Last updated: 26 September 2026
A balanced chemical equation is more than a statement of which substances react. Its coefficients tell us the relative numbers of particles — and therefore the relative numbers of moles — involved in the reaction. This is what allows an experimental measurement such as a mass, gas volume or titre to reveal the amount of another substance.
The important habit is not to memorise a different calculation method for every situation. Instead, think of moles as the bridge. Convert what you know into moles, cross the bridge using the balanced-equation ratio, and then convert the result into whatever quantity the question asks for.
Try these five questions before reading the chapter. They check the mole ideas, unit conversions and reaction ratios that this topic builds on.
- 5 short questions
- Instant explanation for every option
- No marks are lost — this is a starting-point check
The Universal Stoichiometric Route
Most reacting-quantity questions can be solved using the same sequence:
The formula used to enter or leave this sequence depends on the quantity involved.
| Information given | Useful relationship | Important point |
|---|---|---|
| Mass | n = m/Mᵣ | Use mass in g. |
| Solution concentration and volume | n = cV | Use V in dm³ when c is in mol dm⁻³. |
| Gas volume and molar gas volume | n = V/Vₘ | V and Vₘ must use the same volume unit. |
| Gas pressure, volume and temperature | pV = nRT | Use a consistent set of units. |
The balanced equation supplies the mole ratio
Consider:
The equation tells us that:
- 1 mol of N₂ reacts with 3 mol of H₂;
- 1 mol of N₂ produces 2 mol of NH₃; and
- 3 mol of H₂ produces 2 mol of NH₃.
These numbers are mole ratios. They are not normally mass ratios. One gram of nitrogen therefore does not react with three grams of hydrogen simply because the coefficients are 1 and 3.
Masses, solution concentrations and many gas measurements cannot be compared directly through the coefficients. Convert the known quantity into moles first, apply the mole ratio, and only then convert to the quantity required.
Reacting-Mass Calculations
A reacting-mass calculation begins with the mass of one substance and asks for the mass of another substance. Because different substances have different molar masses, the masses cannot be compared directly.
Worked example: forming magnesium oxide
Magnesium burns in oxygen:
Calculate the mass of MgO formed when 4.86 g of Mg burns completely in excess oxygen.
Mg and MgO have a 2 : 2 mole ratio, which is equivalent to 1 : 1.
Answer: 8.06 g of MgO is formed.
Mole ratio does not mean mass ratio
Incorrect: 2 mol Mg forms 2 mol MgO, so 4.86 g Mg must form 4.86 g MgO.
Correct: MgO contains both magnesium and oxygen, so its molar mass is greater. The 2 : 2 ratio applies to moles, not grams.
When both reactants are given: the limiting reagent
If quantities of two reactants are supplied, they may not be present in exactly the stoichiometric ratio. The limiting reagent is the reactant that is completely consumed first. It determines the maximum amount of product that can form.
One reliable method is to convert both reactants into moles and compare:
The smaller value identifies the limiting reagent.
Suppose 4.86 g of Mg reacts with 4.00 g of O₂.
| Reactant | Amount / mol | Coefficient | Amount ÷ coefficient |
|---|---|---|---|
| Mg | 0.200 | 2 | 0.100 |
| O₂ | 0.125 | 1 | 0.125 |
Magnesium gives the smaller value, so Mg is the limiting reagent. Oxygen is present in excess.
Finding how much excess reagent remains
Being in excess does not mean that a substance does not react. Some of it reacts and the rest remains.
Therefore:
From the equation, 2 mol of Mg react with 1 mol of O₂. Therefore, 0.200 mol of Mg consumes:
Initially there were 0.125 mol of O₂.
Here, both 0.125 mol and 0.100 mol refer to O₂. One is the amount initially present and the other is the amount of O₂ that reacted. Their difference is therefore the amount of O₂ remaining.
Reacting Volumes of Gases
At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. Gas volume is therefore proportional to the amount of gas in moles.
