Pearson Edexcel International GCSE in Chemistry · 4CH1

Reacting Masses and Percentage Yield

How much product a reaction should give, how much it actually gives, and why the two are never the same.

Topic 1 · Principles of chemistry — one of 18 lessons in this topic, and one of 61 in Chemistry.

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

  • 1.29 — Calculating reacting masses from experimental data and equations
  • 1.30 — Calculating percentage yield

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

A balanced equation is a recipe written in PARTICLES, not in grams. When it says that two molecules of hydrogen react with one of oxygen, it is counting things — and the whole difficulty of this page is that a balance weighs things instead. The mole is what converts between the two, which is why the last lesson had to come first.

Think of it like a recipe that specifies eggs and a shop that sells them by weight

A recipe says two eggs and one cup of flour. The shop sells eggs by the kilogram. You cannot work directly from the recipe to the shop, because one counts and the other weighs — so you convert: kilograms into eggs, use the recipe's RATIO to find how much flour those eggs need, then convert back into grams for the flour. That three-step shape is every reacting-mass calculation on this specification. The middle step is the one that carries the chemistry, because it is the only place the balanced equation is used at all. Skip it and you are doing arithmetic on masses that happen to work when the ratio is one to one and fail everywhere else.

The three steps, in the order that always works

  1. STEP 1 — convert the mass you were given into MOLES: moles = mass ÷ relative formula mass.
  2. STEP 2 — use the RATIO in the balanced equation to find the moles of the substance you want. This is the only step that uses the chemistry.
  3. STEP 3 — convert those moles back into a MASS: mass = moles × relative formula mass.
  4. Write the balanced equation first, every time, or step 2 has nothing to work from.
  5. Check the answer is sensible: if the ratio is one to one and the products are heavier, the mass should go up.

Reacting masses, with a ratio that is not one to one

What mass of magnesium oxide is formed when 6.0 g of magnesium burns completely in oxygen? (Relative atomic masses: Mg = 24, O = 16.)

  1. Balanced equation: 2Mg + O₂ → 2MgO.
  2. STEP 1 — moles of Mg = mass ÷ Ar = 6.0 ÷ 24 = 0.25 mol.
  3. STEP 2 — the ratio of Mg to MgO in the equation is 2 : 2, which is 1 : 1, so 0.25 mol of MgO is formed.
  4. STEP 3 — relative formula mass of MgO = 24 + 16 = 40. Mass = moles × Mr = 0.25 × 40 = 10 g.
  5. Sensible? The oxygen has added mass, so 6.0 g of magnesium giving 10 g of oxide is exactly what should happen.

Answer: 10 g of magnesium oxide.

That answer is the THEORETICAL YIELD: the mass you would get if the reaction went perfectly and nothing at all was lost. In a real laboratory you never get it. Something is always left behind in the flask, some product is lost when the mixture is filtered or transferred, the reaction may not go to completion, and some of the reactants may take part in a different reaction and produce something else.

PERCENTAGE YIELD compares what you actually obtained with what you theoretically should have: percentage yield = (actual yield ÷ theoretical yield) × 100. A yield of 100% is essentially unheard of, and a yield ABOVE 100% is not a triumph but a mistake — almost always a product that has not been dried, so the mass includes water.

Percentage yield, using the theoretical yield just calculated

The reaction above is carried out and 8.4 g of magnesium oxide is actually collected. Calculate the percentage yield.

  1. Theoretical yield, from the calculation above = 10 g.
  2. Actual yield = 8.4 g.
  3. percentage yield = (actual ÷ theoretical) × 100 = (8.4 ÷ 10) × 100.
  4. = 84%.
  5. The missing 16% is not destroyed. It is unburnt magnesium, oxide lost as smoke, or powder left in the crucible.

Answer: 84% — and the 'missing' mass has been left behind or lost, not annihilated.

It is worth saying plainly why yield is never 100%, because the reasons are examined and they are all practical rather than mysterious. The reaction may be REVERSIBLE and so never fully complete. Some product is always LOST in transferring, filtering or washing. Some reactants may undergo SIDE REACTIONS giving a different product. And the reactants themselves may be IMPURE. None of these breaks conservation of mass: every atom is still somewhere, just not in the beaker you are weighing.

One further point, because it explains why industry cares. A low percentage yield means wasted raw material and therefore wasted money, so improving yield is one of the two things a chemical engineer works on. The other is the rate — and a process with a high yield that takes a week may be worth less than one with a lower yield that takes an hour.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

moles = mass ÷ relative formula mass

mass in grams (g), relative formula mass has no units

Units: mol, g. Rearranged: mass = moles × relative formula mass. This converts between what a balance measures and what an equation counts.

percentage yield = (actual yield ÷ theoretical yield) × 100

actual yield is the mass obtained in the experiment; theoretical yield is the mass calculated from the balanced equation

Units: Both masses in the same unit, which cancels; the answer is a percentage. A value above 100% means the product was not dry.

The three-step reacting-mass route

  1. Write the BALANCED equation.
  2. Convert the given mass to moles: moles = mass ÷ relative formula mass.
  3. Use the RATIO in the equation to find the moles of the substance wanted.
  4. Convert those moles back to a mass: mass = moles × relative formula mass.
  5. Check the answer is sensible against the equation.

