How to balance chemical equations

Mid-September: science lessons restarted a few weeks ago and the teacher is introducing (or about to introduce) the laws of mass and the balancing of chemical equations. The first test arrives in five to eight weeks, so between the second half of October and the very first days of November. It is not yet the time for last-minute revision: it is the time to build the method properly, because a good part of Year 3 chemistry rests on this topic.

Here “Year 2” means the second year of Italian high school (the liceo).

How to balance chemical equations: what the test really asks

In my lessons I see the same pattern every year. The October-November science test in Year 2, both at the standard science high school (liceo scientifico) and in its applied sciences option, almost always asks for three things together:

knowing how to state the laws of Lavoisier, Proust and Dalton and link them to the atomic model (not just repeat them by heart);
knowing how to balance a chemical equation, even a non-trivial one, with a repeatable method — not “by trial and error”;
in the applied sciences option, often also doing the first calculations with moles on equations that are already balanced.

The weak point, the one nearly always flagged by whoever marks the papers, is balancing. Not because it is hard in any absolute sense, but because one student in two treats it as a puzzle of numbers instead of the direct application of a physical law they have just studied. If balancing stays a mechanical, meaningless operation, every slightly more complex equation (brackets, polyatomic groups) becomes a new obstacle instead of an application of the same principle.

The three laws of mass: the why before the how

Before balancing anything, it is worth fixing clearly why we do it.

Lavoisier’s law (conservation of mass). In a chemical reaction the total mass of the reactants equals the total mass of the products: nothing is created and nothing is destroyed, atoms are rearranged. Everything else follows from here: if atoms neither disappear nor are created, the number of atoms of each element on the left of the arrow must be identical to the number of atoms of the same element on the right. Balancing an equation is applying Lavoisier, it is not a separate exercise.

Proust’s law (definite proportions). When two or more elements combine to form a compound, they always do so in a fixed, constant mass ratio. Water is always hydrogen and oxygen in the same mass ratio, however and wherever you prepare it. It is the law that makes it sensible to write a chemical formula as something fixed (H₂O, not “a bit of hydrogen and a bit of oxygen”).

Dalton’s law (multiple proportions). If two elements form several different compounds, the amounts of one element that combine with a fixed amount of the other are in the ratio of small whole numbers. It is the reason why both CO and CO₂ exist, in different but “clean” proportions: 1:1 and 1:2, not 1:1.73.

Taken together, these three laws say: matter is conserved (Lavoisier), compounds have a fixed composition (Proust), and when an element forms several compounds it does so in whole-number ratios (Dalton). Balancing is simply the way, on paper, of respecting the first of the three.

The balancing method, step by step

The method I use with all my Year 2 students always works, even on more complex equations they will meet later. The steps:

Write the equation with the correct formulas (before you even start balancing, check that the formulas of the compounds are right — balancing a wrong formula is wasted time).
Start from the element that appears in only one compound on each side, usually the metal or the carbon, not hydrogen and oxygen.
Leave H and O for last: they almost always appear in several compounds (water, acids, oxides), and balancing them first forces you to redo the sums every time you touch another element.
If you need a fraction halfway through, keep it: balance anyway, then multiply all the coefficients by the denominator to get back to whole numbers.
Check with a two-column table, element by element, atoms on the left against atoms on the right. If even a single element does not match, the equation is not balanced: there are no exceptions.
Reduce to lowest terms: if all the coefficients are divisible by the same number, divide.

The mistakes that cost marks

These are, in order of frequency, the mistakes I most often see marked in red on papers:

Changing the subscripts instead of the coefficients. This is mistake number one, and the most serious because it is not a calculation error: it is conceptually wrong. Writing $H_{2} O_{2}$ instead of $2 H_{2} O$ to “add oxygen” balances nothing — it changes the substance. $H_{2} O_{2}$ is hydrogen peroxide, a compound different from water, with different properties. The subscripts (the small numbers at the bottom, inside the formula) describe the identity of a substance and are untouchable; only the coefficients (the numbers in front of the formula) are changed to balance.
Forgetting that the coefficient multiplies all the atoms of the compound. $2 Ca (OH)_{2}$ does not mean “2 more Ca”: it means 2 whole units of $Ca (OH)_{2}$ , so 2 atoms of Ca, 4 of O and 4 of H in total. A very common mistake is to multiply only the first element of the formula.
Not counting the atoms inside brackets. In $Ca (OH)_{2}$ the bracket with the subscript 2 outside means that the whole OH group is taken twice: 2 O and 2 H, not 1 and 1.
Leaving coefficients that are not reduced to lowest terms. An equation balanced with 4-8-4-8 instead of 1-2-1-2 is formally correct but marked down as imprecise by almost all teachers.
Balancing a polyatomic group piece by piece instead of as a single block, when that group goes through the reaction without breaking up (typically $S O_{4}$ , $P O_{4}$ , $N O_{3}$ , $C O_{3}$ ). If the group appears intact both among the reactants and among the products, treat it as a single entity to balance as a block: it is faster and less prone to counting errors.

