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1
Stoichiometry
© 2009, Prentice-Hall, Inc.
Chapter 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
John D. Bookstaver St. Charles Community College
Cottleville, MO
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten
Stoichiometry
© 2009, Prentice-Hall, Inc.
Chapter 3 Problems
• Problems 11, 13, 19, 21, 23, 33, 45, 51,
59, 73
Stoichiometry
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Chemical Equations
Depict the kind of reactants and
products and their relative amounts
in a reaction.
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
The numbers in the front are called
stoichiometric coefficients The letters (s), (g), and (l) are the
physical states of compounds.
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2
Stoichiometry
© 2009, Prentice-Hall, Inc.
Reaction of Iron with Cl2
Notice the stoichiometric coefficients and the
physical states of the reactants and products.
Stoichiometry
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Chemical Equations 4 Al(s) + 3 O2(g)
---> 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
---give--->
2 “molecules” of Al2O3
4 moles of Al + 3 moles of O2
---give--->
2 moles of Al2O3
Stoichiometry
© 2009, Prentice-Hall, Inc.
Chemical Equations • Because the same atoms are present in a
reaction at the beginning and at the end, the
amount of matter in a system does not
change.
• The Law of the Conservation of Matter
2HgO(s) ---> 2 Hg(liq) + O2(g)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Because of the principle of the
conservation of matter,
an equation must be balanced.
It must have the same
number of atoms of the
same kind on both sides.
Chemical Equations
Lavoisier, 1788
Stoichiometry
© 2009, Prentice-Hall, Inc.
Because of the principle of the
conservation of matter,
an equation must be balanced.
It must have the same
number of atoms of the
same kind on both sides.
Chemical Equations
Lavoisier, 1788
Stoichiometry
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
Reactants appear on the left
side of the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Stoichiometry
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
Products appear on the
right side of the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Stoichiometry
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
The states of the reactants and products are written in parentheses to the right of each compound.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
Coefficients are inserted
to balance the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Stoichiometry
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Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of
molecules.
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Chemical Reaction
Nitrogen monoxide + oxygen → nitrogen dioxide
Step 1: Write the reaction using chemical symbols.
NO + O2 → NO2
Step 2: Balance the chemical equation.
2 1 2
Stoichiometry
© 2009, Prentice-Hall, Inc.
Writing and Balancing
Chemical Equations
• Write a word equation.
• Convert word equation into formula
equation.
• Balance the formula equation by the
use of prefixes (coefficients) to balance
the number of each type of atom on the
reactant and product sides of the
equation.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Example
Hydrogen gas reacts with oxygen gas to
produce water.
Step 1.
hydrogen + oxygen -----> water
Step 2.
H2 + O2 -----> H2O
Step 3.
2 H2 + O2 -----> 2 H2O
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Balancing Equation Strategy
• Balance elements that occur in only one
compound on each side first.
• Balance free elements last.
• Balance unchanged polyatomics as
groups.
• Fractional coefficients are acceptable
and can be cleared at the end by
multiplication.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Example
Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide.
iron(III) oxide + carbon monoxide -----> Fe3O4 + carbon
dioxide
Fe2O3 + CO -----> Fe3O4 + CO2
3 Fe2O3 + CO -----> 2 Fe3O4 + CO2
Stoichiometry
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Balancing Equations
____C3H8(g) + _____ O2(g) ---->
_____CO2(g) + _____ H2O(g)
____B4H10(g) + _____ O2(g) ---->
___ B2O3(g) + _____ H2O(g)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Reaction
Types
Stoichiometry
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Combination Reactions
• Examples:
– 2 Mg (s) + O2 (g) 2 MgO (s)
– N2 (g) + 3 H2 (g) 2 NH3 (g)
– C3H6 (g) + Br2 (l) C3H6Br2 (l)
• In this type of
reaction two
or more
substances
react to form
one product.
Stoichiometry
© 2009, Prentice-Hall, Inc.
• In a decomposition
one substance breaks
down into two or more
substances.
Decomposition Reactions
• Examples:
– CaCO3 (s) CaO (s) + CO2 (g)
– 2 KClO3 (s) 2 KCl (s) + O2 (g)
– 2 NaN3 (s) 2 Na (s) + 3 N2 (g)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Combustion Reactions
• Examples:
– CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
– C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
• These are generally
rapid reactions that
produce a flame.
