What you'll learn
Chemists need to know how much product a reaction will make, and how much of each reactant to use so nothing is wasted — this comes from calculations of reacting masses and limiting reactants. For AQA GCSE Chemistry you need to be able to use balanced equations and relative formula masses to calculate reacting masses, work out the mass of product formed, and identify the limiting reactant. This guide covers moles and reacting masses, the steps of a reacting-mass calculation, the idea of a limiting reactant, and how the amount of product depends on it. By the end you should be able to calculate reacting masses and identify which reactant limits the amount of product.
Key terms and definitions
Mole (mol) — The unit for amount of substance.
Relative formula mass (Mr) — The sum of the relative atomic masses of all atoms in a formula.
Reacting mass — The mass of a substance that takes part in or is produced by a reaction.
Limiting reactant — The reactant that is used up first, which limits the amount of product formed.
Excess reactant — A reactant present in more than enough quantity, so some is left over.
Molar ratio — The ratio of moles of substances shown in a balanced equation.
Balanced equation — An equation with equal numbers of each type of atom on both sides.
Yield — The amount of product actually formed.
Core concepts
Moles and mass
The link between mass and moles is:
moles = mass ÷ relative formula mass (Mr)
This can be rearranged to find any of the three. It lets you convert a mass into moles, and moles back into mass. Because balanced equations work in moles, converting to moles is the key to reacting-mass calculations.
The steps of a reacting-mass calculation
To calculate the mass of a product (or reactant), follow these steps:
- Write the balanced equation.
- Convert the known mass into moles using moles = mass ÷ Mr.
- Use the molar ratio in the equation to find the moles of the substance you want.
- Convert those moles back into mass using mass = moles × Mr.
Setting the work out in these steps keeps it organised and earns method marks even if the arithmetic slips.
Using the balanced equation
The balanced equation gives the ratio of moles in which substances react and are produced. For example:
2Mg + O₂ → 2MgO
means 2 moles of magnesium react with 1 mole of oxygen to make 2 moles of magnesium oxide. So the ratio of Mg to MgO is 2:2 (that is, 1:1), while Mg to O₂ is 2:1. You must use the correct ratio when converting between substances.
The limiting reactant
When two reactants are mixed, they are often not present in the exact ratio needed. One reactant will be used up first — this is the limiting reactant, because once it runs out the reaction stops, even if some of the other reactant is left. The reactant left over is the excess reactant.
The amount of product formed depends on the limiting reactant, because it determines how much can react. To find the maximum product, you must base your calculation on the limiting reactant, not the one in excess.
Identifying the limiting reactant
To identify the limiting reactant:
- Convert the mass of each reactant into moles.
- Use the balanced equation to compare how many moles of each are needed.
- The reactant that provides fewer moles than the ratio requires is the limiting reactant.
Once you know the limiting reactant, you use its moles to calculate the amount of product.
Why the limiting reactant controls the product
The limiting reactant controls the amount of product because the reaction can only continue while all the reactants are available. As soon as the limiting reactant is used up, no more product can form, regardless of how much of the excess reactant remains. So doubling the excess reactant would not make more product — only increasing the limiting reactant would.
Conservation of mass in reacting-mass problems
Reacting-mass calculations always obey the law of conservation of mass: no atoms are created or destroyed, so the total mass of the reactants equals the total mass of the products. This is a useful check on your working. For example, if magnesium and oxygen react, the mass of magnesium oxide formed must equal the mass of magnesium plus the mass of oxygen that reacted. If your calculated product mass is greater than the total reactant mass, or a product mass is larger than the reactant it came from in a way that breaks the ratio, you know a mistake has been made. Keeping conservation of mass in mind helps you spot errors before writing a final answer.
Percentage yield and why it is less than 100%
In practice, the mass of product actually obtained (the yield) is often less than the amount your calculation predicts. This is because not all of the reactants may react, some product may be lost when it is separated or purified, or side reactions may produce other products. The percentage yield compares the actual amount obtained with the maximum possible amount from the calculation. Reacting-mass calculations give the maximum (theoretical) amount that could be made if everything reacted perfectly, so the real amount is usually a bit lower. Understanding this connects the calculation to what really happens in the lab and in industry.
Worked examples
Example 1: Converting mass to moles
How many moles are there in 12 g of carbon? The Ar of carbon is 12, so moles = mass ÷ Mr = 12 ÷ 12 = 1 mol.
Example 2: A reacting-mass calculation
What mass of magnesium oxide is formed when 12 g of magnesium burns completely? The equation is 2Mg + O₂ → 2MgO. Ar: Mg = 24, O = 16. Step 1 — moles of Mg = 12 ÷ 24 = 0.5 mol. Step 2 — the ratio of Mg to MgO is 2:2 (1:1), so moles of MgO = 0.5 mol. Step 3 — Mr of MgO = 24 + 16 = 40, so mass = moles × Mr = 0.5 × 40 = 20 g.
Example 3: Identifying the limiting reactant
In the reaction 2H₂ + O₂ → 2H₂O, 4 mol of hydrogen is mixed with 1 mol of oxygen. Which is the limiting reactant? The equation needs 2 mol of hydrogen for every 1 mol of oxygen. For 1 mol of oxygen, only 2 mol of hydrogen is needed, but 4 mol is present, so hydrogen is in excess. Oxygen is the limiting reactant.
Example 4: Product from the limiting reactant
Using the reaction above with 1 mol of oxygen as the limiting reactant, how many moles of water are formed? The ratio of O₂ to H₂O is 1:2, so 1 mol of oxygen makes 2 mol of water. The excess hydrogen does not increase this.
Common mistakes and how to avoid them
The most common error is using the wrong molar ratio. Always read the balancing numbers from the equation — for 2Mg + O₂ → 2MgO, the Mg to MgO ratio is 1:1, but Mg to O₂ is 2:1.
Students often base the product calculation on the wrong reactant. The amount of product depends on the limiting reactant, so always identify it first and use its moles.
Another mistake is forgetting to convert between mass and moles. You cannot use masses directly in the ratio — convert mass to moles first, apply the ratio, then convert back to mass.
When calculating Mr, take care to add up every atom, including those in brackets or multiplied by a large number in front of the formula.
Finally, remember that adding more of the excess reactant does not make more product — only the limiting reactant controls the yield.
Exam technique for "Reacting masses and limiting reactants"
Always start with the balanced equation and set out the calculation in clear steps: mass to moles, apply the molar ratio, moles back to mass. Show every step to earn method marks.
For limiting-reactant questions, convert both reactants to moles, compare with the ratio in the equation, and identify which runs out first. Then base the product calculation on that limiting reactant.
Take care with units and Mr calculations, and give your final answer to a sensible number of significant figures. Use precise terms — moles, molar ratio, limiting reactant, excess — and always link the amount of product to the limiting reactant.
Quick revision summary
- moles = mass ÷ Mr; rearrange to find mass = moles × Mr.
- Reacting-mass steps: balanced equation → mass to moles → apply molar ratio → moles back to mass.
- Read the ratio from the balancing numbers (e.g. 2Mg + O₂ → 2MgO gives Mg:MgO = 1:1, Mg:O₂ = 2:1).
- The limiting reactant is used up first and controls the amount of product; the other is in excess.
- To find the limiting reactant, convert both reactants to moles and compare with the equation's ratio.
- Adding more excess reactant does not increase the product — only more limiting reactant does.