Equation Stoichiometry and Limiting Reactants

What the numbers in an equation mean

This is Higher Tier content.

A balanced chemical equation shows the proportions in which substances react and products form. These proportions are the reaction’s stoichiometry. The numbers before the chemical formulae are called coefficients: they describe ratios of particles and, therefore, ratios of moles.

For example, a coefficient of 2 before a formula means twice as many particles or moles as a coefficient of 1. It does not necessarily mean twice the mass, because different substances have different masses per mole.

Turning measured masses into an equation

To deduce coefficients from masses, first convert each mass into an amount in moles. A mole is an amount containing a fixed number of particles, so comparing moles lets us compare numbers of particles even when their masses differ.

Use:

$$\text{amount in moles} = \frac{\text{mass in g}}{\text{molar mass in g mol}^{-1}}$$

Calculate each molar mass from the formula and relative atomic masses. Then divide all the mole amounts by the smallest to find their simplest ratio. If the ratio contains fractions, multiply every part by the same number until all coefficients are whole numbers. For instance, $1:2:2.5$ becomes .

The reactant masses used for this method must be the masses that actually reacted. If some reactant was supplied in excess and remains afterwards, its initial mass is not the mass consumed.

Deducing the equation for methanol combustion

Methanol, $\mathrm{CH_3OH}$, reacts with oxygen to form carbon dioxide and water. Suppose the reacting masses are 64 g of methanol and 96 g of oxygen, producing 88 g of carbon dioxide and 72 g of water.

Using relative atomic masses C = 12, H = 1 and O = 16 gives the following conversions:

The mole ratio is $2:3:2:4$. These are already the smallest whole-number coefficients, so the equation is:

$$2\mathrm{CH_3OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 4\mathrm{H_2O}$$

Check the atoms: each side contains two carbon atoms, eight hydrogen atoms and eight oxygen atoms. The total masses also agree: 160 g of reactants gives 160 g of products. Mass conservation is a useful check, but the coefficients come from the mole ratio, not the mass ratio.

Why one reactant can limit the product

Reactants must combine in the proportions shown by the balanced equation. If one runs out while another remains, no more product can form through that reaction. The substance used up first is the limiting reactant; the substance left over is an excess reactant.

Consider the formation of sodium sulfide:

$$2\mathrm{Na} + \mathrm{S} \rightarrow \mathrm{Na_2S}$$

The reaction needs two moles of sodium for every mole of sulfur. With 9.2 g of sodium and 8.0 g of sulfur, using molar masses of 23 g mol⁻¹ and 32 g mol⁻¹:

• Sodium supplied: $9.2/23 = 0.40$ mol.
• Sulfur supplied: $8.0/32 = 0.25$ mol.

The 0.40 mol of sodium needs only $0.40/2 = 0.20$ mol of sulfur. There is more sulfur than this available, so sodium is limiting and sulfur is in excess. Sodium runs out, leaving 0.05 mol of sulfur unreacted. Notice that sodium is limiting despite having both the greater mass and the greater number of moles: the required ratio matters.

How the limiting amount affects yield

The limiting reactant determines the maximum amount of product obtainable. Adding more of an already excess reactant cannot increase this maximum, because the missing reaction partner has been used up.

If the other reactants remain in excess, halving the amount of limiting reactant halves the maximum product amount; doubling it doubles the maximum product amount. This proportional relationship only holds while that same reactant remains limiting. Increasing it far enough can make a different reactant the limiting one.