Why can't I change subscripts when balancing an equation?

Subscripts define what a substance is. Changing H_2O to H_2O_2 or CO_2 to CO replaces the actual product with a different chemical. Coefficients only say how many units of that unchanged substance take part, which is exactly what conservation of atoms lets you adjust.

What does it mean to use the smallest whole-number coefficients?

After the atoms balance, check whether every coefficient shares a common factor. If they do, divide them all by that factor. For example 4\,H_2 + 2\,O_2 arrow 4\,H_2O is correctly balanced but reduces to 2\,H_2 + O_2 arrow 2\,H_2O, which is the accepted form. Coefficients of 1 are never written out.

Is it acceptable to use a fraction as a coefficient?

As a temporary step, yes — fractions like 72 for O_2 are a fast way to finish a combustion equation. But a final answer should have whole numbers, so multiply every coefficient by the denominator before you stop. Fractional coefficients are only left in place in certain thermochemistry contexts, which come later.

How do I handle polyatomic ions like sulfate or nitrate?

If the ion appears unchanged on both sides, count it as one unit rather than counting its atoms separately. In Al_2(SO_4)_3 there are three sulfate groups, so balancing three sulfates at once is far quicker than tracking 3 S and 12 O. If the ion breaks apart during the reaction, fall back on counting individual atoms.

CHEM-5.1

Chemical Equations & Balancing

Learn to turn word descriptions of reactions into balanced chemical equations using conservation of atoms, atom inventories, and smallest whole-number coefficients.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Chemical Equations & Balancing, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

A chemical equation is a sentence written in the language of chemistry. Before you can predict products, calculate yields, or classify reactions, you have to be able to say exactly what went in, what came out, and in what proportion. That last part is where the real chemistry lives: atoms are never created or destroyed in a chemical change, so every atom that enters a reaction has to leave it. Balancing an equation is just bookkeeping that enforces that rule.

In this lesson you will translate word descriptions into symbolic equations with correct formulas and state labels, build an atom inventory to see what is out of balance, and adjust coefficients — never subscripts — until both sides match with the smallest whole numbers possible. These same balanced equations become the foundation for stoichiometry later in the course.

From Words to a Skeleton Equation

Start by identifying reactants (what you begin with) and products (what you end with). Reactants go on the left, products on the right, separated by an arrow that reads "yields" or "produces." Plus signs separate multiple substances on the same side.

Write each substance as a correct chemical formula first. This step is not negotiable: the formula comes from the identity of the substance, not from what would make balancing easier. Water is always

\mathrm{H_2O}

. Calcium chloride is always

\mathrm{CaCl_2}

because calcium forms a

2+

ion and chloride a

1-

ion. Seven elements exist as diatomic molecules in their pure elemental form:

\mathrm{H_2}

\mathrm{N_2}

\mathrm{O_2}

\mathrm{F_2}

\mathrm{Cl_2}

\mathrm{Br_2}

\mathrm{I_2}

. Writing "O" for oxygen gas is one of the most common early errors.

State symbols in parentheses tell the reader the physical form of each substance.

Symbol	Meaning	Typical use
$(s)$	solid	metals, precipitates, ionic solids
$(l)$	pure liquid	water, molten substances
$(g)$	gas	$\mathrm{O_2}$ , $\mathrm{CO_2}$ , $\mathrm{H_2}$
$(aq)$	dissolved in water	solutions, acids, most "aqueous" reagents

A skeleton equation has correct formulas and states but no coefficients yet. For example, "solid magnesium burns in oxygen gas to form solid magnesium oxide" becomes

\mathrm{Mg}(s) + \mathrm{O_2}(g) \rightarrow \mathrm{MgO}(s)

. It is a true statement about identity but not yet a true statement about quantity.

Conservation of Atoms: Why Balancing Is Required

A chemical reaction rearranges atoms; it does not manufacture or destroy them. Bonds break and re-form, but the same collection of nuclei is present before and after. This is the law of conservation of mass: the total mass of the reactants equals the total mass of the products in a closed system.

Look at the magnesium skeleton equation above. On the left there are two oxygen atoms (in

\mathrm{O_2}

); on the right there is only one. As written, the equation claims an oxygen atom vanished. Fixing it requires coefficients — whole numbers placed in front of formulas that tell you how many units of that substance participate:

2\,\mathrm{Mg}(s) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{MgO}(s)

Now both sides have two Mg atoms and two O atoms. A coefficient multiplies every atom in the formula that follows it, so

2\,\mathrm{MgO}

means two Mg and two O.

