Amount of Substance (mol) and Counting Particles

25min Part 1 / Ch2 / Lesson 1

Objectives

Explain the definition of amount of substance (mol)
Convert between number of particles and amount of substance using Avogadro's constant
Convert between mass and amount of substance using molar mass
Convert between gas volume and amount of substance using molar volume

The Core Idea

The mol is a 'dozen for particles' — just as 1 dozen = 12, 1 mol = 6.02×10²³

Amount of substance (mol) counts particles (atoms, molecules, ions, etc.) in groups of $6.022 \times 10^{23}$ . This number $6.022 \times 10^{23}\ \mathrm{/mol}$ is called Avogadro’s constant $N_A$ .

Why Do We Need the Mol?

Atoms and molecules are extremely small. Chemical reactions involve enormous numbers of particles.

Mass of one hydrogen atom: ~ $1.67 \times 10^{-24}\ \mathrm{g}$
Water molecules in a glass of water: ~ $6.7 \times 10^{24}$

Writing such huge numbers every time is impractical. That’s why we use a “grouping unit” = mol.

Common Misconception

“The mol is a unit of mass” → Wrong. The mol is a unit of quantity (number of particles). $1\ \mathrm{mol} = 6.022 \times 10^{23}$ particles. The unit of mass is $\mathrm{g}$ . To convert mol to mass, you need the “molar mass.”

Mol ↔ Number of Particles

The relationship between number of particles $N$ and amount of substance $n\ \mathrm{(mol)}$ :

N = n \times N_A

mol → particles: ×NA; particles → mol: ÷NA

Example: If you have $2.0\ \mathrm{mol}$ of water $\mathrm{H_2O}$ , how many water molecules is that?

N = 2.0 \times 6.022 \times 10^{23} = 1.2 \times 10^{24}\ \text{molecules}

Mol ↔ Mass (Molar Mass)

Molar mass $M\ \mathrm{(g/mol)}$ = mass per mole of substance. Its numerical value equals the relative atomic/molecular/formula mass.

m = n \times M

The triangle of conversions: mol ↔ mass ↔ particles

Example: Carbon $\mathrm{C}$ has an atomic mass of $12.0$ , so molar mass $M = 12.0\ \mathrm{g/mol}$ . For $24.0\ \mathrm{g}$ of carbon:

n = \frac{m}{M} = \frac{24.0\ \mathrm{g}}{12.0\ \mathrm{g/mol}} = 2.00\ \mathrm{mol}

Mol ↔ Gas Volume (Standard Conditions)

At standard conditions ( $0\ \mathrm{°C}$ , $1.013 \times 10^5\ \mathrm{Pa}$ ), all gases occupy $22.4\ \mathrm{L}$ per mol regardless of the type of gas (Avogadro’s law).

V = n \times 22.4\ \mathrm{(L)}

At standard conditions, 1 mol of any gas = 22.4 L (a cube ~28 cm on each side)

Summary: Four Conversions with Mol

Conversion	Formula	Key value
mol → particles	$N = n \times N_A$	$N_A = 6.022 \times 10^{23}\ \mathrm{/mol}$
mol → mass	$m = n \times M$	$M$ : molar mass $\mathrm{(g/mol)}$
mol → gas volume	$V = n \times 22.4$	$22.4\ \mathrm{L/mol}$ at STP
mass → mol	$n = m / M$

Worked Examples

Example 1: Mass → mol → number of particles

Find the number of water molecules and the total number of hydrogen atoms in $36.0\ \mathrm{g}$ of water $\mathrm{H_2O}$ .

( $\mathrm{H} = 1.0,\ \mathrm{O} = 16.0,\ N_A = 6.022 \times 10^{23}\ \mathrm{/mol}$ )

Molar mass of $\mathrm{H_2O}$ :

M = 1.0 \times 2 + 16.0 = 18.0\ \mathrm{g/mol}

n = \frac{m}{M} = \frac{36.0\ \mathrm{g}}{18.0\ \mathrm{g/mol}} = 2.00\ \mathrm{mol}

N_{\mathrm{H_2O}} = 2.00 \times 6.022 \times 10^{23} = 1.20 \times 10^{24}\ \text{molecules}

Each $\mathrm{H_2O}$ molecule contains 2 hydrogen atoms:

N_\mathrm{H} = 1.20 \times 10^{24} \times 2 = 2.41 \times 10^{24}\ \text{atoms}

Example 2: Gas volume → mol → mass

Find the mass of $5.60\ \mathrm{L}$ of $\mathrm{CO_2}$ at STP.

