Balancing Chemical Equations

Here is the error that accounts for most lost marks on this topic, and it is not an arithmetic one. Faced with

H₂ + O₂ → H₂O

a lot of people balance the oxygen by turning the product into H₂O₂. The atoms now tally perfectly. The equation is also completely wrong, because hydrogen peroxide is a different substance from water and the reaction described no longer happens.

Subscripts are part of the identity of a substance. Coefficients are counts of it. You may change the counts freely and you may never touch the identities. Every legitimate move in balancing is a change to a number written in front of a formula. (The correct answer, for the record, is 2H₂ + O₂ → 2H₂O.)

Lavoisier's constraint, written as bookkeeping

The reason a balance is required at all is that matter is conserved. Antoine Lavoisier's careful weighings in the 1770s and 1780s established that the total mass of a closed system does not change during a reaction, and the atomic interpretation is stricter still: every atom present at the start is present at the end, merely rearranged.

An equation is therefore a claim with two sides that must agree on:

• the count of every element, and
• total charge, if the equation shows ions.

A coefficient multiplies everything in the formula it precedes. In 3Ca(OH)₂ there are three calcium, six oxygen and six hydrogen. Reading a coefficient as applying only to the first atom is the second most common source of error, and it hides very well inside a bracket.

Two further pieces of notation carry meaning rather than decoration. State symbols — (s), (l), (g), (aq) — say what physical form each substance is in, and an examiner who asks for them is asking about chemistry, not tidiness. And the arrow itself is informative: a single arrow means the reaction runs essentially to completion, while ⇌ means it settles at an equilibrium with both sides present.

Propane, in the order that makes it easy

Balancing by inspection is trial and error, but the order you try things in decides whether it takes thirty seconds or ten minutes. The heuristic: start with the element that appears in the fewest formulas, and finish with the one that appears in the most. For anything containing carbon, hydrogen and oxygen, that means carbon first, hydrogen second, oxygen last — oxygen almost always appears on both sides in several compounds, so fixing it early guarantees you will have to unfix it.

Propane burning:

C₃H₈ + O₂ → CO₂ + H₂O

Carbon. Three on the left, one on the right. Put a 3 in front of CO₂.

Hydrogen. Eight on the left. Water carries two each, so four molecules: 4H₂O.

Oxygen, last. The right now holds (3 × 2) + (4 × 1) = 10 oxygen atoms. O₂ supplies two at a time, so five of them.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Tally to confirm: carbon 3 and 3, hydrogen 8 and 8, oxygen 10 and 10.

The half that has to be doubled

Ethane makes the same journey and lands somewhere awkward:

C₂H₆ + O₂ → CO₂ + H₂O

Carbon gives 2CO₂. Hydrogen gives 3H₂O. Oxygen on the right is then (2 × 2) + 3 = 7 atoms, and seven is odd, so O₂ cannot supply it in whole molecules. The coefficient is 7/2.

Rather than abandoning the attempt, finish it with the fraction and then multiply every coefficient in the equation by two:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Whenever an odd number of atoms has to be supplied by a diatomic molecule, this doubling is coming, and it is faster to expect it than to discover it. Worth knowing, though: a fractional coefficient is not always wrong. Thermochemical equations are deliberately written per mole of one named substance, so ½O₂ appears constantly in enthalpy tables, and rewriting it as O₂ would change the quoted energy by a factor of two.

Phosphate as a single object

When a polyatomic ion survives a reaction intact, balancing it as one unit rather than atom by atom removes most of the work.

Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O

The phosphate group appears whole on both sides, so treat "PO₄" as if it were a single element. Two of them are needed on the right, so 2H₃PO₄ on the left. Calcium needs three, so 3Ca(OH)₂. That fixes hydrogen: the left now has (3 × 2) from the hydroxides plus (2 × 3) from the acid, which is twelve, so 6H₂O.

3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O

Check the oxygen as a whole, since it is spread across three formulas: left has six in the hydroxides and eight in the phosphates, fourteen; right has eight in the phosphate and six in the water, fourteen. The shortcut fails only when the polyatomic ion is broken apart by the reaction — if phosphate appeared on one side and phosphorus on the other, it would have to be balanced element by element.

When inspection stalls: four unknowns, four equations

Some equations resist trial and error, and redox reactions between a metal and a concentrated acid are the usual offenders. A balanced equation is really the solution to a system of linear equations, and writing that system out is a reliable fallback.

a Cu + b HNO₃ → c Cu(NO₃)₂ + d NO + e H₂O

One equation per element:

• copper: a = c
• hydrogen: b = 2e
• nitrogen: b = 2c + d
• oxygen: 3b = 6c + d + e

Set c = 1, which makes a = 1. From the nitrogen line, b = 2 + d. Substituting into the oxygen line gives 6 + 3d = 6 + d + e, so e = 2d. The hydrogen line then says b = 4d, and combining that with b = 2 + d gives 3d = 2, so d = 2/3.

Fractions are expected here; clear them at the end by multiplying everything by three:

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Verify all four: copper 3 and 3, hydrogen 8 and 8, nitrogen 8 on the left against 6 + 2 on the right, oxygen 24 on the left against 18 + 2 + 4. The equation balancer solves the same system with exact integer arithmetic and prints the elimination, which is worth comparing against your own working when a by-hand attempt refuses to close.

Equations that genuinely have more than one answer

Textbooks imply that every equation has a single balanced form with smallest whole numbers. Almost all do, and there is a well-defined class that does not.

Consider carbon burning to a mixture of both oxides:

C + O₂ → CO + CO₂

Both 3C + 2O₂ → 2CO + CO₂ and 5C + 4O₂ → 2CO + 3CO₂ balance perfectly. The reason is that this is not one reaction but two independent ones sharing a page, and the proportion between them is not determined by conservation of atoms. Mathematically the system is underdetermined — its solution space has more than one dimension.

The rule that follows is a good one to carry: if you find two different balanced forms, you have not made a mistake, you have written down more than one reaction at once. Split it into separate equations, each of which will balance uniquely.

What a balanced equation still does not tell you

Balancing is bookkeeping, and it is silent on everything a reaction is actually like. A perfectly balanced equation makes no claim that the reaction occurs at all, says nothing about how fast it goes, gives no information about the sequence of steps by which atoms actually rearrange, and does not indicate how much energy is absorbed or released.

It does exactly one job, and it does it as a precondition for the next thing: the coefficients are the mole ratios that every stoichiometry calculation depends on. An unbalanced equation produces confidently wrong numbers all the way down.