• Most often involve
hydrocarbons
reacting with oxygen
in the air.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Writing and Balancing an Equation: The Combustion of a
Carbon-Hydrogen-Oxygen Compound.
Liquid triethylene glycol, C6H14O4, is used a a solvent and
plasticizer for vinyl and polyurethane plastics. Write a
balanced chemical equation for its complete combustion.
Stoichiometry
© 2009, Prentice-Hall, Inc.
15 2
6 7 C6H14O4 + O2 → CO2 + H2O 6
2. Balance H.
2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O
4. Multiply by two
Example 4-2
3. Balance O.
and check all elements.
Chemical Equation:
1. Balance C.
6 7
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Formula Weight/Molecular
Weight
Stoichiometry
© 2009, Prentice-Hall, Inc.
Formula Weight (Mass) (FW)
• A formula weight is the sum of the
atomic weights for the atoms in a
chemical formula.
• So, the formula weight of calcium
chloride, CaCl2, would be Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported
for ionic compounds.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Molecular Weight (Mass) (MW)
• A molecular weight is the sum of the
atomic weights of the atoms in a
molecule.
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
30.0 amu
+ H: 6(1.0 amu)
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Molecular Weight (Mass)
HOO
H
H
HO
H
OH
OHHH
OH
Molecular formula C6H12O6
Empirical formula CH2O
Glucose
6 x 12.01 + 12 x 1.01 + 6 x 16.00
Molecular Mass: Use the naturally occurring mixture of isotopes,
= 180.18
Stoichiometry
© 2009, Prentice-Hall, Inc.
Percent Composition
One can find the percentage of the mass
of a compound that comes from each of
the elements in the compound by using
this equation:
% element = (number of atoms)(atomic mass)
(FM of the compound) x 100
Stoichiometry
© 2009, Prentice-Hall, Inc.
Percent Composition
So the percentage of carbon in ethane
is…
%C = (2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu = x 100
= 80.0%
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Stoichiometry
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Moles
Stoichiometry
© 2009, Prentice-Hall, Inc.
Counting Atoms
Mg burns in air (O2) to
produce white magnesium
oxide, MgO.
How can we figure out how
much oxide is produced
from a given mass of Mg?
Stoichiometry
© 2009, Prentice-Hall, Inc.
Counting Atoms
Chemistry is a quantitative
science—we need a
“counting unit.”
1 mole is the amount of
substance that contains as
many particles (atoms,
molecules) as there are in
12.0 g of 12C.
MOLE
518 g of Pb, 2.50 mol
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Particles in a Mole
6.02214199 x 1023
Avogadro’s Number
There is Avogadro’s number of
particles in a mole of any substance.
Amedeo Avogadro
1776-1856
Stoichiometry
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Molar Mass 1 mol of 12C
= 12.00 g of C = 6.022 x 1023 atoms of C
12.00 g of 12C is its
MOLAR MASS
Taking into account all of
the isotopes of C, the
molar mass of C is
12.011 g/mol
Stoichiometry
© 2009, Prentice-Hall, Inc.
Molar Mass
• By definition, a molar mass is the mass
of 1 mol of a substance (i.e., g/mol).
– The molar mass of an element is the mass
number for the element that we find on the
periodic table.
– The formula weight (in amu’s) will be the
same number as the molar mass (in
g/mol).
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Stoichiometry
© 2009, Prentice-Hall, Inc.
One-mole Amounts
Stoichiometry
© 2009, Prentice-Hall, Inc.
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
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Stoichiometry
© 2009, Prentice-Hall, Inc.
PROBLEM: What amount of
Mg is represented by 0.200 g?
How 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 1021 atoms Mg
How many atoms in this piece of Mg?
Stoichiometry
© 2009, Prentice-Hall, Inc.
Finding
Empirical
Formulas
Stoichiometry
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Empirical formula
1. Choose an arbitrary sample size (100g).
2. Convert masses to amounts in moles.
3. Write a formula.
4. Convert formula to small whole numbers.
5. Multiply all subscripts by a small whole
number to make the subscripts integral.