This gives us a useful shortcut. If all the quantities being compared are gaseous and are measured under the same conditions, the balanced-equation coefficients can be used directly as volume ratios.
Worked example: reacting gas volumes
If 25.0 cm³ of N₂ reacts completely, with all gas volumes measured at the same temperature and pressure:
No conversion to moles is required because the volume ratio itself represents the mole ratio under unchanged temperature and pressure.
Use the coefficients directly as volume ratios only when the gases are compared at the same temperature and pressure. If conditions differ, convert the gas quantities into moles first.
Using the molar volume of a gas
When a measured gas volume must be converted into moles, use:
At room temperature and pressure, a commonly used approximate molar gas volume is:
At 273 K and 100 kPa, the molar gas volume is approximately:
Use the conditions or molar gas volume specified in the question. If pressure, volume and temperature are supplied instead, the ideal gas equation may be more appropriate.
Worked example: converting gas volume into mass
Calcium carbonate reacts with hydrochloric acid:
Calculate the mass of CaCO₃ required to produce 480 cm³ of CO₂ at room temperature and pressure.
CaCO₃ and CO₂ are in a 1 : 1 mole ratio.
Answer: 2.00 g of CaCO₃ is required.
Gas mixtures before and after reaction
Gas-volume questions may also give a mixture containing an excess of one gas and ask for the total volume after reaction. In these questions, identify the limiting gas, calculate the excess gas remaining, and then add any gaseous products that remain under the final conditions.
For example, 20.0 cm³ of CH₄ is mixed with 50.0 cm³ of O₂ and ignited.
At the same temperature and pressure, 20.0 cm³ of CH₄ requires 40.0 cm³ of O₂. Methane is therefore completely consumed, while 10.0 cm³ of O₂ remains.
The 20.0 cm³ of CH₄ also produces 20.0 cm³ of CO₂.
| Gas | Initial volume / cm³ | Change / cm³ | Volume after reaction / cm³ |
|---|---|---|---|
| CH₄ | 20.0 | −20.0 | 0 |
| O₂ | 50.0 | −40.0 | 10.0 |
| CO₂ | 0 | +20.0 | 20.0 |
If the mixture is cooled to room temperature, the water formed condenses and is no longer counted as part of the gas volume.
Always check the final physical states
Hot reaction mixture: water produced during combustion may be present as steam.
After cooling: the water condenses, so it is normally excluded from the final gas volume.
Reacting Volumes and Concentrations of Solutions
The molar concentration of a solution tells us how many moles of dissolved solute are present per dm³ of solution.
Rearranging:
If concentration is expressed in mol dm⁻³, volume must be expressed in dm³.
Worked example: reacting two solutions
Sodium carbonate reacts with hydrochloric acid:
Calculate the volume of 0.200 mol dm⁻³ HCl required to react completely with 25.0 cm³ of 0.120 mol dm⁻³ Na₂CO₃.
One mole of Na₂CO₃ reacts with two moles of HCl.
Answer: 30.0 cm³ of HCl is required.
Use n = cV before applying the mole ratio
There is no need to memorise a special formula for reacting solution volumes. Keep the reasoning visible:
Begin with n = cV. Use the balanced equation to find the moles of the other substance. Then use c = n/V or V = n/c as required.
The same mole method works for every reaction ratio. It remains clear whether the reacting ratio is 1 : 1, 1 : 2, 2 : 3 or something else.
Reaction is not the same as dilution
Dilution: the amount of the same solute remains unchanged while more solvent is added. This can lead to c₁V₁ = c₂V₂.
Reaction: different substances are chemically consumed, so the balanced-equation mole ratio must be used.
Titration Calculations
A titration determines the volume of one solution required to react with a known volume of another. At the equivalence point, the reacting substances have been mixed in the stoichiometric ratio shown by the balanced equation.
