Difficulty: Introductory | Prerequisites: ability to write chemical formulas, understanding of subscripts vs. coefficients.
Balancing equations is one of the foundational skills in chemistry. You will use it in every subsequent topic, from stoichiometry to equilibrium. This set of notes walks through the method and provides worked examples drawn from the Lab 10 report sheet and problem set.
Balancing a chemical equation means adjusting the coefficients so that the same number of each type of atom appears on both sides of the arrow. This follows the law of conservation of mass: atoms are not created or destroyed in a chemical reaction. Once you can balance equations reliably, you can do stoichiometry, predict yields, and work through nearly every quantitative problem in general chemistry.
Coefficient
The large number placed in front of a chemical formula in a balanced equation. It multiplies every atom in that formula. Think of it as: the number of "units" of that substance taking part. 2 H₂O means two water molecules, so four H atoms and two O atoms total.
Subscript
The small number written to the lower right of an element symbol within a chemical formula. It tells you how many atoms of that element are in one unit of the substance. In simple terms, subscripts are locked. You never change them when balancing. Changing a subscript changes the substance itself.
Law of conservation of mass
Matter cannot be created or destroyed in a chemical reaction. The total mass of the reactants equals the total mass of the products. Think of it as: whatever atoms you start with, you must end with. That is why equations must balance.
State symbols
(s) = solid, (l) = liquid, (g) = gas, (aq) = dissolved in water (aqueous). Written as subscript labels after each formula in a balanced equation. In simple terms, these tell you the physical form of each substance. They matter for predicting whether a precipitate forms (a solid product from aqueous reactants) and for lab observations.
Reactants and products
Reactants are the starting substances (left side of the arrow). Products are the substances formed (right side of the arrow). Think of it as: ingredients (reactants) go in, the finished dish (products) comes out.
Write the unbalanced equation with correct formulas and state symbols.
List each element and count atoms on each side.
Balance one element at a time by adjusting coefficients. Start with the element that appears in the fewest formulas, or the most complex formula.
Leave hydrogen and oxygen for last (they tend to appear in multiple substances).
For combustion: balance C first, then H, then O last.
Check every element. If any coefficient is a fraction, multiply all coefficients by the denominator to clear it.
Reduce to lowest whole-number coefficients.
Unbalanced: Al(s) + O₂(g) → Al₂O₃(s)
Al: 1 on left, 2 on right. Put a 4 in front of Al: 4 Al.
O: 2 on left, 3 on right. Put a 3 in front of O₂: 3 O₂ = 6 O atoms.
Now right side has 3 in Al₂O₃ = 6 O. Needs 2 Al₂O₃ to get 4 Al. Check: 2 Al₂O₃ = 4 Al, 6 O.
Balanced: 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
Unbalanced: C₅H₁₂(l) + O₂(g) → CO₂(g) + H₂O(g)
C: 5 on left, 1 on right. Put 5 CO₂.
H: 12 on left, 2 on right. Put 6 H₂O.
O: count right side: 5(2) + 6(1) = 16 O atoms needed. Left side: O₂ gives 2 per molecule, so 16/2 = 8 O₂.
Balanced: C₅H₁₂(l) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g)
Unbalanced: Fe₂O₃(s) + H₂O(l) → Fe(OH)₃(s)
Fe: 2 on left, 1 on right. Put 2 Fe(OH)₃.
O: left has 3 (from Fe₂O₃) + 1 per H₂O. Right has 2 × 3 = 6 O in OH groups. Total O on right = 6. Left needs 6 total: 3 from Fe₂O₃ + 3 from H₂O = 3 H₂O.
H: left 3 × 2 = 6. Right 2 × 3 = 6. Balanced.
Balanced: Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s)
Unbalanced: Ca(OH)₂(aq) + HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l)
Ca: 1 and 1. Already balanced.
NO₃: 1 on left, 2 on right. Put 2 HNO₃.
H: left 2(1) from Ca(OH)₂ + 2(1) from 2 HNO₃ = 4. Right 2 in H₂O, so put 2 H₂O.
O: check all O atoms. Left: 2 (from OH) + 6 (from 2 NO₃) = 8. Right: 6 (from 2 NO₃) + 2 (from 2 H₂O) = 8. Balanced.
Balanced: Ca(OH)₂(aq) + 2 HNO₃(aq) → Ca(NO₃)₂(aq) + 2 H₂O(l)
Polyatomic ions (SO₄²⁻, NO₃⁻, PO₄³⁻): if the ion appears intact on both sides, count it as a unit rather than splitting it into individual atoms. This saves time.
If you end up with fractional coefficients, multiply everything through. For example, a coefficient of 3/2 on O₂ means you multiply all coefficients by 2.
When the problem says "write the correct formulas," you must determine the products yourself before balancing. Know your reaction types: that is how you predict what forms.