5 Step approach:
Stoichiometry
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C: 61.31 g x = 5.105 mol C
H: 5.14 g x = 5.09 mol H
N: 10.21 g x = 0.7288 mol N
O: 23.33 g x = 1.456 mol O
1 mol
12.01 g
1 mol
14.01 g
1 mol
1.01 g
1 mol
16.00 g
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C: = 7.005 7
H: = 6.984 7
N: = 1.000
O: = 2.001 2
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol
0.7288 mol
0.7288 mol
1.458 mol
0.7288 mol
Stoichiometry
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
© 2009, Prentice-Hall, Inc.
Determining the Empirical and Molecular Formulas of a
Compound from Its Mass Percent Composition.
Dibutyl succinate is an insect repellent used against household
ants and roaches. Its composition is 62.58% C, 9.63% H and
27.79% O. Its experimentally determined molecular mass is
230 u. What are the empirical and molecular formulas of
dibutyl succinate?
Step 1: Determine the mass of each element in a 100g sample.
C 62.58 g H 9.63 g O 27.79 g
Dibutyl succinate is an insect repellent used against household
ants and roaches. Its composition is 62.58% C, 9.63% H and
27.79% O. Its experimentally determined molecular mass is
230 u. What are the empirical and molecular formulas of
dibutyl succinate?
Example
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Step 2: Convert masses to amounts in moles.
OmolOg
OmolOgn
HmolHg
HmolHgn
CmolCg
CmolCgn
O
H
C
737.1999.15
179.27
55.9008.1
163.9
210.5011.12
158.62
Step 3: Write a tentative formula.
Step 4: Convert to small whole numbers.
C5.21H9.55O1.74
C2.99H5.49O
Example
Stoichiometry
© 2009, Prentice-Hall, Inc.
Step 5: Convert to a small whole number ratio.
Multiply 2 to get C5.98H10.98O2
The empirical formula is C6H11O2
Step 6: Determine the molecular formula.
Empirical formula mass is 115 u.
Molecular formula mass is 230 u.
The molecular formula is C12H22O4
Example
Stoichiometry
© 2009, Prentice-Hall, Inc.
Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have been
determined.
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Combustion analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empirical Formulas from Analyses
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empirical Formulas from Analyses
Cmol
Cg
COmol
Cmol
COg
COmolg
1
011.12
1
1
011.44
12829.0
22
2
Combustion Analysis
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Carbon
mass CO2 moles CO2 moles C grams C
Empirical Formulas from Analyses
CgCmol
Cg
COmol
Cmol
COg
COmolg 07721.0
1
011.12
1
1
011.44
12829.0
22
2
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empirical Formulas from Analyses
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empirical Formulas from Analyses
OHg
OHmolg
2
2
02.18
11159.0
Combustion Analysis
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empirical Formulas from Analyses
OHmol
Hmol
OHg
OHmolg
22
2
1
2
02.18
11159.0
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empirical Formulas from Analyses
Hmol
Hg
OHmol
Hmol
OHg
OHmolg
1
01.1
1
2
02.18
11159.0
22
2
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Hydrogen
mass H2O moles H2O moles H grams H
Empirical Formulas from Analyses
gHmol
Hg
OHmol
Hmol
OHg
OHmolg 01299.0
1
01.1
1
2
02.18
11159.0
22
2
Combustion Analysis
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Oxygen
mass O = mass of sample – (mass C +mass H)
Empirical Formulas from Analyses
gggOxygenmass 01299.007721.01005.0
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Mass of Oxygen
mass O = mass of sample – (mass C +mass H)
Empirical Formulas from Analyses
g
gggOxygenmass
01030.0
01299.007721.01005.0
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Mass of elements:
C 0.07721g
H 0.01299g
O 0.01030g
Empirical Formulas from Analyses
Combustion Analysis
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol =
H 0.01299g/1.01g/mol =
O 0.01030g/16.00g/mol =
Empirical Formulas from Analyses
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.01286 mol
O 0.01030g/16.00g/mol = 0.0006438 mol
Empirical Formulas from Analyses
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.01286 mol
O 0.01030g/16.00g/mol = 0.0006438mol
Empirical Formulas from Analyses
0006438.001286.0006428.0 OHC
Combustion Analysis
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.01286 mol
O 0.01030g/16.00g/mol = 0.0006438 mol
Empirical Formulas from Analyses
0006438.0
0006438.0
0006438.0
01286.0
0006438.0
006428.0 OHC
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
Menthol, the substance we can smell in mentholated cough drops,
is composed of C, H and O. A 0.1005g sample of menthol is
combusted, producing 0.2829g of CO2 and 0.1159g of H2O. What
is the empirical formula for menthol?