A reliable calculation sequence is:
- Obtain the mean concordant titre.
- Convert the titre from cm³ into dm³.
- Calculate the moles of the solution whose concentration is known.
- Use the balanced equation to find the moles of the other substance.
- Use its measured volume to calculate the required concentration.
Worked example: finding an acid concentration
A 25.0 cm³ sample of H₂SO₄ is titrated with 0.150 mol dm⁻³ NaOH. The mean NaOH titre is 24.60 cm³.
Two moles of NaOH react with one mole of H₂SO₄.
This amount of H₂SO₄ was present in 25.0 cm³ = 0.0250 dm³.
Answer: The concentration of H₂SO₄ is 0.0738 mol dm⁻³.
Recording Burette Readings and Titres
A conventional burette is normally graduated in 0.1 cm³ divisions. The position of the meniscus can be estimated between adjacent divisions, so a burette reading is recorded to two decimal places, normally ending in 0 or 5.
For a typical aqueous solution, read the bottom of the concave meniscus with your eye at the same horizontal level. Looking from above or below introduces parallax error.
| Reading / cm³ | Rough | 1 | 2 | 3 |
|---|---|---|---|---|
| Final burette reading | 24.60 | 24.45 | 48.85 | 24.40 |
| Initial burette reading | 0.00 | 0.10 | 24.45 | 0.05 |
| Titre | 24.60 | 24.35 | 24.40 | 24.35 |
The rough titration is used to locate the approximate endpoint and is normally not included in the mean. The accurate titres 24.35 cm³, 24.40 cm³ and 24.35 cm³ agree closely and are concordant.
The mean titre is also written to two decimal places. Unlike an individual burette reading, its second decimal place does not have to be 0 or 5 because the mean is a calculated value.
Record the actual initial reading and actual final reading. The volume delivered is their difference. Starting every titration exactly at zero is unnecessary.
Which titres should be averaged?
Use the concordant accurate titres rather than automatically averaging every titre recorded. The rough titre is excluded, and an anomalous accurate titre should not be included merely because it appears in the table.
Keep measurement precision separate from final-answer rounding
Burette readings and individual titres: two decimal places.
Mean titre: two decimal places.
Intermediate calculations: retain additional calculator digits.
Final calculated answer: normally three significant figures unless the question instructs otherwise.
Back Titration
Sometimes a substance cannot conveniently be titrated directly. It may be insoluble, may react too slowly, or may not give a suitable direct endpoint. In such cases, a back titration can be used.
A known amount of another reagent is first added in excess. The original sample reacts completely, leaving some of the added reagent unreacted. A second titration then determines how much of that excess reagent remains.
The second titration does not directly measure the amount that reacted with the original sample. It measures the amount of excess reagent remaining. Subtract this from the amount originally added to find the amount that actually reacted.
The four-stage logic
- Calculate the amount of excess reagent originally added.
- Use the second titration to determine how much of that reagent remains.
- Subtract to find how much reacted with the original sample.
- Use the original reaction equation to calculate the amount of the substance being analysed.
Worked example: percentage purity of calcium carbonate
A 1.00 g impure sample containing CaCO₃ is treated with 50.0 cm³ of 0.400 mol dm⁻³ HCl. After the reaction is complete, the excess HCl requires 18.60 cm³ of 0.200 mol dm⁻³ NaOH for neutralisation.
Assume that the impurities do not react with HCl. Calculate the percentage by mass of CaCO₃ in the sample.
The reactions are:
First calculate the amount of HCl originally added.
Next use the NaOH titration to determine the HCl remaining.
HCl and NaOH react in a 1 : 1 mole ratio.
Both quantities now refer to the same substance, HCl:
- 0.0200 mol HCl was present initially;
- 0.003720 mol HCl remained after reaction with the sample.
The difference is the amount of HCl that reacted with CaCO₃.
Two moles of HCl react with one mole of CaCO₃.
