In the Classroom
Balancing Chemical Equations by Inspection Zoltán Tóth Team of Chemical Methodology, Lajos Kossuth University, H-4010 Debrecen, P.O. Box 66, Hungary Balancing chemical equations by inspection is often believed to be a trial-and-error process (1–3), and therefore it can be used only for simple chemical reactions. Although a very simple systematic procedure was published and used for balancing rather complicated chemical equations in 1986 by Harjadi (4), this powerful reaction balancing method was not widely used until now. Some general chemistry textbooks use similar methods, but mainly implicitly (3, 5). This paper shows that the balancing of chemical equations by inspection is not a trial-and-error process, because a systematic procedure based on Harjadi’s method can be suggested. This procedure, which I dub the “chain method” (Harjadi called it “Ping-Pong method”) is suitable for balancing simple and more complicated chemical equations without formal charges (oxidation number method) or equations with several unknowns (algebraic method). The starting point for balancing by inspection is the fact that certain atoms in the skeletal chemical equation appear only in one reactant and one product each. So after identifying the atoms that occur in only one substance on each side of the equation, we can balance the skeletal equation first in one of the identified atoms. For example, in reaction 1, only hydrogen occurs in one of the products (H2O) and in one of the reactants (NaOH). S + NaOH → Na2S + Na2S2O3 + H2O
CH4 + NH3 + O 2 → HCN + H2O
(2)
After balancing in C atoms, eq 2 can be written as follows: 1 CH4 + NH3 + O 2 → 1 HCN + H2O
(2a)
We can continue the balancing in N atoms 1 CH4 + 1 NH3 + O 2 → 1 HCN + H2O
(2b)
and after balancing in H atoms 1 CH4 + 1NH3 + O 2 → 1HCN + 3 H2O
(2c)
and finally in O atoms, we get the balanced equation: 1 CH4 + 1NH3 + 3/2 O2 = 1 HCN + 3 H2O
(2d)
After multiplying eq 2d by 2, we obtain 2 CH4 + 2 NH3 + 3O2 = 2 HCN + 6 H2O
(2e)
(1)
If there is more than 1 kind of atom that occurs only in one compound on each side of the equation, it is usually best to select the one that is a component of the compound with the most atoms. For example, in the famous eq 3, published in this Journal (6),
(1a)
P2I4 + P4 + H 2O → PH 4I + H3PO 4
So the equation is balanced in the H atom first: S + 2 NaOH → Na2S + Na2S2O3 + 1H2O
present in compounds only. According to this rule, the balancing of reaction 2 can be started with N or C atoms.
(3)
We can continue the process by balancing the equation in those other atoms that are components of one unbalanced substance only. The next atom suitable for balancing is the O atom, because oxygen is the component of one unbalanced substance (Na2S 2O3):
two kinds of atoms (I and O) occur in only one substance on each side of the equation. Both are present only in compounds, and oxygen is the component of the compound with most atoms (H3PO 4). Therefore we start the balancing with O atoms:
S + 2 NaOH → Na 2S + 1/3 Na2S 2O3 + 1H2O
(1b)
P2I 4 + P4 + 4H 2O → PH 4I + 1 H3PO 4
Then we balance the Na atoms by placing the coefficient 2/3 for Na2S:
To balance the equation in H atoms, we write:
S + 2 NaOH → 2/3Na2S + 1/3Na2S 2O3 + 1 H2O (1c)
P2I4 + P4 + 4 H2O → 5/4PH 4I + 1 H 3PO 4
(3a)
(3b)
After balancing in I atoms we get: Finally, the S atoms are balanced: 4/3S + 2 NaOH = 2/3Na2S + 1/3Na2S2O3 + 1H2O (1d)
5/16P 2I 4 + P4 + 4H 2O → 5/4PH4I + 1 H3PO4
(3c)
Finally, we balance the equation in P atoms: After multiplying the eq 1d by 3, we get the conventional form of balanced equation: 4 S + 6 NaOH = 2 Na2S + 1 Na2S 2O3 + 3H2O
(1e)
In reaction 2, there are three kinds of atoms (O, N, C) that occur in only one substance on each side of the equation. However, O is present on the left side as an element, so after balancing O2 and H2O the chain is broken. The atoms suitable for starting balancing must be
5/16P2I4 + 13/32P4 + 4H2O = 5/4PH4I + 1H3PO4 (3d) After multiplying the eq 4d by 32, we obtain the balanced equation: 10P2I4 + 13P4 + 128H2O = 40PH 4I + 32H3PO4 (3e) Based on the above rules and experiences the following procedure for balancing chemical equations by
Vol. 74 No. 11 November 1997 • Journal of Chemical Education
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In the Classroom inspection can be suggested.
We can continue the balancing with hydrogen atom:
1. Identify those kinds of atoms that occur in only one substance on each side of the equation.
I { + 1 IO3{ + 6 H + → I2 + 3 H2O
2. If the number of identified kinds of atoms is higher than 1, choose those that are present in compounds only, and not in elements.
