moles of I and Sn that combined.

1.534 x 10 -2 mol I

3.83 x 10-3

mol Sn

=4.01 mol I

1.00 mol Sn

Empirical formula isEmpirical formula is SnISnI44

Chapter1 2 Dian

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