1 stoichiometrystoichiometry determiningformulas

11

STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRY

DetermininDeterminingg

FormulasFormulas

22

Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula

Burn 0.115 g of a hydrocarbon, CBurn 0.115 g of a hydrocarbon, CxxHHyy, and , and

produce produce 0.379 g of CO0.379 g of CO22 and and 0.1035 g of 0.1035 g of

HH22OO. .

CCxxHHy y + some oxygen ---> + some oxygen --->

0.379 g CO0.379 g CO22 + + 0.1035 g H0.1035 g H22OO

What is the empirical formula of CWhat is the empirical formula of CxxHHyy??

33


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in and all H in

HH22O is from CO is from CxxHHyy..

CCxxHHy y + some oxygen ---> CO + some oxygen ---> CO22 + H + H22OO

0.379 g0.379 g 0.1035 g0.1035 g

Puddle of CxHy

0.115 g

0.379 g CO0.379 g CO22

+O2

+O2

0.1035 g H2O1 H2O molecule forms for each 2 H atoms in CxHy

1 CO2 molecule forms for each C atom in CxHy

44


First, recognize that all C in COFirst, recognize that all C in CO22 and all H in H and all H in H22O O

is from Cis from CxxHHyy..

1. Calculate amount of C in CO1. Calculate amount of C in CO22

8.61 x 108.61 x 10-3 -3 mol COmol CO22 --> 8.61 x 10 --> 8.61 x 10-3 -3 mol Cmol C

2. Calculate amount of H in H2. Calculate amount of H in H22OO

5.744 x 105.744 x 10-3-3 mol H mol H22O -- >1.149 x 10O -- >1.149 x 10-2 -2 mol Hmol H


0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

55


Now find ratio of mol H/mol C to find values of x and y in Now find ratio of mol H/mol C to find values of x and y in CCxxHHyy..

1.149 x 10 1.149 x 10 -2 -2 mol Hmol H/ / 8.61 x 108.61 x 10-3 -3 mol Cmol C

= = 1.33 mol H1.33 mol H / / 1.00 mol C1.00 mol C

= = 4 mol H4 mol H / / 3 mol C3 mol C

Empirical formula = CEmpirical formula = C33HH44


0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

66

STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRY

LIMITINGLIMITING

REACTANTSREACTANTS

77

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

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

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

88

LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

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

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

99

STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product

1010

Rxn 1: Zn remainingRxn 1: Zn remaining

* * More than enough Zn to use up the 0.100 More than enough Zn to use up the 0.100

mol HCmol HC

Rxn 2: no Zn leftRxn 2: no Zn left* Not enough Zn to use up 0.100 mol HCl* Not enough Zn to use up 0.100 mol HCl

LIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTSLIMITING REACTANTS

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

Zn + 2 HCl ---> ZnClZn + 2 HCl ---> ZnCl22 + H + H22

1111

PROBLEM: PROBLEM: 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 form? can form?PROBLEM: PROBLEM: 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 form? can form?

Mass reactant

Coefficients

Mole ratio

Molesreactant

Moles product

Mass product

1212

Step 1 of LR problem: Step 1 of LR problem: compare compare actualactual mole mole ratio of reactants to ratio of reactants to theoreticaltheoretical mole mole ratio.ratio.

Step 1 of LR problem: Step 1 of LR problem: compare compare actualactual mole mole ratio of reactants to ratio of reactants to theoreticaltheoretical mole mole ratio.ratio.

1313

2 Al + 3 Cl2 Al + 3 Cl22 ---> 2AlCl ---> 2AlCl33

Reactants must be in the mole ratioReactants must be in the mole ratio

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

mol Cl2mol Al

= 32

mol Cl2mol Al

= 32

1414

Deciding on the Limiting Deciding on the Limiting ReactantReactant


IfIf

There is not enough Al to use up all There is not enough Al to use up all

the Clthe Cl22

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

mol Cl2mol Al

> 32

Lim. reagent = AlLim. reagent = Al

1515

IfIf

There is not enough ClThere is not enough Cl22 to use to use

up all the Alup all the Al


mol Cl2mol Al

< 32

Lim reagent = ClLim reagent = Cl22



1616

We have 5.40 g of Al and 8.10 g of ClWe have 5.40 g of Al and 8.10 g of Cl22

Step 2 of LR problem: Calculate Step 2 of LR problem: Calculate moles of each reactantmoles of each reactantStep 2 of LR problem: Calculate Step 2 of LR problem: Calculate moles of each reactantmoles of each reactant

5.40 g Al • 1 mol27.0 g

= 0.200 mol Al

8.10 g Cl2 • 1 mol70.9 g

= 0.114 mol Cl2

1717Find mole ratio of Find mole ratio of reactantsreactants

Find mole ratio of Find mole ratio of reactantsreactants

This This should be 3/2 or 1.5/1 if should be 3/2 or 1.5/1 if reactants are present in the reactants are present in the exact stoichiometric ratio.exact stoichiometric ratio.

Limiting reagent is Limiting reagent is ClCl22

mol Cl2mol Al

= 0.114 mol 0.200 mol

= 0.57

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

1818Mix 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 . What mass of

AlClAlCl33 can form? can form?

Limiting reactant = ClLimiting reactant = Cl22

Base all calcs. on ClBase all calcs. on Cl22

Limiting reactant = ClLimiting reactant = Cl22

Base all calcs. on ClBase all calcs. on Cl22

molesCl2

moles AlCl3

gramsCl2

grams AlCl3

2

3

Cl mol 3

ClAl 2mol


1919

CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected. expected.CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected. expected.

Step 1: Step 1: Calculate moles of AlClCalculate moles of AlCl33 expected expected

based on LR.based on LR.

32

32 AlCl mol 0.0760 =

Cl mol 3

AlCl 2 • Cl 114.0mol

mol

33

3 AlCl g 10.1 = mol

AlCl g 33.21 • AlCl 0760.0 mol

Step 2: Step 2: Calculate mass of AlClCalculate mass of AlCl33 expected expected

based on LR.based on LR.

2020

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

• Therefore, Al was present Therefore, Al was present

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

• First find how much Al was required.First find how much Al was required.

• Then find how much Al Then find how much Al is in excess.is 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?

2121

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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