Chemical Reaction Equations: Balancing Coefficients

When iron burns in oxygen to form iron(III) oxide, the first representation looks like this: formulas of the reactants on the left, formulas of the products on the right, with an arrow in between. This is a reaction scheme.

A scheme is not yet an equation. It states what turns into what, but not how much of each substance is involved: the number of iron and oxygen atoms is different on the left and right sides. According to the law of conservation of mass, this cannot be the case.

The representation becomes a reaction equation when coefficients are placed before the formulas – numbers that equalize the number of atoms of each element. Compare the two representations of the same reaction.

The law of conservation of mass was formulated by M. V. Lomonosov (1748) and later independently confirmed by A. Lavoisier: the mass of the substances that reacted is equal to the mass of the substances that were formed.

The reason is simple: during a chemical reaction, atoms do not disappear or appear out of nowhere. Old bonds break, new ones form, and the atoms themselves merely rearrange. The number of iron atoms before the reaction will be the same as after – they simply become part of a different substance.

Hence, the main requirement for an equation: the number of atoms of each element on the left must equal the number of atoms of that element on the right. The only valid way to achieve this is by adjusting coefficients.

A coefficient is placed before the formula and multiplies all atoms in it, including those within parentheses. For example, the notation \(\mathrm{3Ca(OH)_2}\) means three formula units of calcium hydroxide: 3 atoms of \(\mathrm{Ca}\), 6 atoms of \(\mathrm{O}\), and 6 atoms of \(\mathrm{H}\). A coefficient of 1 is not written but is implied.

Balancing coefficients is not a random guess but a short procedure. The order is important: first, balance elements that appear in exactly one substance on the left and one on the right, and only then hydrogen and oxygen – they are usually distributed among several substances and are adjusted last.

The school order can be remembered by the chain: metal → nonmetal → hydrogen → oxygen. The six steps of the algorithm are below.

If you get a fractional coefficient at any step (e.g., 3/2 before a diatomic gas molecule), it's not an error or a dead end: complete the balancing, and then multiply all coefficients by 2 – the fractions will disappear, and the balance will be maintained.

In higher grades, for redox reactions, a more powerful method is introduced – the electron balance method. It relies on oxidation states, but double displacement and combination reactions are still balanced using this simple algorithm.

This is the number one mistake for eighth graders. In the scheme \(\mathrm{Na + Cl_2 \rightarrow NaCl}\), chlorine doesn't balance: 2 atoms on the left, 1 on the right. The temptation is great – to add a subscript and get \(\mathrm{NaCl_2}\). The atoms will formally balance, but the written substance does not exist: sodium chloride has the composition \(\mathrm{NaCl}\), and no other.

A subscript is part of a formula. It is determined by the composition of the substance, i.e., the valency and oxidation state of the elements, and you cannot change it at will. By changing a subscript, you change the substance itself: \(\mathrm{H_2O}\) is water, while \(\mathrm{H_2O_2}\) is hydrogen peroxide, a completely different compound with different properties.

A coefficient is a multiplier before the formula. It only indicates the number of particles, not their composition, so balancing is only allowed using coefficients.

The correct answer for the scheme above is: \(\mathrm{2Na + Cl_2 \rightarrow 2NaCl}\). Sodium 2 and 2, chlorine 2 and 2, formulas are unchanged.

Equations are classified by the number and type of initial substances and products. In basic school, four types are distinguished: combination, decomposition, displacement, and double displacement.

Determining the type before balancing coefficients is useful: it suggests which substances will be on the right side and how many there will be, thus helping to avoid errors in the scheme itself.

In tests and quizzes, the question is often not about the entire equation but about the sum of coefficients – this is a quick way to check if it's balanced correctly. It's calculated simply: add up all the coefficients in the left and right sides.

For the equation \(\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}\), the sum is 4 + 3 + 2 = 9.

Three common places where points are lost:

• The invisible '1' counts too. In the equation \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}\), all three coefficients are 1, so the sum is 3;
• Subscripts are not included in the sum – only the numbers before the formulas are added;
• Coefficients must be the smallest possible integers. The notation \(\mathrm{4Na + 2Cl_2 \rightarrow 4NaCl}\) is balanced in terms of atoms, but all numbers are divisible by 2, so it's reduced to 2, 1, 2 – and the correct sum is 5, not 10.

Coefficients are needed not only for balancing atoms. They show the ratio of substance amounts (in moles) in which substances react and are formed. This is the bridge to calculation problems.

The equation \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\) is read as: 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. The ratio 1 : 3 : 2 is maintained regardless of the amount of substance.

Hence, the usual calculation scheme based on an equation: convert the given mass or volume to the amount of substance, find the amount of the desired substance using the coefficients, and convert back to mass or volume. The calculation problems themselves are a separate large topic, but it always starts here: if the coefficients are incorrectly balanced, any such calculation will yield an incorrect answer.