Concentration and Solution Calculations — AQA A-level Chemistry

SPEC 3.1.2.2 • AMOUNT OF SUBSTANCE

Moles, Mass & Concentration Calculations

Converting between mass, moles, and concentration in solutions

Introduction

In Lesson 2.01, we learned that the mole is the chemist's counting unit. Now we'll master the calculations that connect mass, moles, and concentration — essential skills for almost every chemistry problem.

Key Relationships

This lesson connects three fundamental quantities:Mass (m)

Measured in grams (g) Moles (n)

Amount of substance (mol) Concentration (c)

mol dm⁻³ or g dm⁻³

Learning Objectives

By the end of this lesson, you will be able to:

Convert between mass and moles using n = m/M

Calculate concentration using c = n/V

Convert between mol dm⁻³ and g dm⁻³

Solve multi-step problems combining all three quantities

The Mole Equation

The fundamental relationship between moles, mass, and molar mass is:

Formula

n = \frac{m}{M}

SymbolQuantityUnit

n

Number of molesmol

m

Massg

M

Molar mass (same as Mᵣ numerically)g mol⁻¹

Using the Triangle

Cover the quantity you want to find. What remains shows the calculation:

Find n: cover n → m ÷ M remains

Find m: cover m → n × M remains

Find M: cover M → m ÷ n remains

Rearranging the Mole Equation

From the base equation

n = \frac{m}{M}

, we can derive two rearrangements:

Find moles

n = \frac{m}{M}

Find mass

m = n \times M

Find molar mass

M = \frac{m}{n}

Why This Works

Molar mass (M) tells us the mass of one mole of a substance. If M(H₂O) = 18.0 g mol⁻¹, then:

1 mol of water has mass 18.0 g

2 mol of water has mass 36.0 g

0.5 mol of water has mass 9.0 g

To find moles, we simply divide the mass by the mass of one mole.

Common Error

Don't confuse Mᵣ (relative molecular mass, no units) with M (molar mass, units of g mol⁻¹). Numerically they're equal, but M has units.

01Worked Example

Calculate the number of moles in 9.80 g of sulfuric acid, H₂SO₄.

Aᵣ values: H = 1.0, S = 32.1, O = 16.0

1

Calculate the molar mass: M(H₂SO₄) = 2(1.0) + 32.1 + 4(16.0) = 2.0 + 32.1 + 64.0 = 98.1 g mol⁻¹

2

Apply the formula:

n = \frac{m}{M} = \frac{9.80}{98.1}

3

Calculate: n = 0.0999 mol ≈ 0.100 mol

Answer: 0.100 mol

02Worked Example

Calculate the mass of 0.350 mol of calcium carbonate, CaCO₃.

Aᵣ values: Ca = 40.1, C = 12.0, O = 16.0

1

Calculate the molar mass: M(CaCO₃) = 40.1 + 12.0 + 3(16.0) = 40.1 + 12.0 + 48.0 = 100.1 g mol⁻¹

2

Rearrange to find mass: m = n × M

3

Calculate: m = 0.350 × 100.1 = 35.0 g

Answer: 35.0 g

Concentration of Solutions

When a solute dissolves in a solvent, we describe the amount of solute using concentration.

Definition

Concentration is the amount of solute dissolved per unit volume of solution.

c = \frac{n}{V}

SymbolQuantityUnit

c

Concentrationmol dm⁻³

n

Number of moles of solutemol

V

Volume of solutiondm³

Understanding mol dm⁻³

A concentration of 2.0 mol dm⁻³ means there are 2.0 moles of solute in every 1 dm³ (1 litre) of solution. This unit is sometimes written as "M" (molar) in older texts.

Volume Conversions

Volumes are often given in cm³ but the formula requires dm³. You must convert!

Conversion

1 dm³ = 1000 cm³To convert cm³ to dm³: divide by 1000

Quick Examples

250 cm³ = 250 ÷ 1000 = 0.250 dm³

50 cm³ = 50 ÷ 1000 = 0.050 dm³

25.0 cm³ = 25.0 ÷ 1000 = 0.0250 dm³

Most Common Exam Error

Forgetting to convert cm³ to dm³ is the single most common error in concentration calculations. Always check your volume units!

