Balancing Chemical Equations

Chemistry studies the changes matter undergoes, and every chemical equation you write has to obey one basic rule: atoms don't appear or disappear during a reaction. Balancing an equation — and using that balanced equation to calculate moles, mass, and yield — is how you make that rule work for you on problems.

Key Takeaways

• A physical change preserves chemical composition (e.g., phase changes); a chemical change rearranges atoms into new substances.

• The law of conservation of mass: atoms are neither created nor destroyed in a reaction, so equations must be balanced — matching numbers of each atom on both sides.

• To balance an equation: identify reactants/products/states, write the unbalanced equation, balance one atom at a time (most complex substance first, simplest last), use the smallest whole-number coefficients, and check every atom.

• A mole is 6.022 × 10²³ particles (Avogadro's number); it acts as a conversion factor, relates to mass via atomic/molar weight, and provides mole-to-mole ratios from a formula's subscripts or a balanced equation's coefficients.

• The stoichiometric conversion pathway: mass of X → moles of X (÷ molar mass) → moles of Y (× mole ratio) → mass of Y (× molar mass) → optionally, particles of Y (× Avogadro's number).

• The limiting reagent is consumed first and caps product formation; the excess reagent is left over.

• Theoretical yield is the stoichiometrically predicted amount; actual yield is what's actually obtained; percent yield = (actual ÷ theoretical) × 100%.

Physical Changes vs. Chemical Changes

A substance that undergoes a physical change keeps its chemical composition. Phase changes are the clearest example: when water freezes into ice or vaporizes into steam, it's still H₂O — it has simply taken on a different phase.

A chemical change is different: the atoms present rearrange to form new substances entirely. Signs that a chemical change has occurred include a color change, bubbling (gas formation), a change in odor or temperature, or the formation of a precipitate.

The Law of Conservation of Mass

Whether a change is physical or chemical, the law of conservation of mass always holds: there is no detectable change in mass over the course of a chemical reaction. Atoms are neither created nor destroyed — they simply rearrange into new arrangements.

This is exactly why chemical equations must be balanced: the number of atoms of each element on the reactant side has to match the number of atoms of that element on the product side. If a reaction has 6 oxygen atoms among the reactants, it needs 6 oxygen atoms among the products, too.

Steps to Balance a Chemical Equation

Balancing is done by placing coefficients in front of chemical formulas to indicate the relative amounts of reactants and products — this is different from the subscripts inside a formula, which are fixed by the substance's identity and can't be changed to balance an equation.

To balance an equation:

1. Determine what reaction is occurring. Identify the reactants, the products, and their physical states.

2. Write the unbalanced equation that summarizes the reaction. (Steps 1 and 2 matter most when you're not handed a ready-made equation — sometimes you're given a description of a reaction and have to use nomenclature to write it out yourself.)

3. Balance one atom at a time, starting with the substance that has the most elements present, and saving the simplest substance for last.

4. Use the smallest whole-number coefficients that balance the equation.

5. Check every atom to confirm the equation is fully balanced before moving on.

Worked Example: Balancing a Combustion Equation

Worked example: Balance the combustion of propane gas (C₃H₈) in oxygen.

Unbalanced: C₃H₈ + O₂ → CO₂ + H₂O

Propane is the substance with the most elements present (carbon and hydrogen), so balance it first, saving the simplest species (O₂) for last. There are 3 carbons on the left, so place a 3 in front of CO₂. There are 8 hydrogens on the left, so place a 4 in front of H₂O (since each H₂O carries 2 hydrogens):

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

Now balance oxygen last. The right side has 3 × 2 = 6 oxygens from CO₂ plus 4 × 1 = 4 oxygens from H₂O, for a total of 10. Since O₂ supplies 2 oxygens per molecule, place a 5 in front of O₂:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Checking every atom: 3 C = 3 C, 8 H = 8 H, and 10 O = 10 O on both sides. Balanced.

Moles and Mole-to-Mole Ratios

A mole is a unit of measurement chemists use to express an amount of a chemical substance. One mole of anything contains 6.022 × 10²³ particles — a value known as Avogadro's number.

Three points about the mole are worth keeping straight:

• The mole is a "number word" — it functions as a conversion factor in dimensional analysis, the same way "dozen" or "pair" would.

• The mole relates to mass: one mole of an element has a mass, in grams, equal to that element's atomic weight.

• The mole concept provides mole-to-mole ratios, which let you relate quantities of reactants and products to each other. There are two places to read a mole-to-mole ratio from: the subscripts within a chemical formula, or the coefficients of a balanced equation.

Stoichiometric Conversions: Mass, Moles, and Particles

Mole-to-mole ratios are what make it possible to move between the mass of one substance in a reaction and the mass (or particle count) of another. The conversion pathway from a given mass of substance X to a related quantity of substance Y runs as follows:

1. Mass of X → moles of X: divide the given mass of X by X's molar mass (in g/mol).

2. Moles of X → moles of Y: multiply by the mole-to-mole ratio read from the coefficients of the balanced equation (e.g., if the equation shows 1 mole of X producing 2 moles of Y, the ratio is 1:2).