Reaction | Balanced Equation | Type |
|---|---|---|
Mg + O₂ | 2 Mg(s) + O₂(g) → 2 MgO(s) | Combination |
Zn + CuSO₄ | Zn(s) + CuSO₄(aq) → Cu(s) + ZnSO₄(aq) | Single replacement |
Zn + HCl | Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g) | Single replacement |
Cu + HCl | Cu(s) + HCl(aq) → NR | No reaction |
Mg + HCl | Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g) | Single replacement |
CaCl₂ + Na₃PO₄ | 3 CaCl₂(aq) + 2 Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6 NaCl(aq) | Double replacement |
FeCl₃ + KSCN | FeCl₃(aq) + 3 KSCN(aq) → Fe(SCN)₃(aq) + 3 KCl(aq) | Double replacement |
Na₂CO₃ + HCl | Na₂CO₃(s) + 2 HCl(aq) → CO₂(g) + H₂O(l) + 2 NaCl(aq) | Double replacement + decomposition |
H₂O₂ decomposition | 2 H₂O₂(l) → 2 H₂O(l) + O₂(g) | Decomposition |
Mg + HCl (Q2a) | Mg(s) + 2 HCl(aq) → H₂(g) + MgCl₂(aq) | Single replacement |
Al + O₂ (Q2b) | 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s) | Combination |
Fe₂O₃ + H₂O (Q2c) | Fe₂O₃(s) + 3 H₂O(l) → 2 Fe(OH)₃(s) | Combination |
Ca(OH)₂ + HNO₃ (Q2d) | Ca(OH)₂(aq) + 2 HNO₃(aq) → Ca(NO₃)₂(aq) + 2 H₂O(l) | Double replacement |
Pentane combustion (Q3a) | C₅H₁₂(l) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g) | Combustion |
Na + H₂O (Q3b) | 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) | Single replacement |
Fe + O₂ (Q3c) | 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) | Combination |
Engineers use balanced equations to calculate how much fuel a rocket needs and how much exhaust it will produce. The stoichiometric ratios come straight from the balanced equation.
Pharmaceutical manufacturing relies on balanced equations to determine the exact masses of reagents needed to synthesise a drug, minimising waste and cost.
Environmental scientists balance combustion equations to estimate CO₂ emissions from burning fossil fuels, which feeds directly into carbon footprint calculations.
Students change subscripts instead of coefficients when balancing. Changing H₂O to H₂O₂ does not balance oxygen; it turns water into hydrogen peroxide. Never touch subscripts.
Students sometimes forget to multiply subscripts by the coefficient. 2 Fe₂O₃ contains 4 Fe atoms and 6 O atoms, not 2 and 3.
A common error is balancing atoms inside a polyatomic ion individually. If SO₄ appears intact on both sides, count it as one unit. Only break it into S and O when it does not survive the reaction intact.
Students occasionally think a coefficient of 1 must be written. It is implied and should not appear in a final balanced equation.
⚠️ Balancing equations is tested on virtually every general chemistry exam, often as a standalone question and as a prerequisite step in stoichiometry problems.
⚠️ State symbols are frequently required for full marks. Know (s), (l), (g), (aq) and when each applies.
⚠️ "Write and balance" questions require you to determine the correct products first (from the reaction type), then balance. Missing the product formula means the entire equation is wrong.
⚠️ Combustion equations are a favourite for timed exams because balancing oxygen at the end can produce fractions that need clearing. Practise these until they are fast.
True or False: You can change a subscript to balance an equation. (False. Only coefficients are changed.)
Fill in the blank: 2 H₂O₂ → 2 H₂O + ______. (O₂)
True or False: The coefficient 1 should be written explicitly in a balanced equation. (False. It is implied.)
Fill in the blank: In the equation 4 Al + 3 O₂ → 2 Al₂O₃, there are ______ aluminium atoms on each side. (4)
True or False: When balancing a combustion reaction, you should balance oxygen first. (False. Balance C first, then H, then O last.)
Q: Balance the following equation: Fe₂O₃(s) + C(s) → Fe(s) + CO₂(g)
A: 2 Fe₂O₃(s) + 3 C(s) → 4 Fe(s) + 3 CO₂(g). Start with Fe (2 on left per formula, 1 on right, so 2 Fe₂O₃ gives 4 Fe), then balance C and O.
Q: Write a balanced equation for sodium reacting with water to form sodium hydroxide and hydrogen gas. Include state symbols.
A: 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
Q: Balance: C₂H₄(g) + O₂(g) → CO₂(g) + H₂O(g)
A: C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(g). C: 2 = 2. H: 4 = 4. O: 6 = 4 + 2 = 6.
Q: A student writes Mg + HCl → MgCl + H. What errors have they made?
A: Two errors. First, the product should be MgCl₂ (magnesium has a 2+ charge, chloride is 1−, so two chlorides are needed). Second, hydrogen gas is diatomic: H₂, not H. The correct balanced equation is Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g).
Q: How many total oxygen atoms are on each side of: 3 CaCl₂(aq) + 2 Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6 NaCl(aq)?
A: Left: 2 Na₃PO₄ contains 2 × 4 = 8 oxygen atoms. Right: Ca₃(PO₄)₂ contains 2 × 4 = 8 oxygen atoms. Both sides have 8.
Balanced equations are the entry point for stoichiometry: once the equation is balanced, the coefficients become mole ratios, which let you calculate how much product forms from a given mass of reactant. This leads directly into limiting reagent problems (which reactant runs out first) and percent yield calculations. Later, in equilibrium and thermodynamics, balanced equations define the equilibrium expression and the enthalpy change for a reaction.
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