Worked Example03

0.500 mol of sodium hydroxide is dissolved in water to make 250 cm³ of solution. Calculate the concentration in mol dm⁻³.

1

Convert volume to dm³: V = 250 ÷ 1000 = 0.250 dm³

2

Apply the formula:

c = \frac{n}{V} = \frac{0.500}{0.250}

3

Calculate: c = 2.00 mol dm⁻³

Answer: 2.00 mol dm⁻³

Worked Example04

Calculate the number of moles of HCl in 50.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid.

1

Convert volume to dm³: V = 50.0 ÷ 1000 = 0.0500 dm³

2

Rearrange to find moles: n = c × V

3

Calculate: n = 0.200 × 0.0500 = 0.0100 mol

Answer: 0.0100 mol (or 1.00 × 10⁻² mol)

Concentration in g dm⁻³

Concentration can also be expressed as mass per unit volume (g dm⁻³) instead of moles per unit volume (mol dm⁻³).

Converting Between Units

mol dm⁻³ → g dm⁻³

g dm^{- 3} = mol dm^{- 3} \times M

g dm⁻³ → mol dm⁻³

mol dm^{- 3} = \frac{g dm ^{- 3}}{M}

Why Two Units?

mol dm⁻³ is more useful for chemical calculations (stoichiometry, reactions).

g dm⁻³ is more practical for making solutions in the lab (weighing solutes).

Memory Aid

To convert mol dm⁻³ to g dm⁻³, multiply by M (making it bigger — grams are bigger numbers than moles).

To convert g dm⁻³ to mol dm⁻³, divide by M (making it smaller).

Making Standard Solutions

A standard solution is a solution of known, precise concentration. To make one, we need to calculate the mass of solute required.

Key Relationship

Combining our equations:

m = c \times V \times M

Mass of solute = concentration × volume × molar mass

Practical Application

To make 250 cm³ of 0.100 mol dm⁻³ NaOH solution:

Calculate moles needed: n = c × V = 0.100 × 0.250 = 0.0250 mol

Calculate mass needed: m = n × M = 0.0250 × 40.0 = 1.00 g

Weigh 1.00 g NaOH, dissolve, make up to 250 cm³

05Worked Example

A solution of NaCl has a concentration of 0.500 mol dm⁻³. Convert this to g dm⁻³.

Mᵣ(NaCl) = 58.5

1

Identify what we have: concentration = 0.500 mol dm⁻³, M = 58.5 g mol⁻¹

2

To convert mol dm⁻³ to g dm⁻³, multiply by M

3

Calculate: 0.500 × 58.5 = 29.3 g dm⁻³

Answer: 29.3 g dm⁻³

06Worked Example

Calculate the mass of potassium hydroxide, KOH, needed to make 500 cm³ of 0.250 mol dm⁻³ solution.

Mᵣ(KOH) = 56.1

1

Convert volume: V = 500 ÷ 1000 = 0.500 dm³

2

Calculate moles needed: n = c × V = 0.250 × 0.500 = 0.125 mol

3

Calculate mass: m = n × M = 0.125 × 56.1 = 7.01 g

Answer: 7.01 g

Multi-Step Calculations

Many exam questions require you to combine multiple formulas in a single problem. The key is to identify what you're given and what you need to find.

Problem-Solving Strategy

List what you know and what you need to find

Plan the pathway (which formulas connect your data?)

Convert all units first (especially cm³ to dm³)

Calculate step by step, showing all working

Check your answer makes sense

Golden Rule

Moles is always the connecting point. Whether you're converting mass to concentration, or concentration to particles, you go through moles.

The Concentration Triangle

Just like the mole triangle, we can use a triangle for concentration calculations:

Find n n = c × V

Find c c = n ÷ V

Find V V = n ÷ c

Essential Check

Before using the concentration triangle, always check:

Is the volume in dm³? (If cm³, divide by 1000)

Is the concentration in mol dm⁻³? (If g dm⁻³, divide by M first)

07Worked Example

4.00 g of sodium hydroxide is dissolved in water to make 200 cm³ of solution. Calculate the concentration in mol dm⁻³.