3. Moles of Y → mass of Y: multiply the moles of Y by Y's molar mass.

4. Moles of Y → number of particles of Y (optional): multiply by Avogadro's number (6.022 × 10²³ particles/mol).

This mass–mole–particle pathway is the systematic backbone for essentially any stoichiometric calculation: given some quantity of one substance in a reaction, it lets you determine the corresponding quantity of any other substance in that same reaction.

Limiting and Excess Reagents

In most real reactions, one reactant runs out completely before the other one does. The reactant that gets completely consumed is the limiting reagent — it caps how much product the reaction can form. The reactant that isn't fully used up is the excess reagent.

A simple analogy: suppose you have four burger buns and three hamburger patties. Even though you have four buns, you can only make three complete burgers, because you only have three patties. The patties are the limiting reagent — they run out first and limit total production to three burgers. The one leftover bun makes buns the excess reagent.

Theoretical, Actual, and Percent Yield

Using stoichiometry to predict how much product a reaction should produce gives you the theoretical yield — the amount you'd get if everything went perfectly, with no losses during preparation or transfer.

The amount of product actually collected is the actual yield, and it's almost always somewhat less than the theoretical yield, due to ordinary human and instrument error.

The percent yield relates the two:

Percent Yield = (Actual Yield ÷ Theoretical Yield) × 100%

Worked Example: Percent Yield in Aspirin Synthesis

Worked example: A student synthesizes aspirin (acetylsalicylic acid) from salicylic acid (SA), where 1 mole of SA produces 1 mole of aspirin. Starting from 2 g of SA, the student isolates 1.6 g of aspirin. What is the percent yield?

Using standard molar masses — salicylic acid, C₇H₆O₃ (138.12 g/mol), and aspirin, C₉H₈O₄ (180.16 g/mol) — the theoretical yield is found by following the mass → moles → moles → mass pathway:

• Moles of SA: 2 g ÷ 138.12 g/mol ≈ 0.0145 mol SA

• Moles of aspirin (1:1 ratio): ≈ 0.0145 mol aspirin

• Theoretical mass of aspirin: 0.0145 mol × 180.16 g/mol ≈ 2.61 g

The student's actual yield was 1.6 g, so:

Percent Yield = (1.6 g ÷ 2.61 g) × 100% ≈ 61.3%

Common MCAT Mistakes

• Adjusting subscripts instead of coefficients to balance an equation. Subscripts are fixed by a substance's identity — changing one changes the substance entirely. Only coefficients (multipliers in front of formulas) can be adjusted to balance an equation.

• Balancing atoms in a suboptimal order. Balancing the most complex substance first and the simplest (often a diatomic element like O₂) last, as shown in the propane example, avoids having to re-balance atoms already set.

• Confusing the limiting reagent with the reagent present in the smallest amount. The limiting reagent is whichever reactant runs out first based on the reaction's mole ratio, not necessarily whichever has the smallest starting mass or volume.

• Assuming percent yield can exceed 100%. Actual yield is almost always less than theoretical yield due to loss during a real procedure, so percent yield is expected to fall below 100% — a value above 100% signals a measurement or calculation error, not a "better than expected" result.

MCAT-Style Concept Check

Question: The unbalanced equation for the combustion of methane is CH₄ + O₂ → CO₂ + H₂O. What is the coefficient of O₂ once the equation is fully balanced with the smallest whole-number coefficients?

• A) 1

• B) 2

• C) 3

• D) 4

Answer: B

Explanation: Balancing carbon and hydrogen first gives CH₄ + O₂ → CO₂ + 2 H₂O (1 carbon and 4 hydrogens now match on both sides). The right side then has 2 oxygens from CO₂ plus 2 oxygens from 2 H₂O, for a total of 4 oxygens — so O₂ needs a coefficient of 2 to supply 4 oxygens: CH₄ + 2 O₂ → CO₂ + 2 H₂O.

FAQ

Why can't you balance an equation by changing subscripts?

Subscripts are fixed by a substance's chemical identity — changing H₂O's subscript to H₂O₂ doesn't balance an equation, it turns water into an entirely different compound (hydrogen peroxide). Only coefficients, placed in front of a formula, can be adjusted to balance an equation.

What's the difference between the limiting and excess reagent?

The limiting reagent is the reactant that runs out first, based on the balanced equation's mole ratio, and it caps how much product can form. The excess reagent is whatever reactant is left over once the limiting reagent is used up.

How do you calculate percent yield?

Percent yield equals actual yield divided by theoretical yield, multiplied by 100%. Theoretical yield comes from a stoichiometric calculation assuming a perfect reaction; actual yield is what's actually collected in the lab.

What's the general pathway for converting mass of one substance to mass of another in a reaction?

Convert the given mass to moles using its molar mass, convert to moles of the target substance using the mole-to-mole ratio from the balanced equation's coefficients, then convert those moles to mass using the target substance's molar mass.