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

Why We Need Moles

Atoms and molecules are incredibly small. A mole provides a convenient way to:

• Count particles (atoms, molecules, ions, etc.)
• Relate macroscopic measurements to atomic-scale quantities
• Perform stoichiometric calculations

Analogy: Just like "dozen" means 12, "mole" means $6.022 \times 10^{23}$.

Converting Between Particles and Moles

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

$\text{moles} = \frac{\text{Number of particles}}{N_A}$

Example: How many atoms are in 2.5 moles of carbon?

$\text{atoms} = 2.5 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.51 \times 10^{24} \text{ atoms}$

Molar Mass

Molar mass ($M$) is the mass of one mole of a substance, expressed in grams per mole (g/mol).

For Elements

The molar mass of an element equals its atomic mass from the periodic table.

Examples:

• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.008 g/mol

For Compounds

Add up the molar masses of all atoms in the molecular formula.

Example: Water (H₂O)

$M_{H_2O} = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.02 \text{ g/mol}$

Example: Glucose (C₆H₁₂O₆)

$M_{C_6H_{12}O_6} = 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.10 + 96.00 = 180.16 \text{ g/mol}$

Converting Between Mass and Moles

$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

$\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Example: How many moles are in 25.0 g of NaCl?

First find molar mass: $M_{NaCl} = 22.99 + 35.45 = 58.44$ g/mol

$\text{moles} = \frac{25.0 \text{ g}}{58.44 \text{ g/mol}} = 0.428 \text{ mol}$

The Mole Road Map

To convert between different quantities, use this road map:

$\text{Mass (g)} \xleftrightarrow[\text{divide by } M]{\text{multiply by } M} \text{Moles} \xleftrightarrow[\text{divide by } N_A]{\text{multiply by } N_A} \text{Particles}$

Percent Composition

Percent composition is the percent by mass of each element in a compound.

$\text{Percent of element} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100\%$

Example: Find the percent composition of H₂O

$\%H = \frac{2(1.008)}{18.02} \times 100\% = \frac{2.016}{18.02} \times 100\% = 11.19\%$

$\%O = \frac{16.00}{18.02} \times 100\% = 88.81\%$

Check: $11.19\% + 88.81\% = 100\%$ ✓

Empirical vs. Molecular Formulas

Empirical formula: Simplest whole-number ratio of atoms Molecular formula: Actual number of atoms in one molecule

Example:

• Empirical formula: CH₂O
• Molecular formula: C₆H₁₂O₆ (glucose)

The molecular formula is always a whole-number multiple of the empirical formula.

Common Calculations

1. Mass to Moles

Given mass, find moles: $n = \frac{m}{M}$

2. Moles to Particles

Given moles, find particles: $N = n \times N_A$

3. Mass to Particles

Combine: $N = \frac{m}{M} \times N_A$

4. Molar Mass from Formula

Add atomic masses from periodic table

Step 2: Convert mass to moles

$n = \frac{m}{M} = \frac{8.50 \text{ g}}{44.01 \text{ g/mol}} = 0.193 \text{ mol}$

Step 3: Convert moles to molecules

$N = n \times N_A = 0.193 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$

$N = 1.16 \times 10^{23} \text{ molecules}$

Answer: $1.16 \times 10^{23}$ molecules of CO₂

Verification:

• Used correct molar mass for CO₂ ✓
• Converted g → mol → molecules ✓
• Answer has correct units ✓