Now we can determine the empirical formula
Moles of elements:
C 0.07721g/12.011g/mol = 0.006428 mol
H 0.01299g/1.01g/mol = 0.01286 mol
O 0.01030g/16.00g/mol = 0.0006438 mol
Empirical Formulas from Analyses
12010 OHC
Combustion Analysis
Stoichiometry
© 2009, Prentice-Hall, Inc.
STOICHIOMETRY
- the study of the quantitative aspects of chemical reactions.
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Chemical Equations and Stoichiometry
• Stoichiometry includes all the
quantitative relationships involving:
– atomic and formula masses
– chemical formulas.
• Mole ratio is a central conversion factor.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products.
Stoichiometry
© 2009, Prentice-Hall, Inc.
GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
Mass
reactant
Stoichiometric
factor Moles
reactant
Moles
product
Mass
product
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Stoichiometry
© 2009, Prentice-Hall, Inc.
PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products?
STEP 1
Write the balanced chemical equation
NH4NO3 ---> N2O + 2 H2O
Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 2 Convert mass reactant (454 g) --> moles
454 g • 1 mol
80.04 g = 5.68 mol NH4NO3
STEP 3 Convert moles reactant
(5.68 mol) --> moles product
Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3 Convert moles reactant --> moles
product
Relate moles NH4NO3 to moles product
expected.
1 mol NH4NO3 --> 2 mol H2O
Express this relation as the
STOICHIOMETRIC FACTOR.
2 mol H2O produced
1 mol NH4NO3 used
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Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
= 11.4 mol H2O produced
5.68 mol NH4NO3 • 2 mol H2O produced
1 mol NH4NO3 used
STEP 3 Convert moles reactant (5.68 mol) --> moles product
Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
11.4 mol H2O • 18.02 g
1 mol = 204 g H2O
STEP 4 Convert moles product (11.4 mol) --> mass product
Called the THEORETICAL YIELD
ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!
Stoichiometry
© 2009, Prentice-Hall, Inc.
Limiting
Reactants
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Stoichiometry
© 2009, Prentice-Hall, Inc.
How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients.
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat).
Stoichiometry
© 2009, Prentice-Hall, Inc.
How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2).
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually
produces and measures.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Theoretical Yield Percent Yield = x 100
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Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 5 How much N2O is formed?
Total mass of reactants =
total mass of products
454 g NH4NO3 = ___ g N2O + 204 g
H2O
mass of N2O = 250. g
Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6
Calculate the percent yield
If you isolated only 131 g of N2O, what
is the percent yield?
This compares the theoretical (250.
g) and actual (131 g) yields.
Stoichiometry
© 2009, Prentice-Hall, Inc.
454 g of NH4NO3 --> N2O + 2 H2O
% yield = actual yield
theoretical yield • 100%
STEP 6 Calculate the percent yield
% yield = 131 g
250. g • 100% = 52.4%
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Stoichiometry
© 2009, Prentice-Hall, Inc.
Determining Limiting Reagent:
Another Example
• The reactant that is completely consumed
determines the quantities of the products
formed.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Determining the Limiting Reactant in a Reaction.
Phosphorus trichloride , PCl3, is a commercially important
compound used in the manufacture of pesticides, gasoline
additives, and a number of other products. It is made by the
direct combination of phosphorus and chlorine
P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)
What mass of PCl3 forms in the reaction of 125 g P4 with
323 g Cl2?
Example
Strategy: Compare the actual mole ratio
to the required mole ratio.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Example
nCl2 = 323 g Cl2 × = 4.56 mol Cl2
1 mol Cl2
70.91 g Cl2
nP4 = 125 g P4 × = 1.01 mol P4
1 mol P4
123.9 g P4
4.56 mol Cl2 x 4PCl3/6 Cl2 = 3.04 mole PCl3
1.01 mol P4 x 4PCl3/1P4 = 4.04 mole PCl3
Chlorine gas is the limiting reagent.