Unit 3: Mole Concept

Understanding the bridge between the microscopic world of atoms and the macroscopic measurements in the laboratory.

3.8 Avogadro's Number

Avogadro's number ($N_A$) is one of the most fundamental constants in chemistry, representing the number of particles (atoms, molecules, or ions) in one mole of any substance. This constant is defined as 6.022 × 10²³ particles per mole. Named after Amedeo Avogadro, this number provides the crucial link between the microscopic world of individual atoms and molecules and the macroscopic world of laboratory measurements.

The significance of Avogadro's number cannot be overstated—it allows chemists to count particles by weighing substances. For example, exactly 12.0 g of carbon-12 contains precisely 6.022 × 10²³ carbon atoms. This relationship forms the foundation of quantitative chemistry and enables us to perform stoichiometric calculations with precision.

Key Formula: Number of particles = Number of moles × Avogadro's number
$N = n × N_A = n × 6.022 × 10^{23}$
Solved Examples:
  1. How many atoms are in 24.0 g of carbon-12?
    Solution:
    • Molar mass of carbon-12 = 12.0 g/mol
    • Number of moles = mass ÷ molar mass = 24.0 g ÷ 12.0 g/mol = 2.0 mol
    • Number of atoms = moles × $N_A$ = 2.0 mol × 6.022 × 10²³ atoms/mol = 1.2044 × 10²⁴ atoms
  2. Calculate the number of molecules in 0.5 mol of water (H₂O).
    Solution: Number of molecules = 0.5 mol × 6.022 × 10²³ molecules/mol = 3.011 × 10²³ molecules
  3. How many moles contain 1.8066 × 10²⁴ atoms of sodium?
    Solution: Number of moles = Number of atoms ÷ $N_A$ = (1.8066 × 10²⁴) ÷ (6.022 × 10²³) = 3.0 mol
  4. Find the number of ions in 0.25 mol of NaCl.
    Solution: NaCl dissociates into Na⁺ and Cl⁻ ions (2 ions per formula unit).
    • Number of formula units = 0.25 mol × 6.022 × 10²³ = 1.5055 × 10²³
    • Total number of ions = 1.5055 × 10²³ × 2 = 3.011 × 10²³ ions
  5. What mass of aluminum contains 9.033 × 10²³ atoms? (Atomic mass of Al = 26.98 g/mol)
    Solution:
    • Number of moles = (9.033 × 10²³) ÷ (6.022 × 10²³) = 1.5 mol
    • Mass = moles × molar mass = 1.5 mol × 26.98 g/mol = 40.47 g

3.9 Relative Molecular and Formula Masses

Relative molecular mass ($M_r$) is the mass of a molecule compared to 1/12th the mass of a carbon-12 atom. It is calculated by summing the relative atomic masses of all atoms present in the molecular formula. For covalent compounds, we use the term "molecular mass," while for ionic compounds, we use relative formula mass since ionic compounds don't exist as discrete molecules but as formula units in a crystal lattice.

Both relative molecular mass and relative formula mass are dimensionless quantities because they represent ratios compared to the carbon-12 standard. However, when we express molar mass (the mass of one mole), we use units of g/mol, which numerically equals the relative molecular or formula mass.

Compound Type Term Used Example Calculation
Covalent Relative Molecular Mass H₂O (2 × 1.0) + (1 × 16.0) = 18.0
Ionic Relative Formula Mass NaCl (1 × 23.0) + (1 × 35.5) = 58.5
Solved Examples:
  1. Calculate the relative molecular mass of CO₂. (C = 12.0, O = 16.0)
    Solution: $M_r$ = 12.0 + (2 × 16.0) = 12.0 + 32.0 = 44.0
  2. Find the relative molecular mass of ethanol (C₂H₅OH). (C = 12.0, H = 1.0, O = 16.0)
    Solution: $M_r$ = (2 × 12.0) + (6 × 1.0) + (1 × 16.0) = 24.0 + 6.0 + 16.0 = 46.0
  3. Calculate the relative formula mass of calcium carbonate (CaCO₃). (Ca = 40.1, C = 12.0, O = 16.0)
    Solution: Formula mass = 40.1 + 12.0 + (3 × 16.0) = 40.1 + 12.0 + 48.0 = 100.1
  4. Determine the relative molecular mass of sulfuric acid (H₂SO₄). (H = 1.0, S = 32.1, O = 16.0)
    Solution: $M_r$ = (2 × 1.0) + 32.1 + (4 × 16.0) = 2.0 + 32.1 + 64.0 = 98.1
  5. Find the relative formula mass of aluminum sulfate [Al₂(SO₄)₃]. (Al = 27.0, S = 32.1, O = 16.0)
    Solution: Formula mass = (2 × 27.0) + (3 × 32.1) + (12 × 16.0) = 54.0 + 96.3 + 192.0 = 342.3

