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Calculations Using Reacting Masses

A focused revision guide to using balanced equations, mole ratios and molar masses to calculate the mass of a reactant or product in a chemical reaction.

AS Level
Topic 2: Atoms, Molecules and Stoichiometry
9701 Papers 1 and 2
Dr. Mohammed Al-Fatah

Written by:
Dr. Mohammed Al-Fatah

Chemistry specialist revision notes for A Level Chemistry.

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Before you start

GCSE Recap: Balanced Equations and Conservation of Mass

Before you start, check that you can balance an equation and use conservation of mass.

1

Why Equations Can Be Used for Mass Calculations

A balanced chemical equation tells you the mole ratio between reactants and products. Once you know how many moles of one substance are involved, the equation lets you work out how many moles of another substance react or form.

Reacting mass calculations connect three ideas: mass, amount in moles and the balanced equation.

Reacting mass calculation: a calculation that uses a balanced equation to find the mass of a reactant used or a product formed.

Key idea: the balancing numbers in the equation are mole ratios, not mass ratios.

Check your understanding

Quick Check: What the Balancing Numbers Mean

Decide whether the statement about this equation is true or false.

2

The Core Formula

The most important formula for reacting mass calculations links mass, amount and molar mass.

amount in mol = mass ÷ Mr
mass = amount in mol × Mr

You usually use the first form when converting the given mass into moles, and the second form when converting the final amount in moles back into a mass.

QuantityMeaningTypical unit
massThe mass of the substance in the questiong, kg or tonnes
amountThe number of moles of particlesmol
MrRelative formula mass or relative molecular massg mol-1 when used as molar mass
3

The Standard Method

Most reacting mass questions can be solved using the same sequence. The method is especially reliable because it keeps the mole ratio separate from the mass calculation.

Step 1: Write or check the balanced equation

The coefficients in the equation give the mole ratio between substances.

Step 2: Convert the given mass into moles

Use amount = mass ÷ Mr.

Step 3: Use the mole ratio

Use the balanced equation to convert from moles of the known substance to moles of the required substance.

Step 4: Convert moles into the required mass

Use mass = amount × Mr.

Check your understanding

Quick Check: Order the Standard Method

Drag the steps of this reacting mass calculation into the right order.

Using moles and a balanced equation to calculate reacting masses

The calculation pathway is mass → moles → equation ratio → moles → mass.

4

Worked Example: Sodium Hydrogencarbonate

Calculate the mass of carbon dioxide produced when 5.50 g of sodium hydrogencarbonate is heated.

2NaHCO3 → Na2CO3 + CO2 + H2O

Step 1: Find moles of NaHCO3

Mr of NaHCO3 = 84
amount = 5.50 ÷ 84 = 0.0655 mol

Step 2: Use the equation ratio

2 mol NaHCO3 gives 1 mol CO2
0.0655 mol NaHCO3 gives 0.0328 mol CO2

Step 3: Find the mass of CO2

Mr of CO2 = 44.0
mass = 0.0328 × 44.0 = 1.44 g

Exam focus: the mass of CO2 is not found by directly comparing 84 and 44. The equation shows that 2 mol NaHCO3 forms only 1 mol CO2.

Check your understanding

Quick Check: Reacting Masses in Grams

Work each calculation out on paper before you turn the card.

5

Using Direct Proportion

Some reacting mass questions can be solved by treating one formula unit or one mole ratio as a direct proportion. This is useful when the units are large, such as kilograms or tonnes.

For example, in a blast furnace, haematite is reduced to iron:

Fe2O3 + 3CO → 2Fe + 3CO2

From the equation, 1 mol Fe2O3 gives 2 mol Fe. Using Ar values Fe = 56 and O = 16:

160 g Fe2O3 gives 112 g Fe
160 tonnes Fe2O3 gives 112 tonnes Fe
16 tonnes Fe2O3 gives 11.2 tonnes Fe

Key idea: if the same mass unit is used throughout, the ratio still works. Do not mix grams and tonnes in the same line unless you deliberately convert units.

Check your understanding

Quick Check: Tonnes and Kilograms

Use direct proportion and type each answer to 3 significant figures.

6

Multi-Step Equation Chains

Sometimes the product of one equation is used as the reactant in the next equation. The safest approach is to follow the mole ratio through each equation before converting back to mass.

