Balancing Chemical Equations

The first chapter of this block established that a chemical reaction only rearranges atoms, never creating or destroying them — the law of conservation of mass. A chemical equation is the written statement of exactly how that rearrangement happens: which substances react, which substances form, and in what whole-number ratio. This chapter is entirely about writing that statement correctly. No calculation of a reacting mass or a reaction yield appears here — both are the subject of the next chapter, which takes a correctly balanced equation as its starting point and asks what quantities of reactant and product it predicts.

Balancing an equation is sometimes treated as a purely mechanical puzzle, a matter of trial and error with coefficients. It is better understood as the direct arithmetic consequence of a physical law: if no atom is created or destroyed, then the equation must show exactly the same number of atoms of every element on both sides. Learning to balance equations fluently, including the trickier cases of polyatomic ions and ionic equations with spectator ions removed, is one of the most load-bearing skills in the rest of the course.

1. From a Description to a Chemical Equation

A chemical reaction is first described in words — "magnesium reacts with oxygen to form magnesium oxide" — and this word equation is then translated into chemical formulas:

The arrow, $\longrightarrow$, is read "reacts to form" or "yields," and separates the reactants (left) from the products (right). Writing the correct formula for every species — $\text{O}_2$ for elemental oxygen, not $\text{O}$; $\text{MgO}$ for magnesium oxide, found from the ionic charges $\text{Mg}^{2+}$ and $\text{O}^{2-}$ — is a prerequisite that must be settled correctly before any balancing is attempted. As written above, the equation is not yet balanced: there are two oxygen atoms on the left and only one on the right, which Section 2 corrects.

Key distinction. A correctly written equation (right formulas, right arrow) is a separate requirement from a correctly balanced equation (equal atom counts on both sides). An equation can fail at either stage independently, and both must be checked.

2. Balancing by Inspection

Balancing an equation means placing a whole-number coefficient in front of each formula so that the number of atoms of every element is identical on both sides, exactly as the law of conservation of mass requires. The standard method, balancing by inspection, is systematic rather than random trial and error:

1. Write the unbalanced equation with correct formulas for every reactant and product.
2. Count the atoms of each element on both sides.
3. Balance one element at a time, usually starting with whichever element appears in the fewest formulas, and generally leaving free (uncombined) elements until last.
4. Adjust coefficients, never subscripts. A coefficient multiplies an entire formula; changing a subscript would change the identity of the substance itself.
5. Check every element again, since balancing one element can unbalance another that was already correct.
6. Reduce all coefficients to the smallest possible whole-number ratio, if they share a common factor.

Warning. Never balance an equation by editing a subscript. Changing $\text{O}_2$ to $\text{O}_3$ to "fix" an atom count does not rebalance the equation — it silently swaps oxygen gas for ozone, an entirely different substance, and breaks the formula that Chapter 1 and Chapter 2 worked to establish correctly.

Worked example. Balance $\text{Mg} + \text{O}_2 \longrightarrow \text{MgO}$.

Step 1 — count atoms on each side as written: left has $1\,\text{Mg}$, $2\,\text{O}$; right has $1\,\text{Mg}$, $1\,\text{O}$.

Step 2 — oxygen is unbalanced; place a coefficient of $2$ on the product to match the two oxygen atoms on the left:

Step 3 — re-check: oxygen is now balanced ($2$ and $2$), but magnesium is not ($1$ on the left, $2$ on the right). Place a coefficient of $2$ on the reactant magnesium:

Step 4 — final check: $2\,\text{Mg}$ and $2\,\text{O}$ on each side. The equation is balanced, and the coefficients $2,1,2$ share no common factor greater than $1$, so this is the simplest form.

3. Balancing Equations Involving Polyatomic Groups

When a polyatomic ion or group (such as $\text{NO}_3^{-}$, $\text{SO}_4^{2-}$ or $\text{OH}^{-}$) survives unchanged from one side of the equation to the other, it is often faster and less error-prone to balance it as a single unit rather than atom by atom.

Worked example. Balance $\text{Ca(OH)}_2 + \text{HNO}_3 \longrightarrow \text{Ca(NO}_3\text{)}_2 + \text{H}_2\text{O}$.

Step 1 — count the units as written: left has $1\,\text{Ca}$, $2\,\text{OH}$ (treated as a unit), $1\,\text{H}$ and $1\,\text{NO}_3$ (from $\text{HNO}_3$); right has $1\,\text{Ca}$, $2\,\text{NO}_3$, $2\,\text{H}$ and $1\,\text{O}$ (inside $\text{H}_2\text{O}$).