3.10 Moles and Molar Masses

A mole is the SI base unit for the amount of substance, defined as the amount containing as many elementary entities as there are atoms in exactly 12 g of carbon-12. This equals approximately 6.022 × 10²³ particles (Avogadro's number). The mole allows chemists to work with manageable quantities while keeping track of individual atoms and molecules.

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). For elements, the molar mass in g/mol numerically equals the relative atomic mass. For compounds, it numerically equals the relative molecular or formula mass. This relationship provides a direct conversion between mass and amount of substance.

Key Relationships:
• Number of moles (n) = mass (m) ÷ molar mass (M)
• Mass (m) = number of moles (n) × molar mass (M)
• Molar mass (M) = mass (m) ÷ number of moles (n)
$$ n = \frac{m}{M} \quad \text{where } n = \text{moles, } m = \text{mass (g), } M = \text{molar mass (g/mol)} $$
Solved Examples:
  1. Calculate the number of moles in 117.0 g of NaCl. (Na = 23.0, Cl = 35.5)
    Solution:
    • Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
    • Number of moles = mass ÷ molar mass = 117.0 g ÷ 58.5 g/mol = 2.0 mol
  2. What is the mass of 0.5 mol of ammonia (NH₃)? (N = 14.0, H = 1.0)
    Solution:
    • Molar mass of NH₃ = 14.0 + (3 × 1.0) = 17.0 g/mol
    • Mass = moles × molar mass = 0.5 mol × 17.0 g/mol = 8.5 g
  3. How many moles are in 80.0 g of oxygen gas (O₂)? (O = 16.0)
    Solution:
    • Molar mass of O₂ = 2 × 16.0 = 32.0 g/mol
    • Number of moles = 80.0 g ÷ 32.0 g/mol = 2.5 mol
  4. Calculate the molar mass of a compound if 3.5 mol has a mass of 245 g.
    Solution: Molar mass = mass ÷ moles = 245 g ÷ 3.5 mol = 70.0 g/mol
  5. How many molecules are in 36.0 g of glucose (C₆H₁₂O₆)? (C = 12.0, H = 1.0, O = 16.0)
    Solution:
    • Molar mass of C₆H₁₂O₆ = (6 × 12.0) + (12 × 1.0) + (6 × 16.0) = 72.0 + 12.0 + 96.0 = 180.0 g/mol
    • Number of moles = 36.0 g ÷ 180.0 g/mol = 0.2 mol
    • Number of molecules = 0.2 mol × 6.022 × 10²³ molecules/mol = 1.2044 × 10²³ molecules

Knowledge Check (20 Questions)

Answer: 6.022 × 10²³ particles per mole. It represents the number of particles in one mole of any substance.

Answer: Number of atoms = 2.5 mol × 6.022 × 10²³ atoms/mol = 1.5055 × 10²⁴ atoms.

Answer: Mr = 12.0 + (4 × 1.0) = 12.0 + 4.0 = 16.0

Answer: Molar mass of CO₂ = 12.0 + (2 × 16.0) = 44.0 g/mol. Number of moles = 44.0 g ÷ 44.0 g/mol = 1.0 mol.

Answer: Molar mass = 40.1 + 12.0 + (3 × 16.0) = 100.1 g/mol. Mass = 0.25 mol × 100.1 g/mol = 25.025 g.

Answer: One mole is the amount of substance containing as many elementary entities as there are atoms in exactly 12 g of carbon-12, approximately 6.022 × 10²³ particles.

Answer: Molar mass of H₂O = (2 × 1.0) + 16.0 = 18.0 g/mol. Moles = 18.0 g ÷ 18.0 g/mol = 1.0 mol. Molecules = 1.0 mol × 6.022 × 10²³ = 6.022 × 10²³ molecules.

Answer: Relative molecular mass is dimensionless (ratio), while molar mass has units of g/mol. Numerically, they are equal.

Answer: Number of moles = (3.011 × 10²³) ÷ (6.022 × 10²³) = 0.5 mol.

Answer: Formula mass = 24.3 + 32.1 + (4 × 16.0) = 24.3 + 32.1 + 64.0 = 120.4.
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