There are no more atoms suitable for balancing, so we must take into account the charge balance as follows. There are no ions on the right side of the eq 5b, but we have 6 positive and 1 negative balanced ions on the left side; therefore we need 5 I{ to make the left side neutral:
3. If the number of remaining kinds of atoms is higher than 1, select the one that is a component of the compound with most atoms or kinds of atoms.
5 I{ + 1 IO3{ + 6 H+ → I2 + 3 H2O
4. Balance the equation first in the selected atom. 5. Continue the process by balancing the equation in the other atoms that are components of one unbalanced substance only.
(4)
According to the proposed procedure the selected atom is hydrogen. After balancing in H atoms we get: Cu + 1 H2SO4 → CuSO4 + SO2 + 1H2O
(4b)
(4c)
After balancing in S atoms we get: x Cu + 1 H2SO4 → x CuSO4 + (1 – x) SO2 + 1H2O (4d) The unknown coefficient can be obtained from the equation of oxygen balance: 4 = 4x + 2 (1 – x) + 1 (for O atoms) x = 1/ 2
1/2Cu + 1 H2SO4 = 1/2 CuSO4 + 1/2 SO2 + 1 H2O (4e) or in conventional form: (4f)
(5)
According to above rules we must start balancing with the oxygen atom: I{ + 1 IO3{ + H+ → I2 + 3 H2O
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(5e)
This nonionic equation can easily be balanced by inspection using the proposed method. First we balance the equation in O atoms; subsequently, in H, Y, and X atoms; and finally in I atoms: (5f)
After elimination of counterions X and Y, we get the balanced ionic form of eq 5d. By this supplement, the proposed method is suitable for balancing all chemical equations that have a single unique solution. To examine the efficiency of this method, all the chemical equations from Ebbing’s General Chemistry (3) were collected. Among the 274 equations examined, 251 (91.6%) were easily balanced by this procedure; in 22 equations (8.0%), one unknown had to be introduced; and only one equation (0.4%) (KMnO4 + H2O2 + H2SO4 → K 2SO4 + MnSO4 + H2O + O2) could not be balanced by this technique. Note that this “odd” reaction is one of the so-called reactions with multiple balanced equations (7); for them the number of independent equations of elemental balance is two less than the number of reactants and products. Recently, it was published in this Journal that a database of 140 inorganic chemical equations culled from standard textbooks contained only 4.5% of equations of this type (8).
I am grateful to my reviewers for their useful suggestions, and to the Hungarian Scientific Research Foundation (OTKA T-023144) for financial support. Literature Cited
This method is useful for balancing ionic equations, too. For example, there are two possible ways to balance ionic eq 5 by inspection. I{ + IO3{ + H+ → I2 + H2O
XI + XIO3 + HY → I2 + H 2O + XY
Acknowledgments
The balanced equation is:
1 Cu + 2 H2SO4 = 1 CuSO4 + 1 SO2 + 2 H2O
(5d)
The other possibility for balancing this equation is to convert the ionic equation to a “molecular” one by introducing X+ and Y{ counterions:
5XI + 1XIO3 + 6HY = 3I2 + 3H2O + 6XY
Now, we can continue the balancing. The coefficient for Cu is x, too. x Cu + 1 H2SO4 → x CuSO4 + SO2 + 1H2O
5 I { + 1 IO3{ + 6 H+ = 3 I 2 + 3H 2O
(4a)
And there are no more atoms suitable for continuing the balancing. We must introduce the unknown coefficient x for CuSO4. Cu + 1 H2SO4 → x CuSO4 + SO2 + 1 H2O
(5c)
Finally, we balance the ionic equation in I atoms:
These examples demonstrate that balancing by inspection may be a powerful method for complicated chemical reactions, too. However, in case of some redox reactions the suggested balancing process can be used only by introducing one or two unknowns, and the complete coefficient set can be obtained only after solution of one or two algebraic equations. Let’s consider the dissolution of copper in concentrated sulfuric acid: Cu + H2SO4 → CuSO4 + SO2 + H2O
(5b)
(5a)
1. Kolb, D. J. Chem. Educ. 1978, 55, 184. 2. Holtzclaw, H. E., Jr.; Robinson, W. R. College Chemistry with Qualitative Analysis, 8th ed.; D. C. Heath: Lexington, MA, 1984; pp 38–40. 3. Ebbing, D. D. General Chemistry; Houghton Mifflin: Boston, 1984; pp 35–37. 4. Harjadi, W. J. Chem. Educ. 1986, 63, 978. 5. Chang, R. Chemistry, 2nd ed.; Random House: New York, 1984; p 60. 6. Carrano, S. A. J. Chem. Educ. 1978, 55, 382. 7. Kolb, D. J. Chem. Educ. 1979, 56, 181. 8. Subramaniam, R.; Goh, N. K.; Chia, L. S. J. Chem. Educ. 1995, 72, 894.
Journal of Chemical Education • Vol. 74 No. 11 November 1997