Three worked examples

1. Combustion of methane

C H_{4} + O_{2} \to C O_{2} + H_{2} O

I start from carbon: 1 on the left, 1 on the right, already fine. I move on to hydrogen: 4 on the left (in $C H_{4}$ ), so I need 2 molecules of water to have 4 H on the right:

C H_{4} + O_{2} \to C O_{2} + 2 H_{2} O

Now oxygen, last: on the right I have 2 O (from $C O_{2}$ ) + 2 O (from $2 H_{2} O$ ) = 4 O. On the left, to have 4 O I need 2 molecules of $O_{2}$ :

C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O

Check in a table:

Element	Left	Right
C	$1$	$1$
H	$4$	$4$
O	$2 \times 2 = 4$	$2 + 2 = 4$

Everything matches, coefficients already in lowest terms.

2. Combustion of ethane

C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O

Carbon: 2 on the left, so I need 2 $C O_{2}$ on the right. Hydrogen: 6 on the left, so 3 $H_{2} O$ on the right (3 × 2 = 6):

C_{2} H_{6} + O_{2} \to 2 C O_{2} + 3 H_{2} O

Oxygen last: on the right I have $2 \times 2 = 4$ O (from $C O_{2}$ ) plus $3 \times 1 = 3$ O (from $H_{2} O$ ), total 7 O. On the left, to get 7 O with $O_{2}$ molecules (which carry 2 at a time), a fractional coefficient is needed:

C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O

Here comes the step many people skip: the fraction is a legitimate result halfway through, but a balanced equation does not leave fractional coefficients. I multiply everything by 2:

2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O

Check:

Element	Left	Right
C	$2 \times 2 = 4$	$4$
H	$2 \times 6 = 12$	$6 \times 2 = 12$
O	$7 \times 2 = 14$	$4 \times 2 + 6 \times 1 = 14$

Everything matches.

3. An equation with a polyatomic group and brackets

Ca (OH)_{2} + H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + H_{2} O

Here it pays to balance the polyatomic groups as a block, because $P O_{4}$ does not break up during the reaction. I start from calcium: on the right I have 3 (in $C a_{3} (P O_{4})_{2}$ ), so on the left I need 3 $Ca (OH)_{2}$ :

3 Ca (OH)_{2} + H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + H_{2} O

Now the $P O_{4}$ group: on the right there are 2 of them (the subscript outside the bracket in $C a_{3} (P O_{4})_{2}$ ), so on the left I need 2 $H_{3} P O_{4}$ :

3 Ca (OH)_{2} + 2 H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + H_{2} O

Hydrogen remains, last. On the left: $3 \times 2 = 6$ H from $Ca (OH)_{2}$ , plus $2 \times 3 = 6$ H from $H_{3} P O_{4}$ , total 12 H. On the right I therefore need 6 molecules of water ( $6 \times 2 = 12$ ):

3 Ca (OH)_{2} + 2 H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 6 H_{2} O

Full check:

Element	Left	Right
Ca	$3$	$3$
O (OH group + $P O_{4}$ group + water)	$3 \times 2 = 6$ (from OH) $+ 2 \times 4 = 8$ (from $P O_{4}$ ) $= 14$	$2 \times 4 = 8$ (from $P O_{4}$ ) $+ 6 \times 1 = 6$ (from $H_{2} O$ ) $= 14$
P	$2$	$2$
H	$3 \times 2 = 6 + 2 \times 3 = 6 \Rightarrow 12$	$6 \times 2 = 12$

Everything matches, and this is exactly the example where treating $P O_{4}$ as a single block (instead of counting P and O separately every time) saves at least two minutes in a test.

Where it leads: the mole arrives in Year 3

This work is not an end in itself. In Year 3, the balanced equation stops being just an exercise in counting atoms and becomes the basis for calculating how many moles, how many grams, how many litres of gas are formed or consumed in a reaction — stoichiometry. This is where a good part of the failing grades of November-December in Year 3 come from, almost always for the same reason: a student who has not made balancing automatic in Year 2 has to redo it from scratch while already learning molar ratios, and the two cognitive loads together cause a crisis. If you want to see already now where it leads, I wrote a dedicated guide to the typical stoichiometry mistakes at high school (also useful for medicine admission tests): typical stoichiometry mistakes.

Staying in Year 2, the other classic stumbling block of the same time of year is in maths: rationalising radicals. I talked about it here: radicals and rationalisation in Year 2.

How I work on this topic in lessons

With Year 2 students tackling the laws of mass and balancing, I always start from the simplest equations (one at a time, with the check table written by hand until it becomes automatic), and I raise the difficulty only when the method is solid — brackets, polyatomic groups, fractional coefficients. The material on my science pages covers exactly this path. To practise outside lessons, on the site you will find the high school test practice papers: at the moment they cover maths and physics (chemistry is not there yet); the correct answer is always free, while the guided step-by-step solution unlocks with the first lesson.

If you want to set the topic up properly before the October-November test arrives, or you already have a marked paper to go over together, write to me from the contact page: we can start with a free 30-minute introductory call, or with a single lesson or the 10-hour package (rates on the pricing page) if you want to follow the whole chemistry module continuously. Payment is always offline — bank transfer, Satispay, PayPal, card, Apple Pay or Revolut.

Balancing chemical equations: a step-by-step method