Chapter 8: The Mole Concept – Study Notes

Chapter 8. The Mole Concept

Avogadro’s Number

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. Avogadro’s number defines the number of particles (atoms, molecules, ions) in one mole of a substance.

• Definition: Avogadro’s number is $6.022 \times 10^{23}$ particles per mole.

• Application: Used to convert between number of particles and moles.

• Example: 1 mole of H2O contains $6.022 \times 10^{23}$ molecules of water.

Mass / Moles / Number of Molecules or Atoms Relations

Understanding the relationships between mass, moles, and number of particles is essential for quantitative chemistry.

• Key Equations:

  • To convert mass to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

  • To convert moles to number of particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• Example: Calculate the number of molecules in 18 g of H2O:

  • Molar mass of H2O = 18 g/mol

  • Moles = $\frac{18}{18} = 1$ mol

  • Number of molecules = $1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

• Calculation: Add the atomic masses of all atoms in a molecule.

• Example: Molar mass of CO2 = 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol

Molar Volume (at STP Conditions)

At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies a specific volume.

• STP Conditions: 0°C (273.15 K) and 1 atm pressure

• Molar Volume: $22.4$ L/mol at STP

• Example: 2 moles of O2 gas at STP occupy $2 \times 22.4 = 44.8$ L

Gas Density (Calculate, Units, Meaning)

Gas density is the mass per unit volume of a gas, typically expressed in g/L.

• Calculation: $\text{Density} = \frac{\text{molar mass}}{\text{molar volume}}$

• Units: grams per liter (g/L)

• Example: Density of O2 at STP: $\frac{32.00\ \text{g/mol}}{22.4\ \text{L/mol}} = 1.43\ \text{g/L}$

Percent Composition

Percent composition describes the percentage by mass of each element in a compound.

• Calculation: $\%\ \text{element} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100\%$

• Example: Percent composition of H in H2O:

  • Mass of H in 1 mol = 2 × 1.01 = 2.02 g

  • Molar mass of H2O = 18.02 g

  • Percent H = $\frac{2.02}{18.02} \times 100\% = 11.2\%$

Empirical Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms of each element in a molecule.

• From Mass: Convert mass of each element to moles, divide by the smallest number of moles, and round to nearest whole number.

• From Percent Composition: Treat percent as grams in a 100 g sample, convert to moles, and determine the simplest ratio.

• Molecular Formula from Empirical Formula: $\text{Molecular formula} = n \times \text{Empirical formula}$, where $n = \frac{\text{molar mass}}{\text{empirical formula mass}}$

• Example: A compound with empirical formula CH2 and molar mass 56 g/mol:

  • Empirical formula mass = 12.01 + 2 × 1.01 = 14.03 g/mol

  • $n = \frac{56}{14.03} \approx 4$

  • Molecular formula = C4H8