Step 2 — balance the nitrate group first, since it appears intact on both sides: the right side has $2\,\text{NO}_3^{-}$ (from $\text{Ca(NO}_3\text{)}_2$), so the left needs $2\,\text{HNO}_3$:

Step 3 — balance hydrogen and oxygen, most simply by balancing water: the left now supplies $2\,\text{OH}$ and $2\,\text{H}$ (from $2\text{HNO}_3$), a total of $4\,\text{H}$ and $2\,\text{O}$ available to form water; placing a coefficient of $2$ on water balances both at once:

Step 4 — final check: $1\,\text{Ca}$, $2\,\text{O}$ (from $\text{OH}$) $+ 4\,\text{H}$ (combined) $+ 2\,\text{N}$, $6\,\text{O}$ (from nitrate) on the left matches $1\,\text{Ca}$, $2\,\text{N}$, $6\,\text{O}$ (nitrate) $+ 4\,\text{H}$, $2\,\text{O}$ (water) on the right. Every element balances.

4. Balancing a Combustion Equation

Combustion of a hydrocarbon or alcohol in excess oxygen, producing only $\text{CO}_2$ and $\text{H}_2\text{O}$, is one of the most frequently balanced equation types, and benefits from a fixed order: balance carbon, then hydrogen, then oxygen last (since oxygen appears in both products, it is easiest to leave until the other two are fixed).

Worked example. Balance the complete combustion of propane, $\text{C}_3\text{H}_8 + \text{O}_2 \longrightarrow \text{CO}_2 + \text{H}_2\text{O}$.

Step 1 — balance carbon: three carbon atoms on the left require a coefficient of $3$ on $\text{CO}_2$:

Step 2 — balance hydrogen: eight hydrogen atoms on the left require a coefficient of $4$ on $\text{H}_2\text{O}$ (since each water molecule carries two hydrogens, $4\times 2 = 8$):

Step 3 — balance oxygen last: the right side now has $3(2) + 4(1) = 10$ oxygen atoms in total, so the left needs five $\text{O}_2$ molecules, each supplying two oxygen atoms:

Step 4 — final check: $3\,\text{C}$, $8\,\text{H}$, $10\,\text{O}$ on each side. The coefficients $1, 5, 3, 4$ share no common factor, so this is the balanced equation in its simplest form.

5. Fractional Coefficients and Clearing Them

Balancing by inspection sometimes produces a coefficient that is most naturally a fraction, typically $\tfrac{1}{2}$. Such a fraction is not a mistake: a fractional coefficient is a fully and correctly balanced equation, and later in the course, when thermochemistry defines an enthalpy change per mole of fuel burned, the fractional form is exactly the one required. For a standalone equation in this chapter, however, the convention — and what an exam mark scheme expects, unless a question specifically asks for the equation "per mole of fuel" — is to report the smallest set of whole-number coefficients, obtained by scaling every coefficient in the equation up together.

Worked example. Balance the combustion of ethanol, $\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \longrightarrow \text{CO}_2 + \text{H}_2\text{O}$.

Balancing carbon and hydrogen first gives $\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \longrightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$. The right side now has $2(2) + 3(1) = 7$ oxygen atoms in total, while the left already supplies one oxygen atom from the ethanol itself, leaving $6$ oxygen atoms still needed from $\text{O}_2$ — exactly $3\,\text{O}_2$ molecules:

This particular example happened not to require a fraction, but had the oxygen requirement instead come out odd (as it does, for instance, in the combustion of methanol, $\text{CH}_3\text{OH}$, before clearing), the standard procedure is to use the fractional coefficient $\tfrac{3}{2}\text{O}_2$ first and then multiply every coefficient in the whole equation by $2$ to obtain the smallest whole-number form — never to round a fraction to the nearest whole number, which would silently break the atom balance.

Warning. A fractional coefficient is not wrong — it is a genuinely balanced equation — but it is also not yet in the whole-number form this course (and most exam mark schemes) expects for a standalone equation. Convert it by multiplying every coefficient in the equation by the fraction's denominator; never round it away, and never discard the fractional form as if it were meaningless, since it is exactly the form needed later for enthalpy changes quoted per mole of fuel.

6. State Symbols

A fully specified equation also records the physical state of every reactant and product using a state symbol in parentheses immediately after its formula: $(s)$ solid, $(l)$ liquid, $(g)$ gas, and $(aq)$ aqueous (dissolved in water).

