What a balanced equation tells you

A chemical equation is a symbolic statement showing the reactants, products and their 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

$2H_2(g) + O_2(g) \to 2H_2O(l)$

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 treat it as the starting mixture. It tells you how amounts change as the reaction occurs.

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. Chemical reactions rearrange atoms; they do not create or destroy them.

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

$C_3H_8 + O_2 \to CO_2 + H_2O$

Balance carbon first, then hydrogen, then oxygen:

$C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O$

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:

$C_3H_8(g) + 5O_2(g) \to 3CO_2(g) + 4H_2O(l)$

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 the reactants and products are given, write the formulae first and 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 $H_2O$ to $H_2O_2$ does not balance water; it changes the substance.

Where half-equations fit

Half-equations help when a reaction involves electron transfer, especially in redox reactions. They let you balance oxidation and reduction separately before combining 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 safest route is usually simple: turn the information you’re given into amount in moles, apply the mole ratio from the balanced equation, then convert to whatever quantity the question wants.

An amount of substance measures how many specified particles are present, expressed in moles. Its symbol is $n$.

For masses:

A molar mass is the mass per mole of a substance. Work it out from relative atomic masses, $A_r$, in the data booklet, using the values to two decimal places unless the question clearly gives other data. Using two-decimal-place values, for example, avoids tiny differences in answers that come from data rounding rather than chemistry.

For gas volumes at a fixed temperature and pressure:

At standard temperature and pressure, STP, the IB value is usually $V_m = 22.7$ dm$^3$ mol$^{-1}$. A molar volume is the volume occupied by one mole of a gas at a specified temperature and pressure.

For solutions:

A molar concentration is the amount of solute per unit volume of solution. One unit conversion catches students every year: 1000 cm$^3$ = 1 dm$^3$, so 25.0 cm$^3$ = 0.0250 dm$^3$.

Reacting masses

Suppose magnesium reacts with hydrochloric acid:

$Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)$

The equation tells you that 1 mol Mg reacts with 2 mol HCl and forms 1 mol $H_2$. If the mass of magnesium is known, convert it to $n(\mathrm{Mg})$, use the $1 : 1$ ratio to find $n(\mathrm{H_2})$, then convert that to a gas volume if needed. If the known value is the mass of one product, go backwards along the same path: mass $o$ amount $o$ mole ratio $o$ amount $o$ mass.

Reacting gas volumes

Avogadro’s law says equal volumes of gases at the same temperature and pressure contain equal numbers of particles. For gases only, and under the same conditions, gas volume ratios therefore match the coefficient ratios.

For example:

$N_2(g) + 3H_2(g) \to 2NH_3(g)$

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 doesn’t explain why gases behave like this. That explanation belongs with particle models and gas laws.

The molar volume of a gas changes when temperature or pressure changes: 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 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:

$H_2SO_4(aq) + 2NaOH(aq) \to Na_2SO_4(aq) + 2H_2O(l)$

1 mol sulfuric acid reacts with 2 mol sodium hydroxide. In a titration, if the concentration and volume of sodium hydroxide are known, calculate $n(\mathrm{NaOH})$, halve it to get $n(\mathrm{H_2SO_4})$, then divide by the acid volume in dm$^3$ 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, the volumes of two aqueous solutions are often treated as additive, so the final volume is the sum of the mixed solution volumes. This is an approximation, but usually it is small compared with normal volumetric uncertainty. Do not add a gas volume to a solution volume; in these calculations, gas absorption usually changes solution volume negligibly.

A reliable calculation layout

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

The 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 that gets 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 left over after the limiting reactant has gone.

Take the formation of water:

If $1.00\ \text{mol}\ H_2$ is mixed with $1.00\ \text{mol}\ O_2$, the equation requires $2\ \text{mol}\ H_2$ for every $1\ \text{mol}\ O_2$. So the $1.00\ \text{mol}\ H_2$ can react with only $0.500\ \text{mol}\ O_2$. Hydrogen is limiting, and $0.500\ \text{mol}\ O_2$ remains.

A good way to check is to work out how much product each reactant could form if all the other reactants were available in unlimited supply. 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 the reaction is complete and no product is lost. An experimental yield is the actual amount or mass of product collected from a laboratory or industrial process.

You calculate the theoretical yield; you measure the experimental yield. Keep those separate. If calcium carbonate precipitate is filtered, washed and dried, the mass read from the balance is the experimental yield. The mass predicted from the moles of 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 higher than the theoretical yield should make you suspicious. It can occur if the product is wet, contaminated with impurities, not dried to constant mass, or if 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. Dry to constant mass, rinse apparatus to transfer all solid, use clean filter paper, and record qualitative observations so you can 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.

You can use moles or masses. Just don’t mix them in the same 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, work through 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 $dm^3$;
6. compare experimental and theoretical yield using percentage yield.

A common trap is calculating the product from the reactant that seems more chemically important, instead of from the limiting reactant. The equation does not 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 mass you measure for the collected product is only part of the total mass picture.

Purity, yield and real laboratory products

In synthesis, a large mass is not automatically a good result. 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 belong alongside practical technique, process design and green chemistry.

Atom economy and green chemistry

Green chemistry is an approach to chemical design that aims to reduce the use and formation of hazardous substances across the whole life cycle of a process. It’s broader than “use a plant-based material” or “use less solvent”; it checks what happens before, during and after the reaction.

Atom economy is a percentage measure of how much of the mass of reactant atoms ends up in the desired product, according to the balanced equation.

The data booklet gives you the equation, but the coefficients still matter. If the equation forms $2 \times$ mol of the desired product, put $2 \times$ its molar mass in the numerator.

Reading atom economy from an equation

For a reaction such as:

$A + B \to C$

where C is the desired product, atom economy can be very high because all reactant atoms are incorporated into C.

For:

$A + B \to C + D$

where C is desired and D is a by-product, atom economy is lower because the atoms in D count as waste for this target product.

Atom economy and wastage therefore move in opposite directions: as a larger fraction of reactant atoms is incorporated into the desired product, a smaller fraction ends up as unwanted material. High atom economy means less built-in waste from the reaction stoichiometry.

Atom economy is not the same as percentage yield

Percentage yield compares what you actually obtained with what you could theoretically obtain. Atom economy looks at the balanced equation before the experiment is even done.

A reaction can have excellent atom economy but poor percentage yield if it is slow, reversible, incomplete or difficult to purify. A reaction can also have a respectable yield but poor atom economy if it produces large amounts of by-product.

Other factors in process efficiency

Atom economy and percentage yield matter, but neither gives the whole judgement. A proper assessment should also consider:

• toxicity of reactants, products, solvents and by-products;
• energy demand, including heating, cooling and pressure requirements;
• renewability and availability of starting materials;
• solvent use and solvent recovery;
• catalyst use, lifetime and recyclability;
• separation and purification steps;
• safety risks, including flammability, corrosion and pressure hazards;
• cost and environmental impact of waste treatment;
• full life-cycle effects, such as land use, transport and production of feedstocks.

Here’s the subtle point: a process branded as green may still need toxic reagents or high energy input upstream. In green chemistry, judge the process, not the slogan.