How Should Equation Balancing be Taught? Spencer K. Porter Capital University, Columbus, OH 43209

A laree literature on methods for eouation balancine has - - evolved over several decades, and a recent interesting addition to it in THIS JOURNAL' advocates matrix methods for solvine equations of balance and for finding the several equations when simultaneous reactions occur. Most general chemistry textbooks use oxidation-number methods even though ev- eryone knows that these numbers bear only faint resemblance to real charges. This paper proposes not to quarrel with the efficacy of either of thesemethods but rather to show that the subject may be introduced by a third method which is related directly to fundamental principle, is easy to learn, and is powerful.

A chemist when asked to balance simple equations like the combustion of methane or photosynthesis will work "hy in- spection" which means that he will look for elements that appear but once on each side and do algebra mentally. Thus if the number of methane molecules on the left-hand side of the combustion equation is set to one we may write

where x is to be determined and may he found by writing oxygen's equation of balance or by doing the same men- tally.

This process is so familiar that few of us think ahout what we are doing. The principle used dates from the great dis- covery of Lavoisier in 1777 a t the very birth of modern chemi~try:~ Mass is conserved, element by element. The de- velopment of atomic theory since that time gives the usual statement that the number of atoms of each element must be the same on each side. The dissection of the atom has also given the other principle needed: Charge is conserved. Stu- dents ought to learn fundamental principles and their appli- cations first, and these are among the most important. The word "conserve" means an equal sign, so we find one equation for each element and one for the charges. These simultaneous equations may be solved by the usual methods of algebra, and this is what is done mentally by inspection. The usual situa- tion, which is shown hy eqn. (11, is for the numher of un- knowns (the coefficients) to exceed the number of balance equations by one, and this requires that one coefficient be set arbitrarily. Since we are after ratios especially when doing stoichiometry problems, this is not a difficulty. Convention- ally, hut not necessarily, we choose the set of coefficients that is the smallest set of whole, positive numbers.

A minority of cases gives a number of balance equations equal to the number of unknowns. Consider the dehydration of sulfuric acid.

This has three elements and three coefficients. If the number of sulfur atoms is set to one on the left-hand side, we find that the number of sulfur trioxide molecules must be one. Further the numher of water molecules must he one as well to balance hydrogen. A check of oxygen shows eqn. (3) to he balanced as written.

' Blakley, G. R., J. CHEM. EOUC., 59, 728 (1982). Garrett. A. B.. "he Flash of Genius." Van Nostrand. Princeton. N.J..

Now consider the sort of thing teachers encounter all too often.

HzS04 - HzO + SO2 (4)

Let the number of sulfur atoms he x , then balance the sulfur, and then the hydrogen.

The oxygen-balance equation provides the rude awakening the student needs.

oxygen: 4.x = r + 2x (6)

The only solution is x = 0, and the reaction is shown to be impossible.

Redox equations are no more difficult. Consider the oxi- dation of copper that follows. The numher of copper atoms has been chosen as one, and the number of H N 0 3 molecules is allowed to be b. Since hydrogen appears just once on each side, we write immediately

Two equations of balance remain:

This system is easily solved by substitution to give b = 813 and and d = 213. Fractions may he cleared to give the conventional solution

I t may be argued that we were lucky in this instance to find a system of simultaneous equations that is simple to solve. The author's experience is that he is always lucky, and he believes that this is due to the happy circumstance that each element appears only a small number of times in even the most com- plex equation.

Reactions in base are no harder. In the following the number of chromium atoms is set to one.

The balance equations are

0 : 4 + o + 2 b = 4 + e H : 4 + a + 2 b = 2 c (11)

charge: -1 - 0 = -2

The charge equation gives a = 1, and the first two together give c = 4. Either the hydrogen or oxygen equation gives b = 3/2. Clearing the fraction gives

Reactions of organic reagents may he done by this method without assigning peculiar charges to carbon atoms as oxi- dation numbers. The oxidation of oxalic acid by acidic per-

Volume 62 Number 6 June 1985 507

manganate is shown? and the number of oxalic acid molecules is set to one.

