Avogadro’s Number, Moles, and Molar Mass: Foundations for Counting Atoms and Molecules

Avogadro’s Number and the Mole

Introduction to the Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. Because atoms and molecules are extremely small, the mole allows chemists to count large numbers of particles by weighing macroscopic amounts.

• Definition: One mole contains exactly 6.022 × 1023 elementary entities (atoms, molecules, ions, etc.). This value is known as Avogadro’s number ($N_A$).

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

• Application: The mole bridges the gap between the atomic scale and the laboratory scale.

Example: 1 mole of H2O contains $6.022 \times 10^{23}$ molecules of water.

Mole Calculations and Avogadro’s Number

Counting Atoms and Molecules

Avogadro’s number allows conversion between moles and number of particles.

• To find number of atoms/molecules: Multiply moles by $N_A$.

• To find moles from number of particles: Divide the number of particles by $N_A$.

Formulas:

• $\text{Number of particles} = \text{moles} \times N_A$

• $\text{Moles} = \frac{\text{Number of particles}}{N_A}$

Example 3.5:

• How many atoms are in 6.8 mol of gold?

• $\text{Number of atoms} = (6.8\, \text{mol}) \times (6.022 \times 10^{23}\, \frac{\text{atoms}}{\text{mol}}) = 4.1 \times 10^{24}\, \text{atoms}$

• How many moles are 3.28 × 1021 methanol molecules?

• $\text{Number of moles} = \frac{3.28 \times 10^{21}\, \text{molecules}}{6.022 \times 10^{23}\, \frac{\text{molecules}}{\text{mol}}} = 5.45 \times 10^{-3}\, \text{mol}$

Molar Mass

Definition and Calculation

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is numerically equal to the atomic or molecular weight in atomic mass units (amu).

• Atomic weight (amu): Mass of one atom in atomic mass units.

• Molar mass (g/mol): Mass of one mole of atoms or molecules in grams.

• For elements: 1 mol Cl = 35.45 g; 1 mol Ni = 58.69 g

Example: The atomic weight of Cl is 35.45 amu, so 1 mol of Cl has a mass of 35.45 g.

Molar Mass of Compounds

The molar mass of a compound is the sum of the molar masses of its constituent elements.

• H2O: 18.02 amu → 1 mol H2O = 18.02 g

• CuO: 79.55 amu → 1 mol CuO = 79.55 g

• N2: 28.01 amu → 1 mol N2 = 28.01 g

Mole Relationships

Comparing Substances, Formula Weights, and Particle Counts

The following table summarizes the relationships between formula, molar mass, and number of particles for various substances:

Name of Substance	Formula	Formula Weight (amu)	Molar Mass (g/mol)	Number and Kind of Particles
Atomic Nitrogen	N	14.0	14.0	6.02 × 1023 N atoms
Molecular Nitrogen	N2	28.0	28.0	6.02 × 1023 N2 molecules 1.204 × 1024 N atoms
Silver	Ag	107.9	107.9	6.02 × 1023 Ag atoms
Silver Ions	Ag+	107.9	107.9	6.02 × 1023 Ag+ ions
Barium Chloride	BaCl2	208.2	208.2	6.02 × 1023 BaCl2 molecules 6.02 × 1023 Ba2+ ions 1.204 × 1024 Cl- ions

Converting between Mass, Moles, and Number of Atoms/Molecules

Conversion Pathways

To convert between mass, moles, and number of atoms or molecules, use the following relationships:

• Mass to moles: $\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}$

• Moles to mass: $\text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)}$

• Moles to particles: $\text{Number of particles} = \text{Moles} \times N_A$

• Particles to moles: $\text{Moles} = \frac{\text{Number of particles}}{N_A}$

Flowchart: Grams → (Molar Mass) → Moles → ($N_A$) → Atoms/Molecules

Worked Example: Mass from Moles

Example 3.7: How many grams does 3.65 mol of sodium chloride weigh?

• Molar mass of NaCl: $22.990\, \text{g/mol} + 35.453\, \text{g/mol} = 58.443\, \text{g/mol}$

• Mass: $3.65\, \text{mol} \times 58.443\, \frac{\text{g}}{\text{mol}} = 213.32\, \text{g}$

General Formula: $\text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)}$

Worked Example: Atoms in a Compound

Example 3.6: Calculate the number of H atoms in 0.350 mol of fructose (C6H12O6).

• Number of molecules: $0.350\, \text{mol} \times 6.022 \times 10^{23}\, \frac{\text{molecules}}{\text{mol}} = 2.1077 \times 10^{23}\, \text{molecules}$

• Number of H atoms: $2.1077 \times 10^{23}\, \text{molecules} \times \frac{12\, \text{H atoms}}{1\, \text{molecule}} = 2.53 \times 10^{24}\, \text{H atoms}$

Summary Table: Key Relationships

Conversion	Formula	Example
Mass to Moles	$\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}$	100 g H2O: $\frac{100}{18.02} = 5.55$ mol
Moles to Particles	$\text{Particles} = \text{Moles} \times N_A$	2 mol Na: $2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ atoms
Particles to Moles	$\text{Moles} = \frac{\text{Particles}}{N_A}$	$3.01 \times 10^{23}$ molecules: $\frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5$ mol

Additional info: These concepts are foundational for understanding chemical reactions, stoichiometry, and laboratory calculations in General Chemistry.