For example, nitrogen can be converted through several steps to nitric acid:

N2 + 3H2 → 2NH3
4NH3 + 5O2 → 4NO + 6H2O
2NO + O2 → 2NO2
2H2O + 4NO2 + O2 → 4HNO3

Tracing the ratios shows that 1 mol N2 eventually gives 2 mol HNO3. Since Mr of N2 = 28 and Mr of HNO3 = 63:

28 tonnes N2 gives 2 × 63 tonnes HNO3
1 tonne N2 gives (2 × 63) ÷ 28 = 4.5 tonnes HNO3
Check your understanding

Quick Check: Follow the Chain

Trace the mole ratio through all three equations before you calculate a mass.

7

Units and Significant Figures

Reacting mass calculations often lose marks through unit errors rather than chemistry errors. Keep the unit consistent and present the final answer to a sensible number of significant figures.

IssueWhat to doExample
Mixed unitsConvert before calculating, or keep the same unit throughout a proportion.Use all grams, all kilograms or all tonnes.
Premature roundingKeep extra figures in intermediate steps.Use 0.03275 mol before rounding the final answer.
Missing equation ratioAlways use the coefficients from the balanced equation.2NaHCO3 : 1CO2, not 1 : 1.
Check your understanding

Quick Check: Sound Working

In each round, choose the statement that shows sound working.

8

Common Exam Points

  • The balancing numbers in an equation represent a mole ratio.
  • Do not compare masses directly unless you have first converted through moles or built a valid proportion.
  • Use amount = mass ÷ Mr to find moles from mass.
  • Use mass = amount × Mr to convert moles back into mass.
  • Use the balanced equation to move between the known substance and the required substance.
  • Check whether the question asks for a reactant mass or a product mass.
  • Keep units consistent, especially when questions use kg or tonnes.
  • Show enough working so the mole ratio and formula mass steps are visible.
9

Limiting and Excess Reagents

Reactants are rarely mixed in exactly the ratio the equation demands. The reactant that runs out first is the limiting reagent; it decides how much product can form. The other reactant is in excess and some of it is left over.

Worked example. 5.6 g of iron is heated with 4.0 g of sulfur: Fe + S → FeS.

  1. n(Fe) = 5.6 / 55.8 = 0.100 mol; n(S) = 4.0 / 32.1 = 0.125 mol
  2. The equation needs 1 : 1, so 0.100 mol of iron needs 0.100 mol of sulfur. There is 0.125 mol of sulfur, so sulfur is in excess and iron is limiting.
  3. Product from the limiting reagent: n(FeS) = 0.100 mol, mass = 0.100 × 87.9 = 8.79 g
  4. Sulfur left over: 0.125 − 0.100 = 0.025 mol = 0.80 g

Always divide each reactant’s moles by its coefficient in the equation before comparing; the smaller result identifies the limiting reagent.

Exam sentence: The limiting reagent is the reactant that is completely used up; the amount of product is calculated from its moles, not from the reactant in excess.

Check your understanding

Quick Check: Which Reactant Runs Out?

Find the limiting reagent, then use it to calculate the mass of product.

QuickSnap

This text summary condenses the page into the essential exam ideas.

  • Balanced equations: show mole ratios between reactants and products.
  • First conversion: use amount = mass ÷ Mr.
  • Equation step: use the coefficients to convert from one substance to another.
  • Final conversion: use mass = amount × Mr.
  • Direct proportion: works when the equation ratio and units are handled consistently.
  • Common mistake: treating the equation numbers as mass ratios instead of mole ratios.

FAQs

These questions address the most common points students confuse when calculating reacting masses.

What is a reacting mass calculation?

It is a calculation that uses a balanced chemical equation to work out the mass of a reactant used or a product formed.

Why do I need the balanced equation?

The balanced equation gives the mole ratio between substances. Without this ratio, the calculation may use the wrong number of moles.

Can I compare masses directly?

Not usually. The equation gives mole ratios, so you should convert through moles unless you have built a valid direct proportion from molar masses.

Does the method work with tonnes?

Yes. The same ratio works with grams, kilograms or tonnes as long as the unit is used consistently throughout the proportion.

What is the most common mistake?

The most common mistake is forgetting to use the coefficient ratio from the balanced equation, such as treating 2NaHCO3 → CO2 as a 1 : 1 mole relationship.

Related Topics

Use these topics to strengthen the equation writing and mole calculation skills that support reacting mass questions.

Writing Chemical Equations Equations and Reaction Types Ionic Equations Calculations Involving Moles RAM Calculations Topic 2 Overview

Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.