Moles and Chemical Equations

Introduction to the Mole Concept

• Mole: A crucial unit in chemistry utilised to quantify the amount of a substance.
• Key for comprehending chemical reactions, stoichiometry, and equation balancing.
• Avogadro's number is critical, representing $6.022 \times 10^{23}$ entities per mole.
• Practical application: Determines the amount of reactant needed or product formed in reactions.

Definition of a Mole

• Mole: Defined by Avogadro's number $6.022 \times 10^{23}$, linking the macroscopic mass in grams with the microscopic scale of atoms or molecules.

Importance of the Mole

• Serves as a link between mass (grams) and particles (atoms/molecules).
• Vital in laboratories for efficiently scaling reactions.

Formula and Calculations

• Conversion Formula: $n = \frac{m}{M}$
  • n: Number of moles.
  • m: Mass in grams.
  • M: Molar mass in g/mol.

Worked Examples

• Example 1: Converting mass to moles (water).
  • Given: 50g of water
  • Molar mass of $\mathrm{H_2O}$ is approximately 18 g/mol.
  • Calculate the number of moles:
    • $n = \frac{m}{M} = \frac{50\text{ g}}{18\text{ g/mol}} \approx 2.78$ moles.
• Example 2: Converting moles to mass (carbon dioxide).
  • Given: 3 moles of carbon dioxide
  • Molar mass of $\mathrm{CO_2}$ is approximately 44 g/mol.
  • Calculate the mass:
    • $m = n \times M = 3\text{ mol} \times 44\text{ g/mol} = 132$ grams.

Balancing Chemical Equations

• Balanced equations: Ensure the mass conservation where the mass of reactants equals the mass of products, essential in industrial processes like the Haber Process.

• Law of Conservation of Mass: Ensures atoms in reactants equal those in products, maintaining mass balance.

Techniques for Balancing Equations

1. Skeleton Equations: Start with: $\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$
2. Counting Atoms: Ensure equal atoms for each element.
3. Adjusting Coefficients: Focus on complex molecules first to simplify balancing.

• Worked Example: Balancing $\mathrm{C_2H_6 + O_2 \rightarrow CO_2 + H_2O}$
  • Steps:
    • Balance C: $\mathrm{2C_2H_6 + O_2 \rightarrow 4CO_2 + H_2O}$
    • Balance H: $\mathrm{2C_2H_6 + O_2 \rightarrow 4CO_2 + 6H_2O}$
    • Balance O: $\mathrm{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O}$

Practice Examples

• Combustion Example: Balance $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$

  Solution: The equation is already balanced.

• Decomposition Example: Balance $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$

  Solution: The equation is already balanced.

Stoichiometry

• Stoichiometry: Analysis of quantitative relationships between reactants and products.

Calculating Stoichiometric Ratios

• Apply mole ratios from balanced equations for conversions.

Example:

$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ implies a mole ratio of $2:1:2$.

Practice Questions

Balancing Chemical Equations

1. Balance: $\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$

   Solution: $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$

Stoichiometry Calculations

1. Mole Conversion: Calculate moles in 25g of NaCl using its molar mass of 58.44 g/mol.

   Solution: $n = \frac{m}{M} = \frac{25\text{ g}}{58.44\text{ g/mol}} = 0.428$ moles of NaCl

Summary of Concepts

• Ensure comprehension through diverse problem-solving, equation balancing, and stoichiometric conversions.
• Utilise the foundational principles of reactions and mass conservation.