The single most damaging misconception in this unit is changing a subscript to force a balance. Turning

\mathrm{MgO}

into

\mathrm{MgO_2}

would balance the oxygen, but

\mathrm{MgO_2}

is a different substance that is not the product of the reaction. Subscripts describe what a compound is; coefficients describe how much of it reacts.

Change	What it alters	Allowed when balancing?
Coefficient (in front)	number of formula units	Yes
Subscript (inside formula)	identity of the compound	No

When a reaction seems to "lose mass," as when a candle burns, mass has escaped as gas, not disappeared. Seal the system and the balance reading does not change.

A Reliable Balancing Procedure

Balancing by inspection works almost every time if you follow a consistent order instead of guessing randomly.

First, build an atom inventory: list every element and tally how many atoms appear on each side of the skeleton equation. Second, balance elements that appear in only one compound on each side. Third, treat a polyatomic ion that survives the reaction intact — such as

\mathrm{NO_3^-}

\mathrm{SO_4^{2-}}

— as a single unit and balance it as a group instead of counting N and O separately. Fourth, balance pure elements such as

\mathrm{O_2}

\mathrm{H_2}

, or a bare metal last, because a coefficient on a lone element changes only one entry in your inventory. Fifth, recount everything.

A useful trick for combustion: if oxygen refuses to balance with a whole number, allow a fraction temporarily, then multiply every coefficient in the equation by the denominator. For example,

\mathrm{C_2H_6} + \tfrac{7}{2}\mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + 3\,\mathrm{H_2O}

becomes

2\,\mathrm{C_2H_6} + 7\,\mathrm{O_2} \rightarrow 4\,\mathrm{CO_2} + 6\,\mathrm{H_2O}

after multiplying through by 2.

Finally, reduce to the smallest whole-number set. The equation

4\,\mathrm{H_2} + 2\,\mathrm{O_2} \rightarrow 4\,\mathrm{H_2O}

conserves atoms perfectly, but every coefficient shares a factor of 2, so the accepted answer is

2\,\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{H_2O}

. A coefficient of 1 is understood and never written.

Where students go wrong most often: changing a coefficient early in the process and forgetting to update every element it affects. Recount the full inventory after each change, not just at the end.

Reading a Balanced Equation and Checking Your Work

Once balanced, an equation carries quantitative meaning. For

\mathrm{N_2}(g) + 3\,\mathrm{H_2}(g) \rightarrow 2\,\mathrm{NH_3}(g)

, the coefficients say that one molecule of nitrogen reacts with three molecules of hydrogen to make two molecules of ammonia. Scale that up by Avogadro's number and it also reads: one mole of

\mathrm{N_2}

reacts with three moles of

\mathrm{H_2}

to make two moles of

\mathrm{NH_3}

. That mole ratio is the bridge to every stoichiometry calculation you will do later.

Coefficients do not give mass ratios directly. One mole of

\mathrm{N_2}

has a mass of about 28 grams while three moles of

\mathrm{H_2}

have a mass of about 6 grams, so the masses are 28 and 6, not 1 and 3. Total mass is still conserved: about 34 grams of reactants produce about 34 grams of ammonia.

To verify a balanced equation, make a final two-column tally.

Element	Reactant atoms	Product atoms
N	$1 \times 2 = 2$	$2 \times 1 = 2$
H	$3 \times 2 = 6$	$2 \times 3 = 6$

Both rows match, and the coefficients 1, 3, 2 share no common factor, so the equation is complete. Also confirm that charge is conserved if you are writing ionic equations, and that no formula was quietly altered during the process — compare your final formulas letter by letter against the skeleton equation you started with.

Special Cases That Trip People Up

Hydrates, polyatomic ions, and compounds that appear on both sides create predictable trouble.

When an element shows up in more than two compounds on one side, save it for last. In the combustion of a compound containing carbon, hydrogen, and oxygen — say

\mathrm{C_2H_6O}

— oxygen appears in the fuel, in

\mathrm{O_2}

, in

\mathrm{CO_2}

, and in

\mathrm{H_2O}

. Balance carbon, then hydrogen, then count all oxygen atoms on the product side, subtract the oxygen already supplied by the fuel, and divide the remainder by 2 to get the coefficient for

\mathrm{O_2}

.

With polyatomic ions, count the whole group. In

\mathrm{Al_2(SO_4)_3}

the subscript 3 outside the parentheses multiplies everything inside: three sulfate groups, so 3 S and 12 O. Students frequently read this as 4 oxygen atoms and stall.

