What a balanced equation tells you

A chemical equation is a symbolic statement showing the reactants, products and relative amounts in a chemical reaction. It looks compact, but it carries plenty of information: formulae, reacting proportions, and often physical states.

A reactant is a chemical species consumed during a reaction. A product is a chemical species formed during a reaction. A chemical species is an atom, ion, molecule or formula unit treated as a single chemical entity in an equation.

A stoichiometric coefficient is a number placed before a formula in a chemical equation that gives the relative amount of that species. In the equation

2H₂(g) + O₂(g) → 2H₂O(l)

the coefficients 2, 1 and 2 show that hydrogen, oxygen and water are involved in a 2 : 1 : 2 ratio. We normally leave out the coefficient 1.

A mole ratio is a ratio of stoichiometric coefficients from a balanced chemical equation. Don’t read it as the starting mixture. It tells you how amounts change as the reaction happens.

Balancing equations: conservation of atoms

A balanced chemical equation is a chemical equation in which the number of atoms of each element is the same on both sides. This is possible because chemical reactions rearrange atoms; they do not create or destroy them.

For example, complete combustion of propane forms carbon dioxide and water:

C₃H₈ + O₂ → CO₂ + H₂O

Balance carbon first, then hydrogen, then oxygen:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

The final equation says 1 mol of propane reacts with 5 mol of oxygen to form 3 mol of carbon dioxide and 4 mol of water. Doubling every coefficient would still give a balanced equation, but IB convention is to use the simplest whole-number ratio.

State symbols

A state symbol is a symbol written after a formula to show the physical state of that species in a chemical equation. Use state symbols when the states are known or requested.

Chemical state symbols used in balanced equations.

So the combustion equation above, with states, is:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Aqueous, (aq), is not the same as liquid. Sodium chloride solution is NaCl(aq), not NaCl(l), because the ions are dissolved in water.

Deducing equations from named reactants and products

When reactants and products are specified, write the formulae first, then balance. A good routine is:

1. write correct formulae for all species;
2. place reactants on the left and products on the right;
3. balance atoms by changing coefficients only;
4. check every element;
5. add state symbols if known.

Never balance by changing subscripts inside formulae. Changing H₂O to H₂O₂ does not balance water; it changes the substance.

Where half-equations fit

Half-equations are useful when a reaction involves electron transfer, especially in redox reactions. They let you balance oxidation and reduction separately before you combine them into one overall equation. For ordinary mole-ratio calculations in this topic, though, start with the overall balanced equation.

The central idea: convert to amount first

Stoichiometry is the quantitative study of reactants and products in chemical reactions. The safe route is usually this: turn the information you are given into amount in moles, apply the mole ratio from the balanced equation, then convert to the quantity the question asks for.

An amount of substance measures how many specified particles are present. It is expressed in moles, and its symbol is n.

For masses:

n = m / M, where n is amount of substance (mol), m is mass (g in IB calculations) and M is molar mass (g mol⁻¹ in IB calculations).

A molar mass is the mass per mole of a substance. Work it out from relative atomic masses, Aᵣ, in the data booklet, using values to two decimal places unless the question clearly provides different data. For example, two-decimal-place values prevent tiny answer differences caused by data rounding, rather than chemistry.

For gas volumes at a fixed temperature and pressure:

n = V / Vₘ, where V is gas volume (dm³ in IB calculations) and Vₘ is molar volume of a gas (dm³ mol⁻¹ at the stated temperature and pressure).

At standard temperature and pressure, STP, the IB value is usually Vₘ = 22.7 dm³ mol⁻¹. A molar volume is the volume occupied by one mole of a gas at a specified temperature and pressure.

For solutions:

n = cV, where c is molar concentration (mol dm⁻³) and V is solution volume (dm³).

A molar concentration is the amount of solute per unit volume of solution. Watch the unit conversion that catches students every year: 1000 cm³ = 1 dm³, so 25.0 cm³ = 0.0250 dm³.

Reacting masses

Suppose magnesium reacts with hydrochloric acid:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

The equation tells you that 1 mol Mg reacts with 2 mol HCl and forms 1 mol H₂. If you know the mass of magnesium, convert it to n(Mg), use the 1 : 1 ratio to find n(H₂), then convert to a gas volume if the question needs it. If you know the mass of one product, work backwards: mass → amount → mole ratio → amount → mass.

Reacting gas volumes

Avogadro’s law says equal volumes of gases at the same temperature and pressure contain equal numbers of particles. So, for gases only and under the same conditions, the ratio of gas volumes follows the ratio of coefficients.

For example:

N₂(g) + 3H₂(g) → 2NH₃(g)

At the same temperature and pressure, 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia. This gives the volume relationship; it does not explain why gases behave like this. That explanation belongs with particle models and gas laws.

The molar volume of a gas changes with temperature and pressure: increasing temperature increases gas volume for a fixed amount at constant pressure, while increasing pressure decreases gas volume for a fixed amount at constant temperature. Use a stated molar volume only under the conditions for which it is valid.

Reactions in solution

For aqueous reactions, don’t try to use the volume ratio unless the concentrations are identical and the equation allows it. Convert each solution volume and concentration into moles first.

