Edited by Charles D. Mickey 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 reasons for teaching students how t o balance equations: (1) to enable them to balance an eauation for stoichiometric calculations and (2) to understand the chemistry involved. When 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 eofs the reactants and products. Likewise, half-reactions are useful when learning the intricacies of electrochemistry. However, if the former is the main purpose, then a straightforward mechanical approach which leads uuicklv to a properlv balanced eauation would seem to he preferred. or that p&pose, the foilowing method is proposed. A hnlnnred equation is one in which the number of atoms of each element on the left side equals the numljer on the right side. Therefore.. whv" not ust: this basir definition to halance equations? For example, consider equation 1:
The set of three equations is now trivial to solve: a = 1, b = 514, c = 1, d = 312. The whole number expression is found by multiplying each coefficient by 4, leading to the following equation: The problem with this approach for more complicated equations is that often we are confronted with a matrix of five, six, or more simultaneous euuations. However. i t is clear that we are experimcing a minor revolution in the computing cnpahilities nvailal~leto mlr students. T o solve a set of five s i ~ u l t a n e o u equations s requires only the ability to invert a 5 X 5 matrix, readily carried out by many desktop calculators. The coefficients are ground out entirely from a mechanical approach based on the definition of the balanced equation. I t teaches no chemistry other than the conservation of mass hut is very effective in balancing equations. Let me present three more examples. Kolb cited eqn. 2 as the most complicated tackled by the method she proposed.'
(1)
The balanced equation will have the following properties for each of the elements involved:
+
aPb(Nsh + bCr(Mn0a)ze cCr203 + dMnOn + ePbsOa fNO (2)
N: a = c H: 3a = 2d 0: 2 b = c + d
By setting a = 1, the set of six simultaneous equations is
Thus, we have three algebraic equations and four unknowns Element
assume Pb
Equation o=1
a=3earo-3e=O
a 1 1
so that an infinite number of chemical equations will satisfy the balancing conditions. A balanced equation normally will have the smallest whole number coefficients, which is one particular expression meeting the balancing requirements stated ahove. We can also look a t eqn. (1)in terms of combining 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:
b 0 0
e 0 0
Matrix Elements d e 0 0 0 -3
f
0 0
-
K
=
1 0
-
Alternatively, a 5 X 5 matrix can he set up to solve for b through f knowing that a = 1.A desktop calculator quickly generates the following solution matrix values:
' Kolb, D., J. CHEM.
EDUC., 58,
Volume 59
642 (1981).
Number 6 June 1982
523
q
,
Coefficient
Multiply Multiply Coefficients X 3 Coefficients X 5 3 15 8.79999 44 4.40001 22 17.60000 88 1 5 18 90
Value 1 2.93333 1.46667 5.86667 0.33333 6.00000
a
b e d e
f
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. However, if one wishes an equation with whole number coefficients, 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 that this equation will have some unusual whole number coefficients. Part of the simple program written for balancing equations allows the operator to multiply the matrix element8 by a chosen value. After multiplying by 3, it is now clear from the new valuesfor b, c, and d that a factor of 5 should also he used. The final whole number coefficients are now available, and the balanced equation is
The solution is given as:
a b C
d e
f The final equation is
In all the examples, it required a minute to generate the equations, a minute to enter the matrix elements, and a blink of the eye for the calculator to generate the proper coefficients. At first glance i t might seem that there should he chemical equations in which the number of algebraic equations would he more or less than the number of coefficients. Consider the chemical equation aHz
For a second example let us consider an ionic equation also cited by Kolb: oZn
+ bN03- + cH+ e dZn2++ eNH4+ + fHzO
Element
assume Zn N 0 H charge
Equation o=l a=d b=e 3b=f e=4e+2f -b e = 2d e
+
+
a 1 1 0 0 0 0
b 0 0 1 3 0 -1
Matrix Elements e f = K c d 0 0 0 0 = 1 0 - 1 0 o = o 0 0 - 1 O = O 0 0 0 - l = O 1 0 - 4 - 2 = 0 1 -2 -1 0 = 0
The solution is given as: Coefficient a b e d e
f
Value 1.00 0.25 2.50 1.00 0.25 0.75
Motriz Coefficients X 4 4 1 10 4 1 3
+
2H2 0 2 t 2H2O H z + 0 2 s H202
They, of course, could he added together in any combination. This method for halancine-eauations auicklv indicates that . a proposed equation is not a single equation with a unique solution when the number of coefficients is ereater than the number of algebraic equations. On the other hand, consider the equation which generates four equations and only has three coefficients: Element assume
Equation a =1 o=b a =b 3a = 2e
K CI 0
Thus, the final equation is Organic reactions-complicated to solve with oxidation state assignments because of the frequent fractional numhers-are no challenge to this straightforward mass balance approach. For the third example consider eqn. 4: aCaHaOa
+ bO2 t cHzO + dHz02
which has three algebraic equations (a = 1plus atom balance equations for 0 and H) and 4 coefficients. A unique solution is not possible. This equation is actually the sum of two separate equations:
(3)
One additional equation is needed, charge balance, making a total of six equations to solve for the six coefficients.
Value 1 14 20 3 14 14
Coefficient
In this case, two of the equations are identical, and one must he deleted from the set. Programs for solving a set of simultaneous 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 based on the definition of a balanced equation and the availability of a desktop calculator offers a rapid method to balance chemical equations.
+ bMnO4- + cOH- e dC0a2- + e M n O P + fH20 (4)
Proceeding as in the example above we write the following: Element
assume C H 0 Mn charge
524
Equation a=1 3a = d 8a e = 2f 3a+4b+c=3d+4e+f b=e -b - c = -2d - 2e
+
Journal of Chemical Education
a
1 3 8 3 0 0
b 0 0 0 4 1 -1
e
0 0 1 1 0 -1
Matrix Elements d e 0 0 -1 0 0 0 -3 -4 0 -1 +2 +2
f 0 0 -2 -1 0 0
--
K 1 0 0 0 0 0