Some equations are already balanced as written, such as

\mathrm{CaCO_3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO_2}(g)

. Do not add coefficients out of habit — check the inventory first.

Watch also for water of hydration, written with a dot as in

\mathrm{CuSO_4 \cdot 5H_2O}

. That dot means five water molecules are part of the crystal, so this formula contributes 10 H and 9 O total.

Finally, remember that a balanced equation says nothing about whether the reaction actually happens, how fast it goes, or whether energy is absorbed or released. Balancing is a statement about atom counts only; the rest of the unit fills in those other questions.

Key terms

Chemical equation.: A symbolic representation of a reaction in which reactant formulas on the left are separated from product formulas on the right by an arrow meaning "yields."
Skeleton equation.: An equation with correct formulas and state symbols but no coefficients, so atom counts are not yet equal on both sides.
Coefficient.: A whole number written in front of a formula that multiplies every atom in that formula; it indicates how many formula units or moles react.
Subscript.: A small number inside a chemical formula giving the number of atoms of the preceding element in one unit of the compound; it defines the substance's identity and may never be changed to balance an equation.
Law of conservation of mass.: In a chemical reaction, matter is neither created nor destroyed, so the total mass of reactants equals the total mass of products in a closed system.
Atom inventory.: A tally of how many atoms of each element appear on the reactant side and the product side, used to identify what still needs balancing.
Diatomic element.: An element that exists as two bonded atoms in its pure form: $\mathrm{H_2}$ , $\mathrm{N_2}$ , $\mathrm{O_2}$ , $\mathrm{F_2}$ , $\mathrm{Cl_2}$ , $\mathrm{Br_2}$ , and $\mathrm{I_2}$ .
Mole ratio.: The proportion between substances given by the coefficients of a balanced equation, used to relate amounts of reactants and products.

Worked example

Ethane gas,

\mathrm{C_2H_6}

, burns completely in oxygen gas to produce carbon dioxide gas and liquid water. Write the balanced chemical equation with state symbols and the smallest whole-number coefficients.

Step 1 — Write the skeleton equation with correct formulas. Ethane is

\mathrm{C_2H_6}

, oxygen gas is diatomic, and complete combustion of a hydrocarbon always gives

\mathrm{CO_2}

and

\mathrm{H_2O}

\mathrm{C_2H_6}(g) + \mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + \mathrm{H_2O}(l)

Step 2 — Take an atom inventory. Left: 2 C, 6 H, 2 O. Right: 1 C, 2 H, 3 O. Nothing matches.

Step 3 — Balance carbon first, since it appears in one compound on each side. Two carbons on the left require

2\,\mathrm{CO_2}

on the right.

Step 4 — Balance hydrogen next. Six hydrogens on the left require

3\,\mathrm{H_2O}

on the right, since each water has 2 H and

3 \times 2 = 6

.

Step 5 — Balance oxygen last. The product side now holds

2 \times 2 = 4

oxygen atoms in

\mathrm{CO_2}

plus

3 \times 1 = 3

in water, for 7 oxygen atoms total. Since

\mathrm{O_2}

supplies atoms two at a time, the coefficient must be

\tfrac{7}{2}

\mathrm{C_2H_6}(g) + \tfrac{7}{2}\,\mathrm{O_2}(g) \rightarrow 2\,\mathrm{CO_2}(g) + 3\,\mathrm{H_2O}(l)

Step 6 — Clear the fraction by multiplying every coefficient by 2:

2\,\mathrm{C_2H_6}(g) + 7\,\mathrm{O_2}(g) \rightarrow 4\,\mathrm{CO_2}(g) + 6\,\mathrm{H_2O}(l)

Step 7 — Verify. Carbon:

2 \times 2 = 4

on the left,

4 \times 1 = 4

on the right. Hydrogen:

2 \times 6 = 12

on the left,

6 \times 2 = 12

on the right. Oxygen:

7 \times 2 = 14

on the left,

(4 \times 2) + (6 \times 1) = 14

on the right. The coefficients 2, 7, 4, 6 share no common factor greater than 1, so this is the final answer.

Practice questions

Which set of coefficients correctly balances

\mathrm{Fe} + \mathrm{O_2} \rightarrow \mathrm{Fe_2O_3}

using the smallest whole numbers?