For neutralisation:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

1 mol sulfuric acid reacts with 2 mol sodium hydroxide. In a titration, if you know the concentration and volume of sodium hydroxide, find n(NaOH), halve it to get n(H₂SO₄), then divide by the acid volume in dm³ to find its concentration.

A titration is an analytical technique in which a measured volume of one solution is reacted with another solution of known concentration to determine an unknown concentration. The balanced equation gives the mole ratio; the colour change of an indicator only shows when the reaction is complete.

In school calculations, volumes of two aqueous solutions are often treated as additive, so the final volume is the sum of the mixed solution volumes. That is an approximation, but it is usually small compared with normal volumetric uncertainty. Do not add a gas volume to a solution volume; gas absorption usually changes solution volume negligibly in these calculations.

A reliable calculation layout

For longer problems, an initial-change-final table keeps the chemistry visible:

nfinal = ninitial + Δn, where nfinal is final amount of substance (mol), ninitial is initial amount of substance (mol) and Δn is change in amount of substance during reaction (mol).

Signs matter: reactants have negative changes, products have positive changes. The change row follows the coefficients; the initial and final rows do not have to.

Limiting and excess reactants

A limiting reactant is the reactant used up first, so it sets the maximum amount of product that can form. An excess reactant is present in more than the mole ratio requires; some of it is still left after the limiting reactant has gone.

Take the formation of water:

2H₂(g) + O₂(g) → 2H₂O(l)

If 1.00 mol H₂ is mixed with 1.00 mol O₂, the equation requires 2 mol H₂ for every 1 mol O₂. So 1.00 mol H₂ can react with only 0.500 mol O₂. Hydrogen is the limiting reactant, and 0.500 mol oxygen remains.

A useful way to check is to work out how much product each reactant could form, assuming the other reactants were unlimited. Whichever reactant gives the smaller amount of product is the limiting reactant.

Theoretical yield and experimental yield

A theoretical yield is the maximum amount or mass of product predicted from the limiting reactant, assuming complete reaction and no product loss. An experimental yield is the actual amount or mass of product obtained from a laboratory or industrial process.

Theoretical yield is calculated; experimental yield is measured. Keep that difference clear. If a calcium carbonate precipitate is filtered, washed and dried, the mass shown on the balance is the experimental yield. The mass predicted from the moles of the limiting reactant is the theoretical yield.

Why experimental yield may be lower or higher

Experimental yields are often lower than theoretical yields because:

• the reaction may not go to completion;
• side reactions may consume reactants;
• product may be lost during transfer, filtration, washing, drying or recrystallisation;
• some product may remain dissolved in the filtrate or solvent;
• the limiting reactant may contain impurities.

An experimental yield above the theoretical yield is a warning sign. It can occur when the product is wet, contaminated with impurities, not dried to constant mass, or when the balance/container mass has been recorded incorrectly. In gravimetric work, a damp precipitate is the classic culprit: water adds mass but is not product.

Practical technique is part of the chemistry here. Drying to constant mass, rinsing apparatus to transfer all solid, using clean filter paper, and recording qualitative observations all help you judge whether the measured yield is trustworthy.

Calculating percentage yield

A percentage yield is the experimental yield written as a percentage of the theoretical yield.

% yield = (Yexp / Ytheo) × 100, where Yexp is experimental yield (mol or g), and Ytheo is theoretical yield in the same unit (mol or g).

You can work in moles or in masses. Just don’t mix them in one fraction. If the experimental yield is in grams and the theoretical yield is in moles, convert one of them first.

The full problem-solving route

For questions involving reacting quantities, limiting reactants and yields, follow this route:

1. write and balance the equation, including state symbols if known;
2. convert all given masses, gas volumes or solution data into amounts;
3. identify the limiting reactant;
4. use the mole ratio to calculate the theoretical amount of product;
5. convert the theoretical amount to the requested unit, usually grams or dm³;
6. compare experimental and theoretical yield using percentage yield.

A common trap is to calculate the product from the reactant that seems more chemically important, rather than from the limiting reactant. The equation doesn’t care which reactant is named first.

Conservation of mass as a check

The law of conservation of mass is a principle stating that the total mass of substances is unchanged during a chemical reaction in a closed system. Atoms are rearranged, so the total mass of reactants consumed equals the total mass of products formed.

That does not mean one isolated product must have the same mass as the reactants you started with. If a gas escapes, a solvent evaporates, an excess reactant remains, or by-products form, the measured mass of your collected product is only one part of the total mass picture.

Purity, yield and real laboratory products

In synthesis, a larger mass is not automatically better. A crude product may show a high apparent yield because it contains solvent, unreacted reactant or by-products. Purification, such as recrystallisation, often lowers yield because some desired product is lost, but it can increase purity.

Melting-point data can help with this judgement: impurities usually lower a melting point and broaden the melting range. A product with a smaller mass but a sharper melting range closer to the literature value may be the better product.

Impossible yields

A percentage yield above 100% is not a chemical success; it shows that something in the measurement or sample is wrong. Check first for wet product, contamination, incomplete drying, uncalibrated apparatus, transcription errors, or the wrong limiting reactant calculation.

In industry, percentage yield affects cost and waste. A small percentage loss on a tonne scale is not small at all. That is why yield calculations fit naturally with practical technique, process design and green chemistry.