Definition: One mole contains exactly $6.022 \times 10^{23}$ particles.

This number is called Avogadro's number ($N_A$).

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

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 3: Calculate percent composition

$\%Ca = \frac{40.08}{164.10} \times 100\% = 24.42\%$

$\%N = \frac{28.02}{164.10} \times 100\% = 17.07\%$

$\%O = \frac{96.00}{164.10} \times 100\% = 58.50\%$

Answer:

• Molar mass: 164.10 g/mol
• Percent composition: 24.42% Ca, 17.07% N, 58.50% O

Verification:

• $24.42 + 17.07 + 58.50 = 99.99 \approx 100\%$ ✓ (rounding)

$\text{mol H} = \frac{6.7 \text{ g}}{1.008 \text{ g/mol}} = 6.65 \text{ mol}$

$\text{mol O} = \frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ mol}$

Step 3: Find simplest ratio (divide by smallest)

Smallest = 3.33

$C: \frac{3.33}{3.33} = 1$ $H: \frac{6.65}{3.33} = 2$ $O: \frac{3.33}{3.33} = 1$

Empirical formula: CH₂O

Step 4: Find empirical formula mass

$M_{CH_2O} = 12.01 + 2(1.008) + 16.00 = 30.03 \text{ g/mol}$

Step 5: Find multiplier

$n = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{180}{30.03} = 6$

Step 6: Multiply empirical formula by n

$\text{Molecular formula} = (CH_2O) \times 6 = C_6H_{12}O_6$

Answer: C₆H₁₂O₆ (glucose)

Verification:

• Molar mass of C₆H₁₂O₆ = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol ✓
• Matches given molar mass ✓