Chemical Reactions Quantitative Calculations

Introduction to Key Concepts

• Law of Definite Proportions: A chemical compound consistently contains elements in the same mass proportion.
  • Example: In the synthesis of ammonia for fertilisers, the mass ratio of nitrogen to hydrogen remains constant.

Concept of Molar Mass

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).
• Facilitates conversion between atomic and large-scale measurements using Avogadro's Number: $6.022 \times 10^{23}$ particles/mole.

Examples:

• Hydrogen:
  • Atomic Mass: 1.01 g/mol
• Sodium Chloride (NaCl):
  • Sodium (Na): 22.99 g/mol
  • Chlorine (Cl): 35.45 g/mol
  • Total Molar Mass: 58.44 g/mol

Practical Investigations

• Objective: Conduct experiments to determine molar mass.
• Procedure:
  • Weigh the piece of magnesium.
  • Heat it to allow reaction with oxygen.
  • Weigh Again to assess the change in mass.

Understanding the Mole and Avogadro's Constant

• Definition of the Mole: The mole is a standard unit for quantifying atoms or molecules.
• Avogadro's Contribution (1811): Established that gas volumes are proportional to the number of molecules.
• Avogadro's Constant: $6.022 \times 10^{23}$ particles per mole.

Percent Composition and Empirical Formula

• Percent Composition Formula:

  • $\text{Percent Composition} = \frac{\text{mass of element}}{\text{total mass of compound}} \times 100$

• Empirical Formula: The simplest whole-number ratio of elements in a compound.

Visual Aids

• Empirical vs. Molecular Formulae:
  • Example:
    • Glucose Molecular Formula: C₆H₁₂O₆
    • Empirical Formula: CH₂O

Limiting Reagent

• Limiting Reagent: The reactant that is entirely consumed first, determining the amount of product formed.

Example Analysis: Reaction of Hydrogen and Oxygen

• Balanced Equation:
  $2H_2 + O_2 \rightarrow 2H_2O$

Formulae and Calculation Techniques

• Molar Mass: Molar Mass = $\sum \text{Atomic Masses of elements}$
• Number of Moles: $n = \frac{m}{M}$
• Avogadro's Number: $N_A = 6.022 \times 10^{23}$
• Percentage Composition: $\% \text{Composition} = \frac{\text{Mass of Element}}{\text{Mass of Compound}} \times 100$

Worked Example

Calculate the molar mass of glucose (C₆H₁₂O₆):

1. First, identify the atomic mass of each element:

   • Carbon (C): 12.01 g/mol
   • Hydrogen (H): 1.01 g/mol
   • Oxygen (O): 16.00 g/mol

2. Multiply each atomic mass by the number of atoms:

   • Carbon: 12.01 g/mol × 6 = 72.06 g/mol
   • Hydrogen: 1.01 g/mol × 12 = 12.12 g/mol
   • Oxygen: 16.00 g/mol × 6 = 96.00 g/mol

3. Add all contributions to find the total molar mass:

   • Total: 72.06 + 12.12 + 96.00 = 180.18 g/mol

Therefore, the molar mass of glucose is 180.18 g/mol.