The balance equations are

H : 2 + 3 n = 2 p 0 : 4 + 4 m + n = 4 + p (14)

charge: -m + n = 2m

These are solved by substitution to yield rn = y5, n = %,and p = 14l5. Clearing fractions yields the conventional solution

Most of us learned to do the equations above by some oxi- dation-number method. These work (though ~ e r h a o s not as easily), but consider the following take; frAm a-popular problem-solving text3 in which several oxidation numbers change or must be assigned arbitrarily. The number of chro- mium atoms is set a t two.

2CrSCN2+ + aCl2 + bHaO+ - 2aC1- + 2N03-

F r o m all thiswe find only two equations that are not identi- ties.

charge: 4 + b = -2a - 2 - 4 - 2

The oxygen equation gives b = -50, and the charge equation then gives a = 19. The balanced equation is, therefore

2CrSCN2+ + 19C12 + 75H20 - 38CI- + 2N03- + 2C02 + 2SOa2- + C r 2 0 F + 50H30C (18)

The negative b puts the hydroxonium ion on the right-hand side and the water on the left. The equation can never be balanced with H30+ on the left even though the problem was eiven to he "in acidic solution." ,, ~

Another instructive exnmple is the oxidation of a sulfide of molvbdmum I n nitric acid. The sulfide coefficient is set to one.

Two interesting algebraic equations remain

charge: p - q = -8

These give p = 0 and q = 8. We write

Therefore, acid is not a reactant. I t may be a catalyst necessary for the reaction's progress at an appreciable rate, hut catalysts do not appear in balanced equations.

Electrochemistry is an important and much-discussed topic,

SONm, C. H.. and Boikess. R. S., "How to Solve General Chemistry Problems," 6th ed. Prentics-Hall, Engiewood Cliffs, N.J., 1981, chap. 17.

a feature of which is that the oxidation process is separated in space from the reduction process. This means that we must he able to writ,e and balance half-cells. When confronted by somethina like the oxidation of Cr3+ to C r O P . the student taught bycurrent methods will apply the rule; fdr calculating oxidation numbers to find that three electrons are ~roduced even though the charge on the chromium atom inthe chro- mate ion is far from six.

The methods of this article may be applied to the problem. Say we have an inert electrode in a basic solution of Cr3+ ions. The chemistry is

Cr(0H)a- +OH- - C r O F + Hz0 + ne- (22)

Set the number of chromium atoms to one and write

Cr(0H)d- + rOH- - C r 0 F + (r - 4, H%O + ne- (23)

The oxygen- and charge-balance equations give r = 4 and n = 3, and

Cr(0H)d- + 40H- - Cr0d2- + 4H20 + 3e- (24)

Thus the correct number of electrons is found without any artifices.

Consider now an example from organic chemistry where reactions are seldom clean and side reactions the rule. We may chlorinate toluene using chlorine gas and the catalyst FeC13. Two main products will form, the o- and p-chlorotoluenes. Isomerization aside we can write the single, balanced equa- tion

Fccg CGHSCHB + C12 d CH~CSHICL + HCI (25)

If, however, we write the isomers separately, we will not find a unique balance. The number of coefficients exceeds the number of balance equations by two, a sure sign of multiple reactions. Only chemical analysis can reveal the ratio of products formed; the mass-conservation law cannot.

Finally an example from inorganic chemistry shows the problem of competing reactions in a redox system. Suppose we observe that

C1O2- + H30+ - C12 + C103- + C102 + H20 (26)

By writing and balancing thc three ~r~ssihle half-crlls in\.ulving the element chlorine, we can show that eqn. (26) i i an autw redox reaction with two oxidation products

2C10z- + 8H30+ + 6e- - Cln + 12H20 C102- + 3Hz0 - C103- + 2H30f + 2e- (27)

ClOz- - ClOz + e-

We have two simultaneous reactions, and we balance each separately. The two possihlities that place products and reactants on the proper sides are

5C102- + 2H30f - C12 + 3C103- + 3H20 (28)

8C102- + 8H30+ - Clz + 6C102 + 12Hz0 (29)

This is as far as we can go; the temptation simply toadd these eouations must be resisted. Just as in the chlorination of tol- uene only experiment can give the proportions of chlorate and chlorine dioxide formed.

This article has presented a method for balancing chemical equations that derives directly from laws of nature, is easy to learn and use, and is powerful in application. The author has seen his students benefit from its use and believes that others would as well.

508 Journal of Chemical Education