Download Ch 7 lecture notes, part 1

4/1/2014
Chapter 7: Stoichiometry - Mass
Relations in Chemical Reactions

How can we used balanced chemical equations to relate
the quantities of substances consumed and produce in
chemical reactions?
 How can we used balanced chemical equations to relate
the quantities of substances consumed and produce in
chemical reactions?
 How can we determine a compound’s elemental
composition and chemical formula?
→
The law of conservation of mass states that the sum
of the masses of the reactants of a chemical equation
is equal to the sum of the masses of the products.
+
12 g
O2(g)
→
1 mole
1 mole
+
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7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
→
Law of Conservation of Mass
C(s)
Chapter Outline
Law of Conservation of Mass
2 C(s)
+
O2(g)
→
2 CO(g)
CO2(g)
1 mole
32 g
44 g
=
2 mole
2 x 12 = 24 g +
Moles and Chemical Equations: Stoichiometry –
mole relations between reactants and products
1 mole
32 g
2 mole
=
Stoichiometric Coefficients:
the ratios between reactants and products
2
4
Molecular
Ratio
SO3(g)
+
Mole Ratio
1 molecule
+
Mass Ratio
80.06 g
+
18.02 g
→
98.08 g
General Case
(moles)
x moles
+
x moles
→
x moles
General Case
(masses)
x(80.06 g)
+
H2O(g)
2 x 28 = 56 g
→ H2SO4(l)
1 molecule → 1 molecule
x(18.02 g) → x(98.08 g)
4 mol NO2
2 mol N2O5
1 mol O2
2 mol N2O5
2 mol N2O5
4 mol NO2
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Balancing Chemical Equations
Chapter Outline
SO2(g) +
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7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
H2O(l)
→
H2O(l)
→
+
SO2(g) +
H2O(l)
→
2 HNO3(aq)
+
NO(aq)
2 HNO3(aq)
+
NO(aq)
Step 3: Atoms = 3 N + 2 H + 7 O → 3 N + 2 H + 7 O
Combustion Reactions
The reaction of an organic compound
with oxygen to produce CO2 + H2O
CH4(g) +
O2(g)
→
CO2(g)
+
H2O(g)
Step 1: Atoms = 1 C + 4 H + 2 O → 1 C + 2 H + 3 O
CH4(g) +
O2(g)
→
CO2(g)
+
2 H2O(g)
Step 1: Atoms = 1 C + 4 H + 2 O → 1 C + 4 H + 4 O
CH4(g) +
2 O2(g)
→
CO2(g)
O2(g)
→
2 SO3(g)
Step 2: Atoms = 1 S + 4 O → 2 S + 6 O
2 SO2(g) +
O2(g)
→
2 SO3(g)
Step 3: Atoms = 2 S + 6 O → 2 S + 6 O
NO(aq)
Step 2: Atoms = 1 N + 2 H + 3 O → 3 N + 2 H + 7 O
3 NO2(g) +
SO3(g)
Chapter Outline
HNO3(aq)
Step 1: Atoms = 1 N + 2 H + 3 O → 2 N + 1 H + 4 O
NO2(g) +
→
Step 1: Atoms = 1 S + 4 O → 1 S + 3 O
A more challenging example:
NO2(g) +
O2(g)
+
2 H2O(g)
Step 1: Atoms = 1 C + 4 H + 4 O → 1 C + 4 H + 4 O
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7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Procedure for Balancing
Combustion Reactions
C2H6 (ethane) reacts with oxygen to form CO2 and H2O
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
C 2 H 6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2 C2H6 NOT C4H12
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Procedure for Balancing
Combustion Reactions
Procedure for Balancing
Combustion Reactions
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbons
on left
C2H6 + O2
multiply CO2 by 2
2 CO2 + H2O
6 hydrogens
on left
2 hydrogens
on right
C2H6 + O2
multiply H2O by 3
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by
7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
2 CO2 + 3 H2O
Procedure for Balancing
Combustion Reactions
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
The reaction of CH4 with O2
actually follows a series of
steps due to incomplete
combustion:
4CO2 + 6H2O
Step 1: CO is the product when O2 is deficient:
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
Step 2: the H2 is combusted to form H2O
Step 3: the CO reacts further to form CO2
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The individual steps can be added together
to get the full reaction -
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