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Lecture Presentation
Chapter 3
Stoichiometry:
Calculations with
Chemical Formulas
and Equations
John D. Bookstaver
St. Charles Community College
Cottleville, MO
© 2012 Pearson Education, Inc.
Law of Conservation of Mass
“We may lay it down as
an incontestable axiom
that, in all the operations
of art and nature, nothing
is created; an equal
amount of matter exists
both before and after the
experiment. Upon this
principle, the whole art of
performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
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Stoichiometry
Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Stoichiometry
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Anatomy of a Chemical Equation
CH4(g) + 2O2(g)
Reactants appear
on the left side of the
equation.
CO2(g) + 2H2O(g)
Products appear on
the right side of the
equation.
Stoichiometry
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Anatomy of a Chemical Equation
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The states of the reactants and products are
written in parentheses to the right of each
compound.
Stoichiometry
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Anatomy of a Chemical Equation
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Coefficients are inserted to balance the
equation.
Stoichiometry
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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
Sample Exercise 3.1 Interpreting and Balancing Chemical
Equations
Practice
In the following diagram, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen
atoms.
To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right
(products) box?
Stoichiometry
Reaction Types
Stoichiometry
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Combination Reactions (Synthesis)
• In combination
reactions two or
more substances
react to form one
product.
• Examples:
– 2Mg(s) + O2(g)  2MgO(s)
– N2(g) + 3H2(g)  2NH3(g)
– C3H6(g) + Br2(l)  C3H6Br2(l)
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Stoichiometry
Decomposition Reactions
• In a decomposition
reaction one
substance breaks
down into two or
more substances.
• Examples:
– CaCO3(s)  CaO(s) + CO2(g)
– 2KClO3(s)  2KCl(s) + O2(g)
– 2NaN3(s)  2Na(s) + 3N2(g)
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Stoichiometry
Sample Exercise 3.3 Writing Balanced Equations for Combination
and Decomposition Reactions
Write a balanced equation for (a) solid mercury(II) sulfide
decomposing into its component elements when heated and (b)
aluminum metal combining with oxygen in the air.
Stoichiometry
Combustion Reactions
• Combustion reactions
are generally rapid
reactions that produce
a flame.
• Combustion reactions
most often involve
hydrocarbons reacting
with oxygen in the air.
• Examples:
– CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
– 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g)
Stoichiometry
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Sample Exercise 3.4 Writing Balanced Equations for Combustion
Reactions
Write the balanced equation for the reaction that occurs when
ethanol, C2H5OH(l), burns in air.
Stoichiometry
Formula
Weights
Stoichiometry
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Formula Weight (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(40.08 amu)
+ Cl: 2(35.45 amu)
110.98 amu
• Formula weights are generally reported for
ionic compounds.
Stoichiometry
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Molecular Weight (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(12.01 amu)
+ H: 6(1.01 amu)
30.08 amu
Stoichiometry
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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:
(number of atoms)(atomic weight)
% Element =
(FW of the compound)
x 100
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane is
Stoichiometry
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Moles
Stoichiometry
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Avogadro’s Number
• 6.022 x 1023
• 1 mole of 12C has a
mass of 12.000 g.
Stoichiometry
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Sample Exercise 3.7 Estimating Numbers of Atoms
Without using a calculator, arrange these samples in order of
increasing number of O atoms: 1 mol H2O,
1 mol CO2, 3  1023 molecules O3.
Stoichiometry
Sample Exercise 3.8 Converting Moles to Number of Atoms
How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b)
1.50 mol of sodium carbonate?
Stoichiometry
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).
Stoichiometry
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Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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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.
Stoichiometry
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Calculations
• Mass, in grams, of 1.50x10-2 mol CdS?
• Mass, in grams, of 1.50x1021 molecules of
aspirin, C9H8O4?
Stoichiometry
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Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms
from Mass
How many nitric acid molecules are in 4.20 g of HNO3? (b)
How many O atoms are in this sample?
Stoichiometry
Finding
Empirical
Formulas
Stoichiometry
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Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
Stoichiometry
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Sample Exercise 3.14 Determining a Molecular Formula
Ethylene glycol, used in automobile antifreeze, is 38.7% C,
9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol.
(a) What is the empirical formula of ethylene glycol? (b) What
is its molecular formula?
Stoichiometry
Combustion Analysis
• Compounds containing C, H, and O are routinely
analyzed through combustion in a chamber like the
one shown in
Figure 3.14.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
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Stoichiometry
Sample Exercise 3.15 Determining an Empirical Formula by
Combustion Analysis
(a) Caproic acid, responsible for the odor of dirty socks, is
composed of C, H, and O atoms. Combustion of
a 0.225-g sample of this compound produces 0.512 g CO2 and
0.209 g H2O.What is the empirical formula
of caproic acid? (b) Caproic acid has a molar mass of 116
g/mol. What is its molecular formula?
Stoichiometry
Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products.
– O is determined by difference after the C and H
have been determined.
Stoichiometry
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Stoichiometric Calculations
Starting with the
mass of Substance
A, you can use
the ratio of the
coefficients of A and
B to calculate the
mass of Substance
B formed (if it’s a
product) or used (if
it’s a reactant).
Stoichiometry
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Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
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Stoichiometry
Calculations
• 4 KO2 + 2 CO2  K2CO3 + 3O2
• How many moles of O2 are produced when
0.400 mol of KO2 reacts?
• How many grams of KO2 are needed to form
7.50 g of O2?
• How many grams of CO2 are used when 7.50
g of O2 are produced?
Stoichiometry
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Limiting
Reactants
Stoichiometry
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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).
Stoichiometry
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Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
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Calculations
• 2 Al(OH)3(s) + 3 H2SO4(aq)  Al2(SO4)3(aq) + 6 H2O(l)
• Which is the limiting reactant when 0.500 mol
Al(OH)3 and 0.500 mol H2SO4 react?
• How many moles of Al2(SO4)3 can form?
• How many moles of the excess reactant
remain after the completion of the reaction?
Stoichiometry
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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
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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):
Percent yield =
actual yield
theoretical yield
x 100
Stoichiometry
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Calculations
• SA
AcAn
Asp
AcA
• C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
• What is the theoretical yield of Asp if
185 kg of SA reacts 125 kg of AcAn?
• What is the percentage yield if the the
reactions produces 182 kg of Asp?
Stoichiometry
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