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Lecture #8
Balancing Chemical Equations:
The Atom Conservation Approach
Chemistry 142 B
James B. Callis, Instructor
Autumn Quarter, 2004
Chemical Equations
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
But also Quantitative Information!
2 H2O (g)
Information Contained in a Balanced Equation
Viewed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules
Amount (mol)
Mass (amu)
Mass (g)
Total Mass (g)
Balancing Chemical Equations by Explicit
Consideration of the Atom Conservation Equations
We start by considering the general algebraic expression
for a chemical equation:
xC  yC  ...  uC  vC  ...
x
y
u
v
Where Cx, Cy, …. represent the chemical formulas of the
reactants and Cu, Cv, … represent the chemical formula of the
products.
The coefficients x, y, …. represent the (sought for)
coefficients of the reactants and the coefficients u, v, ….
represent the (sought for) coefficients of the reactants
We assume that the molecular formula for each
of the reactants and products is known and is
represented in the format:
C  An x Bn x ...
x
A
B
Where Cx is the reactant whose chemical
balance coefficient is x. The integers nxA are the
number of atoms of type A in the compound Cx.
For each elemental specie in the reaction, there
must be the same number of atoms on both sides
of the chemical equation. We express this as a
series of atom conservation equations, one for
each element. For the element A:
xn  yn  ...  un  vn  ....
x
A
y
A
u
A
v
A
Similar equations are written for each one of
the atom types. All of these equations must be
simultaneously obeyed.
There are N of the above equations, one for each element
(atom type) in the reaction. Generally there are M coefficients
to find using the N equations. Unfortunately, in most chemical
equations, M > N. Usually, we have the case that M = N+1.
Thus, we need to find one additional equation.
One simple way to solve the problem is to arbitrarily set one
of the coefficients to one and then solve the system of
equations. We find this approach works best when the
coefficient of the most complicated chemical species (the
one with the largest number of elements) is set to 1.
After all of the coefficients have been solved, we then
multiply the equation by the smallest integer that will
eliminate any fractions. Thus our balanced equation
contains the set of minimum integers.
Problem 8-1: Balance the following chemical equation
using the atom conservation method.
x XeF4 + y H2O -> u Xe + v O2 + w HF
Atom Conservation
Equations:
Solution:
Xe:
F:
H:
O:
Final Result:
Balancing Chemical Equations- Example #2
Problem 8-2: The hydrocarbon hexane is a component of Gasoline that
burns in an automobile engine to produce carbon dioxide and
water as well as energy. Write the balanced chemical
equation for the combustion of hexane (C6H14).
Plan: Write the skeleton equation from the words into chemical
compounds with blanks before each compound. Assign
unknowns x, y, … etc. to each blank
Skeleton equation:
Assign unknowns:
Problem 8-2: Balance the following:
x C6H14 (l) +
y O2 (g)
Atom Conservation
Equations:
->
u CO2 (g) + v H2O(g) + Energy
Solution:
C:
H:
O:
Final Result:
C6H14 (l) +
O2 (g)
->
CO2 (g) +
H2O(g) + Energy
Answers to Problems in Lecture #8
1. XeF4 + 2 H2O -> Xe + O2 +4 HF
2. 2 C6H14 (l) + 19 O2 (g) -> 12 CO2 (g) + 14 H2O(g) + Energy