$2\,\mathrm{Fe} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Fe_2O_3}$
$4\,\mathrm{Fe} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Fe_2O_3}$
$2\,\mathrm{Fe} + \mathrm{O_2} \rightarrow \mathrm{Fe_2O_3}$
$8\,\mathrm{Fe} + 6\,\mathrm{O_2} \rightarrow 4\,\mathrm{Fe_2O_3}$

Answer: $4\,\mathrm{Fe} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Fe_2O_3}$

Balance oxygen first by finding a common multiple of 2 (from

\mathrm{O_2}

) and 3 (from

\mathrm{Fe_2O_3}

), which is 6. That gives

3\,\mathrm{O_2}

and

2\,\mathrm{Fe_2O_3}

. Two units of

\mathrm{Fe_2O_3}

contain

2 \times 2 = 4

iron atoms, so the coefficient on Fe must be 4. Check: 4 Fe and 6 O on each side. The first option leaves iron unbalanced (2 versus 4), the third leaves both elements unbalanced, and the last one is correctly balanced but every coefficient is doubled, so it is not the smallest whole-number set.

A student is asked to balance

\mathrm{H_2} + \mathrm{O_2} \rightarrow \mathrm{H_2O}

and writes

\mathrm{H_2} + \mathrm{O_2} \rightarrow \mathrm{H_2O_2}

, explaining that now both sides have 2 H and 2 O. Explain what is wrong with this reasoning and give the correct balanced equation.

Answer: The student changed a subscript, which changes the product from water into hydrogen peroxide — a completely different substance. Subscripts may never be altered when balancing; only coefficients may be adjusted. The correct equation is $2\,\mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{H_2O}(l)$ .

Atom counts are only half the requirement. A balanced equation must also describe the reaction that actually occurs, and the formula of a compound is fixed by its chemical identity.

\mathrm{H_2O_2}

is hydrogen peroxide, a corrosive liquid, not water. Adjusting coefficients instead: putting 2 in front of

\mathrm{H_2O}

gives 4 H and 2 O on the right, and putting 2 in front of

\mathrm{H_2}

gives 4 H and 2 O on the left. Both sides match and 2, 1, 2 have no common factor.

Translate the following into a balanced chemical equation with state symbols: solid sodium reacts with liquid water to produce aqueous sodium hydroxide and hydrogen gas.

Answer: $2\,\mathrm{Na}(s) + 2\,\mathrm{H_2O}(l) \rightarrow 2\,\mathrm{NaOH}(aq) + \mathrm{H_2}(g)$

Sodium metal is written as the single atom

\mathrm{Na}

because metals are not diatomic; hydrogen gas must be

\mathrm{H_2}

. The skeleton equation

\mathrm{Na} + \mathrm{H_2O} \rightarrow \mathrm{NaOH} + \mathrm{H_2}

has 1 Na, 2 H, and 1 O on the left but 1 Na, 3 H, and 1 O on the right. Treat hydroxide as a unit: putting 2 in front of both

\mathrm{H_2O}

and

\mathrm{NaOH}

gives 2 O on each side and 2 OH groups accounted for. That leaves 4 H on the left, and the right side has 2 H in the hydroxides plus 2 H in

\mathrm{H_2}

, which is 4. Finally set the Na coefficient to 2. Final check: 2 Na, 4 H, 2 O on both sides.

FAQ

Why can't I change subscripts when balancing an equation?: Subscripts define what a substance is. Changing $\mathrm{H_2O}$ to $\mathrm{H_2O_2}$ or $\mathrm{CO_2}$ to $\mathrm{CO}$ replaces the actual product with a different chemical. Coefficients only say how many units of that unchanged substance take part, which is exactly what conservation of atoms lets you adjust.
What does it mean to use the smallest whole-number coefficients?: After the atoms balance, check whether every coefficient shares a common factor. If they do, divide them all by that factor. For example $4\,\mathrm{H_2} + 2\,\mathrm{O_2} \rightarrow 4\,\mathrm{H_2O}$ is correctly balanced but reduces to $2\,\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{H_2O}$ , which is the accepted form. Coefficients of 1 are never written out.
Is it acceptable to use a fraction as a coefficient?: As a temporary step, yes — fractions like $\tfrac{7}{2}$ for $\mathrm{O_2}$ are a fast way to finish a combustion equation. But a final answer should have whole numbers, so multiply every coefficient by the denominator before you stop. Fractional coefficients are only left in place in certain thermochemistry contexts, which come later.
How do I handle polyatomic ions like sulfate or nitrate?: If the ion appears unchanged on both sides, count it as one unit rather than counting its atoms separately. In $\mathrm{Al_2(SO_4)_3}$ there are three sulfate groups, so balancing three sulfates at once is far quicker than tracking 3 S and 12 O. If the ion breaks apart during the reaction, fall back on counting individual atoms.

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