Moles & Molarity Lecture Notes

Module Objectives

• Key concepts: mass, moles, molar mass, percent composition, solution concentration, and dilution problems.

The Mole

• A mole (mol) is a collective unit in chemistry, defined as $1 \text{ mole} = 6.022140857 \times 10^{23} \text{ units}$ (Avogadro’s Constant). It’s crucial for counting reactive units like atoms and molecules, as mass alone is insufficient for this purpose. Atomic mass is expressed in grams per mole (g/mol).

Conversions

• The notes cover conversions between atoms and moles, and grams and moles, using Avogadro's Constant and atomic mass (e.g., $1.01928 \text{ mole He} \times \frac{6.022140857 \times 10^{23} \text{ atoms He}}{1 \text{ mole He}} = 6.13825 \times 10^{23} \text{ atoms He}$ and $1.0000 \text{ g He} \times \frac{1 \text{ mole He}}{4.002602 \text{ g He}} = 0.24984 \text{ moles He}$).

Molecular Mass Calculations

• Molecular mass (molar mass) is the sum of the atomic masses of all atoms in a molecule. Formula: \text{Mol Mass} = \text{(a \times Atomic mass of A)} + \text{(b \times Atomic mass of B)} + \dots

  • Example: For $\text{FeCl}_3$ it is $(1 \times 55.845 \text{ g/mol}) + (3 \times 35.453 \text{ g/mol}) = 162.204 \text{ g/mol}$.

Percent Composition

• Calculated by dividing the total mass of an element in a compound by the compound's molecular mass, then multiplying by 100. For $\text{Fe}<em>2\text{S}</em>3$, \frac{111.690 \text{ g Fe}}{207.885 \text{ g}} \times 100 = 53.7268 \text{% Fe} .

Solution Basics & Molarity

• Solutions have a solvent (larger amount) and solute (smaller amount).

• Molarity (M) defines concentration as moles of solute per liter of solution: $M = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$.

  • This formula can be rearranged to find moles ($\text{Moles} = M \times L$) or volume ($L = \text{Moles} \div M$).

  • Example: 0.500 moles Li in 1.000 L is 0.500 M.

Dilution Concept

• Dilution problems use the formula $M<em>1V</em>1 = M<em>2V</em>2$. This allows calculating unknown concentrations or volumes during dilution.

  • For example, to make 500.0 mL of a 0.1250 M solution from a 2.000 M solution, $V_1 = \frac{0.1250 \text{ M} \times 500.0 \text{ mL}}{2.000 \text{ M}} = 31.25 \text{ mL}$.

Other Concentration Expressions

• Other concentration types like mass-to-mass, volume-to-volume, and mass-to-volume percent concentrations are also noted as context-dependent.