# Balancing chemical equations with a calculator

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```					                                                                                                                                        Edited by
Charles D. Mickey      q   ,
Texas A8M at Galveston
Galveston. TX 77553

Balancing Chemical Equations with a Calculator
John H. Kennedy
University 07 California, Santa Barbara, CA 93106

Recently this journal published a paper describing a new
approach to balancing chemical equatams hy inspection.' Thc
.    .
method proposed, like others which are commonly used, is
somewhat mechanical, allowing students to balance equations
by following a set of rules. There appear to he two basic rea-         The set of three equations is now trivial to solve: a = 1, b =
sons for teaching students how t o balance equations: (1) to           514, c = 1, d = 312. The whole number expression is found by
enable them to balance an eauation for stoichiometric calcu-           multiplying each coefficient by 4, leading to the following
lations and (2) to understand the chemistry involved. When             equation:
the latter is considered. then the use of oxidation states for
balancing is certainly advantageous, allowing students to
appreciate the chemical ~ r o ~ e r t i e s the reactants and
of                                The problem with this approach for more complicated
products. Likewise, half-reactions are useful when learning            equations is that often we are confronted with a matrix of five,
the intricacies of electrochemistry. However, if the former is         six, or more simultaneous euuations. However. i t is clear that
the main purpose, then a straightforward mechanical ap-                we are experimcing a minor revolution in the computing
proach which leads uuicklv to a properlv balanced eauation             cnpahilities nvailal~leto mlr students. T o solve a set of five
would seem to he preferred. or that p&pose, the foilowing                                    s
s i ~ u l t a n e o u equations requires only the ability to invert a
method is proposed.                                                    5 X 5 matrix, readily carried out by many desktop calculators.
A hnlnnred equation is one in which the number of atoms             The coefficients are ground out entirely from a mechanical
of each element on the left side equals the numljer on the right       approach based on the definition of the balanced equation.
side. Therefore.. whv not ust: this basir definition to halance
"                                                 I t teaches no chemistry other than the conservation of mass
equations? For example, consider equation 1:                           hut is very effective in balancing equations. Let me present
three more examples.
(1)           Kolb cited eqn. 2 as the most complicated tackled by the
The balanced equation will have the following properties for           method she proposed.'
each of the elements involved:                                         aPb(Nsh + bCr(Mn0a)ze cCr203 + dMnOn + ePbsOa fNO (2)         +
N: a = c                                      By setting a = 1, the set of six simultaneous equations is
H: 3a = 2d
0: 2 b = c + d
Thus, we have three algebraic equations and four unknowns

Matrix Elements
Element                      Equation                   a          b           e           d           e                f            -        K
assume            o=1                              1          0           0           0           0                0            -         1
Pb              a=3earo-3e=O                     1          0           0           0        -3                  0            =         0

so that an infinite number of chemical equations will satisfy          Alternatively, a 5 X 5 matrix can he set up to solve for b
the balancing conditions. A balanced equation normally will            through f knowing that a = 1.A desktop calculator quickly
have the smallest whole number coefficients, which is one              generates the following solution matrix values:
particular expression meeting the balancing requirements
stated ahove. We can also look a t eqn. (1)in terms of com-
bining ratios, i.e., how many molecules (moles) of 0 2 react per
molecule (mole) of NH3. For that situation we may arbitrarily
assign one coefficient the value of one. Thus, for a = 1:                ' Kolb, D., J. CHEM.EDUC., 58,   642 (1981).

