S1.4 Counting particles by mass: The mole

Mole (mol)

Definition:The SI unit for amount of substance.One mole contains exactly 6.02 × 10²³ elementary entities (particles).

Analogy:Just like a dozen = 12 items,1 mole = 6.02 × 10²³ items.

Question:Q: What does 1 mole of water molecules represent?A: 6.02 × 10²³ H₂O molecules

Avogadro's Constant (Nₐ)

Definition:The number of particles in 1 mole of a substance.Nₐ = 6.02 × 10²³ mol⁻¹

Unit: mol⁻¹

Question:Q: What does the unit mol⁻¹ mean?A: "Per mole" — used to convert moles to number of particles

Mole-Particle Conversion

Formula:

No. of particles=n×N_A

Where:

n = amount in moles

N_A = Avogadro’s constant (6.02 × 10²³ mol⁻¹)

Electrons as Particles

Definition:Electrons can be counted in moles too.

Example:1 mol e⁻ = 6.02 × 10²³ electrons

Application:Useful in redox and electrolysis calculations.

Relative Atomic Mass Ar

Definition:The weighted average mass of a naturally occurring isotope of an element compared to 1/12 of the mass of a carbon-12 atom.

Key fact:

It is a relative value.

No units are used.

Carbon-12 as a Standard

Carbon-12 (¹²C) is the reference isotope used for defining Ar and Mr.

Value:Carbon-12 is assigned a mass of exactly 12.00.

Molar Mass (M)

Definition: The mass of one mole of a substance.It is numerically equal to the relative atomic or formula mass, but its units are g mol⁻¹.Example:

M of Na = 22.99 g mol⁻¹

M of H₂O = 18.02 g mol⁻¹

Formula (from Data Booklet):

n=m/M

where:

n = amount in moles (mol)

m = mass in grams (g)

M = molar mass (g mol⁻¹)

Units for Mass, Moles, and Molar Mass

Mass (m) → grams (g)

Moles (n) → mol

Molar mass (M) → g mol⁻¹

Empirical Formula

Definition:The empirical formula shows the simplest whole number ratio of atoms of each element in a compound.

Example:Glucose:

Molecular formula = C₆H₁₂O₆

Empirical formula = CH₂O

Molecular Formula

Definition:The molecular formula shows the actual number of atoms of each element in a molecule.

Formula:

Molecular formula=n×Empirical formula

Empirical Formula from % Composition

Assume 100 g sample → % becomes mass in grams

Convert mass → moles using:

moles=mass (g)/Ar

Divide all mole values by the smallest number of moles

Adjust to whole numbers (×2, ×3, etc. if needed)

Concentration

Definition:Concentration is the amount of solute dissolved in a certain volume of solution.

It can be expressed in:

g dm⁻³ (grams per cubic decimetre)

mol dm⁻³ (moles per cubic decimetre or molarity)

Formula for Molar Concentration

Formula (from data booklet):

n=C×V

n = amount of solute in mol

C = molar concentration (mol dm⁻³)

V = volume of solution in dm

Converting Units for Concentration

From g dm⁻³ to mol dm⁻³:

C(mol dm−3)=C(g dm−3)/M

Where:

M = molar mass of solute (g mol⁻¹)

From mol dm⁻³ to g dm⁻³:

C(g dm−3)=C(mol dm−3)×M

Avogadro's Law

Definition:At constant temperature and pressure, equal volumes of all gases contain the same number of molecules (or moles).

V∝n (at constant T and P)

Implication of Avogadro's Law

Key Concept:1 mole of any gas occupies the same volume at the same T and P.

At standard temperature and pressure (STP):

Volume = 22.7 dm³ mol⁻¹

At room temperature and pressure (RTP):

Volume = 24.0 dm³ mol⁻¹ (if specified in question)

Factors Affecting Number of Gas Particles

The number of gas particles in a container depends on:

Volume of container – more volume, more particles

Temperature – higher temperature, faster particles, more space needed

Pressure – higher pressure, more particles in same space

↑ T⇒↑ KE⇒↓ particle density

Using Mole Ratios with Gas Volumes

For a balanced equation:

2H2(g)+O2(g)→2H2O(g)

You can treat mole ratios as volume ratios:

2 mol H₂ : 1 mol O₂ → 2 mol H₂O

So,2 dm³ H₂ + 1 dm³ O₂ → 2 dm³ H₂O(at same T & P)

Gases and Molar Volume at STP

Standard Temperature and Pressure (STP):

T = 273 K

P = 100 kPa

At STP,

1 mol of any gas=22.7 dm³

Use this for converting between moles and volume.

Solving Volume-Mole Problems

Formula:

n=V/V_m

Where:

n = number of moles

V = volume of gas (dm³)

V_m​ = molar volume (22.7 or 24.0 dm³ mol⁻¹)

1⃣ – What Counts as an “Elementary Entity”

Definition: An elementary entity can be an atom, molecule, ion, electron, or any specified group of particles.
Example: 1 mol Na+ = 6.02 x 10^23 sodium ions.

Relationship Between Ar, Mr, and M

• Ar = relative atomic mass (no units)

• Mr = relative formula mass (no units)

• M = molar mass (g mol^-1)
  Note: M has the same number as Mr, but with units (g mol^-1).

Empirical and Molecular Formula Relationship

Formula: Molecular formula = n × Empirical formula
Where: n = molar mass ÷ empirical formula mass

Volume Unit Conversion for Concentration

1 dm^3 = 1000 cm^3 = 1 L
If volume is given in cm^3, divide by 1000 before using n = C × V.

Gas Volume Ratio Example
Question: What volume of O2 is needed to react with 4.0 dm^3 of H2

Equation: 2H2 + O2 → 2H2O
Ratio H2:O2 = 2:1
So O2 volume = 4.0 ÷ 2 = 2.0 dm^3.

Relative molecular mass

Definition: The weighted average mass of one molecule compared to one-twelfth of the mass of a carbon-12 atom. It is calculated by adding together the relative atomic masses (Ar) of all the atoms in the molecular formula.
Used for: Covalent (molecular) substances.
No units.

Relative formula mass

Definition: The weighted average mass of a formula unit of a substance compared to one-twelfth of the mass of a carbon-12 atom. It is calculated by adding together the relative atomic masses (Ar) of all the atoms shown in its formula.
Used for: Ionic compounds and giant structures.
No units.

formula for concentration

c(g dm³) = mass of solute (g) / volume of solution

c(mol dm ³) = amount of solute ( mol) / volume of solution (dm³)

ppm = mass of solute (g) / mass of solution (g)

solution

a homogenous mixture of two or more substances in relative amounts

solvent

a substance that dissolves a solute to form a homogeneous mixture called a solution
in a salt solution, water is the solvent and the salt (sodium chloride) is the solut

solute

solute is the substance that gets dissolved in another substance (the solvent) to form a solutio

dilution formula and is used to calculate the unknown quantity (concentration or volume) when diluting a solution. 

The equation C₁V₁ = C₂V₂ (or M₁V₁ = M₂V₂