State symbols carry real chemical information beyond the atom count — the same two formulas can describe different physical situations, and a precipitation reaction (Section 7) is identified specifically by a solid state symbol appearing on the product side of an equation whose reactants were both dissolved.

7. Ionic Equations and Spectator Ions

When a reaction occurs between aqueous ionic compounds, writing out the full formulas can obscure which ions actually participate in forming a new substance. A full ionic equation instead writes every soluble ionic compound as its separated ions, while any insoluble solid, liquid, or gas is still written as a whole formula.

Worked example. Silver nitrate solution is mixed with sodium chloride solution, forming a precipitate of silver chloride.

Full (molecular) equation:

Full ionic equation, separating every aqueous ionic compound into its ions:

Inspecting this equation shows that $\text{Na}^{+}(aq)$ and $\text{NO}_3^{-}(aq)$ appear, unchanged, on both sides. Such an ion is called a spectator ion: present in solution, but taking no part in forming the product. Cancelling spectator ions from both sides leaves the net ionic equation, which states only the chemistry that actually happened:

The net ionic equation is still subject to the same balancing requirement as any other equation — but it must additionally balance charge, not only atoms: the left side carries a net charge of $(+1) + (-1) = 0$, matching the neutral solid on the right. Checking that the net charge is equal on both sides is an essential, additional check specific to ionic equations, with no counterpart in a purely molecular equation.

Key distinction. A full ionic equation separates soluble ionic compounds into ions; solids, liquids, gases and weak (largely undissociated) substances are written as whole formulas throughout, in both the full and the net ionic equation.

8. Interpreting the Coefficients of a Balanced Equation

Once balanced, the coefficients of a chemical equation state the mole ratio in which reactants combine and products form — directly, because a coefficient simply counts how many formula units (or molecules) of a substance are involved, and the mole, from Chapter 1, is nothing more than a fixed-size bundle of such units.

states that two moles of magnesium react with exactly one mole of oxygen gas to produce two moles of magnesium oxide — not that two grams of one react with one gram of the other. This mole-ratio reading of a balanced equation is the single idea that the next chapter builds into full stoichiometric calculations of reacting masses and reaction yield; this chapter's job is only to make sure the ratio itself is written correctly.

9. Common Exam Pitfalls

• Changing a subscript instead of a coefficient. Only the number placed in front of an entire formula may be adjusted; altering a subscript changes which substance is being described.
• Balancing an element that appears as a free element too early. Leaving $\text{O}_2$, $\text{H}_2$, or another elemental reactant until last usually avoids having to re-balance it repeatedly as other elements are fixed.
• Forgetting to re-check every element after each adjustment. Balancing one element frequently unbalances another that was previously correct; the check in step 5 of the inspection method is not optional.
• Rounding a fractional coefficient instead of converting it to whole numbers. A fractional coefficient such as $\tfrac{3}{2}$ is genuinely balanced, not wrong, but for a standalone equation it should be converted to whole numbers by doubling every coefficient — never rounded to $1$ or $2$, which would break the balance.
• Treating an insoluble solid or a gas as if it should be split into ions. Only soluble ionic compounds in aqueous solution are separated into ions in an ionic equation; solids, liquids and gases are always written as whole formulas.
• Balancing atoms in an ionic equation while forgetting to balance charge. A net ionic equation must have the same total charge on both sides, in addition to the same number of atoms of each element.

Summary

• A chemical equation shows reactants, an arrow read "reacts to form," and products, each written with a chemically correct formula — a separate requirement from the equation being balanced.
• Balancing by inspection adjusts whole-number coefficients, never subscripts, one element at a time, re-checking every element after each change, until atom counts match on both sides and the coefficients share no common factor.
• Polyatomic ions that survive a reaction unchanged, and the carbon–hydrogen–oxygen order of a combustion equation, are both strategies that make balancing faster and less error-prone.
• A fractional coefficient is a valid, fully balanced equation (and is the required form later for enthalpy changes per mole of fuel); for a standalone equation it is converted to the smallest whole-number form, never rounded, by multiplying every coefficient by its denominator.
• State symbols $(s)$, $(l)$, $(g)$, $(aq)$ record the physical state of every species and are needed to identify, for instance, a precipitate forming from solution.
• A full ionic equation separates soluble ionic compounds into ions; cancelling the spectator ions that appear unchanged on both sides gives the net ionic equation, which must balance both atoms and net charge.
• The coefficients of a balanced equation state a mole ratio between reactants and products — the exact idea the next chapter extends into full calculations of reacting mass and reaction yield.