1 stoichiometrystoichiometry determiningformulas
TRANSCRIPT
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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??
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Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula
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
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Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula
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
CCxxHHy y + some oxygen ---> + some oxygen --->
0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO
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Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula
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
CCxxHHy y + some oxygen ---> + some oxygen --->
0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO
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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.
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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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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
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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
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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.
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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
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Deciding on the Limiting Deciding on the Limiting ReactantReactant
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
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IfIf
There is not enough ClThere is not enough Cl22 to use to use
up all the Alup all the Al
2 Al + 3 Cl2 Al + 3 Cl22 ---> 2AlCl ---> 2AlCl33
mol Cl2mol Al
< 32
Lim reagent = ClLim reagent = Cl22
Deciding on the Limiting Deciding on the Limiting ReactantReactant
Deciding on the Limiting Deciding on the Limiting ReactantReactant
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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
2 Al + 3 Cl2 Al + 3 Cl22 ---> 2AlCl ---> 2AlCl33
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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.
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• 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?
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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