( $\mathrm{C} = 12.0,\ \mathrm{O} = 16.0$ )

n = \frac{V}{22.4} = \frac{5.60}{22.4} = 0.250\ \mathrm{mol}

M_{\mathrm{CO_2}} = 12.0 + 16.0 \times 2 = 44.0\ \mathrm{g/mol}

m = n \times M = 0.250 \times 44.0 = 11.0\ \mathrm{g}

Check Your Understanding

Q1 How many particles are in 1 mol?

Q2 What is the molar mass of water $\text{H}_2\text{O}$? (H=1.0, O=16.0)

Q3 What is the volume of 2.0 mol of $\text{O}_2$ at STP?

Q4 What does the mol measure?

Exercises

Q1. How many moles is $111.6\ \mathrm{g}$ of iron $\mathrm{Fe}$ ? ( $\mathrm{Fe} = 55.8$ )

Solution

$n = \frac{111.6}{55.8} = 2.00\ \mathrm{mol}$

Divide mass by molar mass $M = 55.8\ \mathrm{g/mol}$ .

Q2. Find the number of nitrogen molecules in $11.2\ \mathrm{L}$ of $\mathrm{N_2}$ at STP.

Solution

$n = \frac{11.2}{22.4} = 0.500\ \mathrm{mol}$ $N = 0.500 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}\ \text{molecules}$

Q3. Find the mass and number of $\mathrm{Na^+}$ ions in $1.00\ \mathrm{mol}$ of $\mathrm{NaCl}$ . ( $\mathrm{Na} = 23.0,\ \mathrm{Cl} = 35.5$ )

Solution

Formula mass $= 23.0 + 35.5 = 58.5$ , so $M = 58.5\ \mathrm{g/mol}$

$m = 1.00 \times 58.5 = 58.5\ \mathrm{g}$

$\mathrm{NaCl}$ contains $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ in a 1:1 ratio, so $\mathrm{Na^+}$ is also $1.00\ \mathrm{mol}$ :

$N_{\mathrm{Na^+}} = 1.00 \times 6.022 \times 10^{23} = 6.02 \times 10^{23}\ \text{ions}$

Q1. $8.00\ \mathrm{g}$ of methane $\mathrm{CH_4}$ is completely combusted at STP. Find the volume of $\mathrm{CO_2}$ produced and the mass of $\mathrm{H_2O}$ formed. ( $M_{\mathrm{CH_4}} = 16.0$ , $\mathrm{H}=1.0,\ \mathrm{C}=12.0,\ \mathrm{O}=16.0$ )

Solution

Reaction: $\mathrm{CH_4 + 2O_2 \to CO_2 + 2H_2O}$

Amount of methane: $n_{\mathrm{CH_4}} = \frac{8.00}{16.0} = 0.500\ \mathrm{mol}$

From stoichiometry:

$n_{\mathrm{CO_2}} = 0.500\ \mathrm{mol}$
$n_{\mathrm{H_2O}} = 0.500 \times 2 = 1.00\ \mathrm{mol}$

Volume of $\mathrm{CO_2}$ at STP: $V = 0.500 \times 22.4 = 11.2\ \mathrm{L}$

Mass of $\mathrm{H_2O}$ : $m = 1.00 \times 18.0 = 18.0\ \mathrm{g}$

Q2. A gas has a density of $1.964\ \mathrm{g/L}$ at STP. Find its molar mass.

Solution

At STP, $1\ \mathrm{mol} = 22.4\ \mathrm{L}$ , so the mass of $1\ \mathrm{mol}$ is:

$M = 1.964 \times 22.4 = 44.0\ \mathrm{g/mol}$

The molar mass is 44.0 g/mol (corresponds to $\mathrm{CO_2}$ ).

Key pattern: density × 22.4 = molar mass (for gases at STP).