The Mole Concept

Overview

The mole is the SI unit for the amount of substance. It provides a bridge between the atomic world (atoms, molecules) and the macroscopic world (grams, liters) that we can measure.

Avogadro's Number

$$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$

One mole of any substance contains exactly $6.022 \times 10^{23}$ particles (atoms, molecules, ions, etc.)

Key Relationships

Moles, Mass, and Molar Mass

$$n = \frac{m}{M}$$

Where:

• $n$ = number of moles (mol)
• $m$ = mass (g)
• $M$ = molar mass (g/mol)

Moles and Number of Particles

$$N = n \times N_A$$

Where:

• $N$ = number of particles
• $n$ = moles
• $N_A$ = Avogadro's number

Moles and Volume (Gases at STP)

$$V = n \times 22.4 \text{ L}$$

At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L.

Molar Mass

The mass of one mole of a substance in grams.

Atomic Molar Mass

From the periodic table:

• C: 12.01 g/mol
• O: 16.00 g/mol
• H: 1.008 g/mol

Molecular Molar Mass

Sum of atomic masses:

$$\text{H}_2\text{O}: 2(1.008) + 16.00 = 18.02 \text{ g/mol}$$ $$\text{CO}_2: 12.01 + 2(16.00) = 44.01 \text{ g/mol}$$ $$\text{C}_6\text{H}_{12}\text{O}_6: 6(12.01) + 12(1.008) + 6(16.00) = 180.16 \text{ g/mol}$$

Conversion Map

                    ÷ M          × Nₐ
        Mass (g) ←——————→ Moles ←——————→ Particles
                    × M          ÷ Nₐ
                              ↕
                           × 22.4
                           ÷ 22.4
                              ↕
                        Volume at STP (L)

Examples

Example 1: Mass to Moles

How many moles are in 54 g of water?

$$n = \frac{m}{M} = \frac{54 \text{ g}}{18.02 \text{ g/mol}} = 3.0 \text{ mol}$$

Example 2: Moles to Particles

How many molecules are in 2.5 mol of CO₂?

$$N = n \times N_A = 2.5 \text{ mol} \times 6.022 \times 10^{23} = 1.51 \times 10^{24} \text{ molecules}$$

Example 3: Mass to Atoms

How many atoms are in 10 g of calcium?

$$\text{Step 1: } n = \frac{10 \text{ g}}{40.08 \text{ g/mol}} = 0.249 \text{ mol}$$ $$\text{Step 2: } N = 0.249 \times 6.022 \times 10^{23} = 1.50 \times 10^{23} \text{ atoms}$$

Example 4: Molecules to Mass

What is the mass of $3.01 \times 10^{23}$ molecules of glucose (C₆H₁₂O₆)?

$$\text{Step 1: } n = \frac{N}{N_A} = \frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.50 \text{ mol}$$ $$\text{Step 2: } m = n \times M = 0.50 \text{ mol} \times 180.16 \text{ g/mol} = 90.08 \text{ g}$$

Example 5: Atoms in a Compound

How many oxygen atoms are in 0.5 mol of Ca(NO₃)₂?

1 formula unit of Ca(NO₃)₂ contains 6 O atoms

$$0.5 \text{ mol Ca(NO}_3\text{)}_2 \times \frac{6 \text{ mol O}}{1 \text{ mol Ca(NO}_3\text{)}_2} = 3.0 \text{ mol O}$$ $$N = 3.0 \times 6.022 \times 10^{23} = 1.81 \times 10^{24} \text{ O atoms}$$

Percent Composition

Mass percent of each element in a compound:

$$\% \text{ Element} = \frac{n \times \text{atomic mass}}{\text{molar mass}} \times 100\%$$

Example: H₂O

$$\% \text{ H} = \frac{2 \times 1.008}{18.02} \times 100\% = 11.2\%$$ $$\% \text{ O} = \frac{16.00}{18.02} \times 100\% = 88.8\%$$

Standard Temperature and Pressure (STP)

Condition	Value
Temperature	0°C (273.15 K)
Pressure	1 atm (101.325 kPa)
Molar Volume	22.4 L/mol

Quick Conversions

Given	Find	Formula
Mass	Moles	$n = m/M$
Moles	Mass	$m = n \times M$
Moles	Particles	$N = n \times N_A$
Particles	Moles	$n = N/N_A$
Moles (gas)	Volume at STP	$V = n \times 22.4$
Volume at STP	Moles	$n = V/22.4$