Volume 59      Number 6 June 1982                     523
Multiply         Multiply              The solution is given as:
Coefficient         Value        Coefficients X 3 Coefficients X 5
a             1                 3                   15                           Coefficient                              Value
b             2.93333           8.79999             44                                a                                     1
e             1.46667           4.40001             22                                b                                    14
d             5.86667          17.60000             88                                C                                    20
e             0.33333           1                    5                                d                                     3
f             6.00000          18                   90                                e                                    14
f                                     14
The values represent one valid solution to the equation, i.e.,
one mole of Ph(N& reacts with 2.93333 moles of Cr(MnOa)p.                   The final equation is
However, if one wishes an equation with whole number coef-
ficients, it is clear from the value of 0.333 fore that a multiplier
of 3 is involved even though the coefficients forb and c show                 In all the examples, it required a minute to generate the
that this equation will have some unusual whole number                      equations, a minute to enter the matrix elements, and a blink
coefficients. Part of the simple program written for balancing              of the eye for the calculator to generate the proper coeffi-
equations allows the operator to multiply the matrix element8               cients.
by a chosen value. After multiplying by 3, it is now clear from               At first glance i t might seem that there should he chemical
the new valuesfor b, c, and d that a factor of 5 should also he             equations in which the number of algebraic equations would
used. The final whole number coefficients are now available,                he more or less than the number of coefficients. Consider the
and the balanced equation is                                                chemical equation
aHz   + bO2 t cHzO + dHz02
which has three algebraic equations (a = 1plus atom balance
For a second example let us consider an ionic equation also              equations for 0 and H) and 4 coefficients. A unique solution
cited by Kolb:                                                              is not possible. This equation is actually the sum of two sep-
arate equations:
oZn   + bN03- + cH+ e dZn2++ eNH4+ + fHzO               (3)
+
2H2 0 2 t 2H2O
One additional equation is needed, charge balance, making                                              H z + 0 2 s H202
a total of six equations to solve for the six coefficients.
They, of course, could he added together in any combination.
Matrix Elements                                            - .
This method for halancine eauations auicklv indicates that
Element           Equation       a    b     c d      e   f = K             a proposed equation is not a single equation with a unique
assume              o=l           1     0    0   0     0    0 = 1           solution when the number of coefficients is ereater than the
Zn                a=d           1     0   0 - 1      0    o = o           number of algebraic equations.
N                  b=e          0     1    0   0 - 1      O = O             On the other hand, consider the equation
0                 3b=f          0     3    0   0     0 - l = O
H               e=4e+2f         0     0    1   0 - 4 - 2 = 0
charge         -b +  e = 2d e +   0   -1     1 -2 -1        0 = 0
which generates four equations and only has three coeffi-
The solution is given as:                                                   cients:
Coefficient            Value             Motriz Coefficients X 4                    Element                                Equation
a                   1.00                       4                                assume                                   a =1
b                   0.25                       1                                   K                                     o=b
e                   2.50                      10                                   CI                                    a =b
d                   1.00                       4                                   0                                    3a = 2e
e                0.25                       1
f                0.75                       3                     In this case, two of the equations are identical, and one must
he deleted from the set. Programs for solving a set of simul-
Thus, the final equation is
taneous equations will normally test that they are indeed a
proper set; i.e., the determinant #O.
In conclusion, the use of a set of simultaneous equations
Organic reactions-complicated to solve with oxidation                     based on the definition of a balanced equation and the
state assignments because of the frequent fractional num-                   availability of a desktop calculator offers a rapid method to
hers-are no challenge to this straightforward mass balance                  balance chemical equations.
approach. For the third example consider eqn. 4:
aCaHaOa      + bMnO4- + cOH- e dC0a2- + e M n O P + fH20
(4)
Proceeding as in the example above we write the following:
Matrix Elements
Element                 Equation                      a               b        e          d          e                  f             -
-   K
assume                   a=1                         1               0        0          0          0               0                -   1
C                     3a = d                       3               0        0        -1           0               0                -   0
H                        +
8a e = 2f                        8               0        1          0          0             -2                 -   0
0              3a+4b+c=3d+4e+f                     3               4        1        -3         -4              -1                 -   0
Mn                     b=e                         0               1        0          0        -1                0                -   0
charge              -b - c = -2d - 2e                0             -1       -1         +2          +2               0                -   0

524            Journal of Chemical Education

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