Keep the unrounded value during the calculation and round only the final result.
Answer: The sample contains 81.5% CaCO₃ by mass.
Track the HCl through the experiment
Initially added: 0.0200 mol HCl.
Remaining: 0.003720 mol HCl, found using the NaOH titration.
Reacted with CaCO₃: initial amount − remaining amount.
Why must the first reagent be in excess?
The purpose of the excess reagent is to ensure that the original sample reacts completely. If insufficient reagent were added, some of the sample could remain unreacted and the analysis would give an incorrect result.
Linking Mass, Solution and Gas Data
Examination questions often combine several types of quantitative information. A question may begin with a mass, produce a gas, and then ask you to determine an unknown molar mass or composition.
The method does not change:
Worked example: identifying a Group 2 carbonate
A 0.500 g sample of a pure Group 2 carbonate, MCO₃, reacts with excess acid and produces 120 cm³ of CO₂ at room temperature and pressure.
Calculate the relative atomic mass of M and identify the metal.
MCO₃ and CO₂ are in a 1 : 1 mole ratio.
Answer: Aᵣ(M) ≈ 40.0, so M is calcium.
Do not round too early
Multi-step stoichiometric calculations often involve several ratios, conversions and subtractions. If every intermediate answer is rounded, the small errors can accumulate.
Keep the full calculator value during the working and normally round only the final calculated answer to three significant figures, unless instructed otherwise.
Common Mistakes and Why They Are Wrong
| Common mistake | Why it is wrong |
|---|---|
| Applying equation coefficients directly to masses | The coefficients represent mole ratios, not mass ratios. Convert mass to moles first. |
| Using an unbalanced equation | The coefficients will not give the correct stoichiometric ratio. |
| Using cm³ directly in n = cV when c is in mol dm⁻³ | The volume must first be converted into dm³ by dividing by 1000. |
| Using gas-volume ratios when the temperatures or pressures differ | Gas volume depends on temperature and pressure. Direct volume ratios apply only under the same conditions. |
| Counting water vapour in a gas mixture after it has been cooled | Water normally condenses on cooling and is no longer part of the gaseous mixture. |
| Using c₁V₁ = c₂V₂ as a general titration formula | That relationship describes dilution of the same solute. A chemical reaction requires the balanced-equation mole ratio. |
| Writing conventional burette readings to one decimal place | Burette readings are recorded to two decimal places, normally ending in 0 or 5. |
| Assuming a burette must start at 0.00 cm³ | The titre is determined from final reading minus initial reading, whatever the starting reading. |
| Including the rough titre in the mean | The rough titre is used to locate the endpoint and is normally excluded from the mean concordant titre. |
| Subtracting amounts of two different substances | Initial, reacted and remaining quantities must refer to the same substance before they can be related by subtraction. |
| Thinking the back titration directly measures what reacted with the sample | The second titration measures the excess reagent left over. This must be subtracted from the initial amount. |
| Forgetting that a back titration involves two reaction ratios | One ratio is used for the second titration and another for the original sample reaction. |
| Comparing reactant moles directly to identify the limiting reagent | The stoichiometric coefficients may differ. Compare each amount relative to its coefficient. |
| Rounding after every calculation step | Premature rounding can accumulate error. Retain additional digits until the final answer. |
Final Revision Checklist
Stoichiometric route
I can convert a known mass, solution volume or gas volume into moles, apply the balanced-equation ratio, and convert to the required quantity.
Reacting masses
I can use n = m/Mᵣ and understand why equation coefficients represent mole ratios rather than mass ratios.
Limiting reagent
I can identify which reactant is completely consumed and calculate how much of an excess reagent remains.
Gas volumes
I can use gas-volume ratios when gases are measured at the same temperature and pressure and use a molar gas volume when conversion to moles is required.
Solutions
I can convert cm³ to dm³, use n = cV, apply the reaction ratio, and calculate an unknown concentration or volume.