Mᵣ(NaOH) = 40.0

1

Calculate moles of NaOH: n = m ÷ M = 4.00 ÷ 40.0 = 0.100 mol

2

Convert volume: V = 200 ÷ 1000 = 0.200 dm³

3

Calculate concentration: c = n ÷ V = 0.100 ÷ 0.200 = 0.500 mol dm⁻³

Answer: 0.500 mol dm⁻³

Worked Example08

What volume of 0.150 mol dm⁻³ sulfuric acid contains 0.0300 mol of H₂SO₄?

1

Identify: n = 0.0300 mol, c = 0.150 mol dm⁻³, find V

2

Use: V = n ÷ c = 0.0300 ÷ 0.150 = 0.200 dm³

3

Convert to cm³ if required: 0.200 × 1000 = 200 cm³

Answer: 0.200 dm³ (or 200 cm³)

09Worked Example — Challenging

A student dissolves 7.36 g of copper(II) sulfate, CuSO₄, in water to make 400 cm³ of solution. Calculate the concentration in both mol dm⁻³ and g dm⁻³.

Aᵣ values: Cu = 63.5, S = 32.1, O = 16.0

1

Calculate molar mass: M(CuSO₄) = 63.5 + 32.1 + 4(16.0) = 63.5 + 32.1 + 64.0 = 159.6 g mol⁻¹

2

Calculate moles: n = m ÷ M = 7.36 ÷ 159.6 = 0.0461 mol

3

Convert volume: V = 400 ÷ 1000 = 0.400 dm³

4

Concentration in mol dm⁻³: c = n ÷ V = 0.0461 ÷ 0.400 = 0.115 mol dm⁻³

5

Concentration in g dm⁻³: c = 0.115 × 159.6 = 18.4 g dm⁻³

Answer: 0.115 mol dm⁻³ and 18.4 g dm⁻³

10

Worked Example — Most Challenging

A chemist has a 0.250 mol dm⁻³ solution of sodium chloride. What volume of this solution contains 3.01 × 10²² formula units of NaCl?

Nₐ = 6.02 × 10²³ mol⁻¹

1

Convert particles to moles: n = N ÷ Nₐ = (3.01 × 10²²) ÷ (6.02 × 10²³)

2

Calculate: n = 0.0500 mol

3

Use V = n ÷ c to find volume: V = 0.0500 ÷ 0.250 = 0.200 dm³

4

Convert to cm³: V = 0.200 × 1000 = 200 cm³

Answer: 0.200 dm³ (or 200 cm³)

Practice

Question 1:

Aᵣ values: Na = 23.0, N = 14.0, O = 16.0

1

Calculate Mᵣ: M(NaNO₃) = 23.0 + 14.0 + 3(16.0) = 23.0 + 14.0 + 48.0 = 85.0 g mol⁻¹

2

Apply formula: n = m ÷ M = 8.50 ÷ 85.0 = 0.100 mol

Answer: 0.100 mol

Practice

Question 2:

Mᵣ(H₂SO₄) = 98.2

1

To convert g dm⁻³ to mol dm⁻³, divide by M

2

Concentration = 4.91 ÷ 98.2 = 0.0500 mol dm⁻³

Answer: 0.0500 mol dm⁻³

Practice

Question 3:

Aᵣ values: Na = 23.0, C = 12.0, O = 16.0

1

Calculate Mᵣ: M(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 46.0 + 12.0 + 48.0 = 106.0 g mol⁻¹

2

Calculate moles: n = m ÷ M = 5.30 ÷ 106.0 = 0.0500 mol

3

Convert volume: V = 250 ÷ 1000 = 0.250 dm³

4

Calculate concentration: c = n ÷ V = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³

Answer: 0.200 mol dm⁻³

Summary

Key Formulas

PurposeFormulaUnits

Moles from mass

n = \frac{m}{M}

mol, g, g mol⁻¹

Mass from moles

m = n \times M

g, mol, g mol⁻¹

Concentration

c = \frac{n}{V}

mol dm⁻³, mol, dm³

Moles from solution

n = c \times V

mol, mol dm⁻³, dm³

Volume conversiondm³ = cm³ ÷ 1000—

Exam Tips

Always convert cm³ to dm³ before using concentration formulas

Moles is the hub — convert to moles first, then to your target quantity

Check units — mol dm⁻³ vs g dm⁻³ require different approaches

Show all working — method marks are often available even if the final answer is wrong