The three yields, distinguished

Theoretical yield
The mass of product calculated from the balanced equation, assuming the reaction is complete and nothing is lost.
Actual yield
The mass of product actually obtained and weighed in the experiment.
Percentage yield
The actual yield as a percentage of the theoretical yield.

Why the yield is never 100%

  • The reaction may be REVERSIBLE and so does not go to completion
  • Product is LOST on transferring, filtering or washing
  • SIDE REACTIONS produce a different product from some of the reactants
  • The reactants may be IMPURE
  • None of this breaks conservation of mass — every atom is still somewhere
  • A yield ABOVE 100% means an error, almost always a product that was not dried

2Mg + O₂ → 2MgO

Conditions: Heated in air. The ratio of Mg to MgO is 2 : 2, so one mole of magnesium gives one mole of oxide.

Model answer [4 marks]

Calculate the mass of calcium oxide formed when 25 g of calcium carbonate is completely decomposed. (Ca = 40, C = 12, O = 16.) [4]

The equation is CaCO₃ → CaO + CO₂. The relative formula mass of CaCO₃ is 40 + 12 + 48 = 100, so moles of CaCO₃ = 25 ÷ 100 = 0.25 mol. The ratio of CaCO₃ to CaO is 1 : 1, so 0.25 mol of CaO is formed. The relative formula mass of CaO is 40 + 16 = 56, so the mass is 0.25 × 56 = 14 g.

Model answer [4 marks]

A student calculates a theoretical yield of 12.5 g but obtains 9.0 g. Calculate the percentage yield and give two reasons why it is below 100%. [4]

Percentage yield = (9.0 ÷ 12.5) × 100 = 72%. Two reasons for a yield below 100% are that some of the product is lost during transfer, filtering or washing, and that the reaction may not have gone to completion because it is reversible. Other acceptable reasons are side reactions producing a different product, or impurities in the reactants.

Not this: A percentage yield below 100% means some of the mass has been destroyed.

This: Mass is always conserved. The missing product is unreacted starting material, product lost on the glassware, or atoms that went into a different product in a side reaction — every atom is still somewhere, just not where it was weighed.

Mark-losing trap. ALWAYS go through moles. Working directly on masses fails as soon as the ratio is not 1 : 1.

Mark-losing trap. Balance the equation FIRST — step 2 uses its ratio and has nothing to work from otherwise.

Mark-losing trap. A yield over 100% is an ERROR, not a success. The product almost certainly was not dry.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

  1. Grade 6 · Calculate [2 marks] — Calculate the number of moles in 20 g of calcium carbonate, CaCO₃. (Relative formula mass of CaCO₃ = 100.)
  2. Grade 7 · Calculate [2 marks] — A student calculates a theoretical yield of 8.0 g of a salt but actually obtains 6.0 g. Calculate the percentage yield.
  3. Grade 8 · Calculate [4 marks] — Calculate the mass of magnesium oxide formed when 12 g of magnesium burns completely: 2Mg + O₂ → 2MgO. (Mg = 24, O = 16.)
  4. Grade 9 · Explain [6 marks] — Select every statement that belongs in a full-mark explanation of why the percentage yield of a laboratory preparation is almost always below 100%, and of why this does not break the conservation of mass.
  5. 9+ · Analyse [6 marks] — A student heats 5.0 g of copper in air to form copper oxide and obtains 6.5 g of black solid. They calculate a percentage yield of 130% and conclude the reaction 'worked better than expected'. (Cu = 64, O = 16.) Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one who established that the masses have to balance at all, and the one who made a formula a statement about countable particles. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Antoine Lavoisier — the one who established that the masses have to balance at all

Every calculation on this page assumes conservation of mass, and that assumption is Lavoisier's. He weighed sealed vessels before and after reactions with a precision nobody had troubled with, and showed that a metal gaining mass when it burns has taken that mass FROM the air rather than acquiring it from nowhere. That killed the phlogiston theory, which had explained combustion by a substance leaving — and which had been forced into claiming phlogiston had negative weight. He is the right person to ask why a percentage yield below 100% cannot mean matter has been destroyed.

  • “Why must the mass of the products equal the mass of the reactants?”
  • “Where does the extra mass come from when a metal burns?”
  • “What was wrong with the phlogiston theory?”
  • “Why did you insist on weighing a sealed vessel?”
  • “If a yield is only 70%, where has the rest actually gone?”

John Dalton — the one who made a formula a statement about countable particles

Step two of every calculation on this page uses the ratio in the balanced equation, and that ratio only means anything because of Dalton. He argued that a compound contains atoms combined in fixed, simple, whole-number ratios — which converts a formula from a summary of proportions by mass into a statement about particles that can be counted. He is also worth asking about being wrong: he assumed the simplest possible formula for any compound, so he took water to be one hydrogen and one oxygen, and the atomic weights that followed were wrong for exactly that reason.

  • “Why must a compound have a fixed whole-number ratio of atoms?”
  • “What changes when a formula counts particles rather than proportions?”
  • “How did you work out relative weights without being able to see an atom?”
  • “Why did you decide water was one hydrogen and one oxygen?”
  • “What made you certain atoms are neither created nor destroyed?”

Then defend it

On Incandio a lesson is not finished when the questions come out right. You teach the idea back to Ember, an AI apprentice who asks the awkward question, and then you argue it against John Dalton in a structured debate marked against descriptors you can read before you enter. Learn it, teach it, then defend it — all three happen on this page once the app loads.

Carry on through the course