Titration readings
I can record conventional burette readings and titres to two decimal places and calculate a mean from suitable concordant titres.
Back titration
I can distinguish the amount of excess reagent initially added, the amount remaining and the amount that reacted with the sample.
Final answers
I can retain unrounded intermediate values and normally report the final calculated answer to three significant figures unless instructed otherwise.
Now apply the chapter to unfamiliar calculations. These questions combine reaction ratios, masses, gases, solutions, excess reagents and titration data.
- 10 application-focused questions
- Feedback explains every answer choice
- Use your score to identify what to revise
Frequently Asked Questions
Questions Students Often Ask About Reacting Quantities
These questions clarify the calculation choices that often cause confusion when mass, gas, solution and titration data appear in the same problem.
Do I need a special formula for titration calculations?
No. Use the normal stoichiometric method. Calculate the amount of the known solution using n = cV, apply the mole ratio from the balanced equation, and then use c = n/V or V = n/c for the required solution.
This is safer than memorising a combined expression because the reaction ratio remains visible throughout the calculation.
When exactly should I subtract in a stoichiometry question?
Subtract when you are comparing the same substance at different stages.
For example, if 0.0200 mol HCl was initially added and 0.00372 mol HCl remains:
Do not subtract simply because the question uses the word “excess”. First identify what was initially present, what reacted and what remained.
Why does a back titration require subtraction?
The reagent added to the original sample is deliberately supplied in excess. The second titration measures the portion of that reagent that remains unreacted.
Subtracting the amount remaining from the amount originally added tells you how much actually reacted with the sample.
Must a burette start at exactly 0.00 cm³?
No. Record the actual initial reading and the actual final reading. The titre is calculated as final reading minus initial reading.
For example, an initial reading of 1.25 cm³ and a final reading of 26.10 cm³ gives a titre of 24.85 cm³.
Why are burette readings written to two decimal places?
A conventional burette is normally graduated every 0.1 cm³, and the position of the meniscus is estimated between adjacent divisions. Readings are therefore recorded to two decimal places, normally ending in 0 or 5.
Can the mean titre end in a digit other than 0 or 5?
Yes. Individual conventional burette readings and titres normally end in 0 or 5 in the second decimal place, but the arithmetic mean is a calculated value.
For example, 24.35 cm³, 24.40 cm³ and 24.35 cm³ have a mean of 24.37 cm³ when stated to two decimal places.
Should I always average every accurate titre?
No. Use suitable concordant accurate titres. The rough titre is normally excluded, and an accurate titre that is clearly anomalous should not be included merely because it was obtained after the rough titration.
Must every gas volume be converted into moles?
No. When gaseous quantities are measured at the same temperature and pressure, their volume ratio is the same as their mole ratio. The balanced coefficients may therefore be used directly as gas-volume ratios.
Convert to moles when the conditions differ, when a molar gas volume or pV = nRT is needed, or when the gas quantity must be linked to a mass or solution quantity.
Why is water sometimes counted as a gas and sometimes not?
It depends on the conditions. In a very hot combustion mixture, water may be present as steam. After the gases are cooled to room temperature, the water normally condenses and is no longer included in the gas volume.
How should I round my answers?
Keep measurement conventions separate from final-answer rounding. Conventional burette readings, titres and mean titres are stated to two decimal places.
During calculations, retain additional calculator digits rather than rounding at every stage. Unless instructed otherwise, normally give the final calculated mass, amount, concentration or percentage to three significant figures.
Bringing the Topic Together
Reacting masses, gas volumes and solution volumes are all versions of the same stoichiometric problem. Convert measurable quantities into moles, use the balanced equation to move from one substance to another, and then convert the result into the quantity required. When excess reagents or back titrations appear, label the amounts as initial, reacted and remaining before doing any subtraction. With that framework secure, even long multi-stage calculations become a sequence of short, logical steps.