Lecture 2: The Mole, Calculations, and Equations

Introduction to The Mole

• A mole is a standardized unit in chemistry representing a specific number of particles; specifically, (6.022 \times 10^{23}) particles, known as Avogadro’s number (N_A).

• This number can be rounded to (6.022 \times 10^{23}) for practical purposes.

Understanding Moles

• Particles in a Mole:

  • 1 mole of carbon (C): (6.022 \times 10^{23}) atoms

  • 1 mole of carbon dioxide (CO(_2)): (6.022 \times 10^{23}) molecules

• Calculating Number of Particles:

  • Formula: [ \text{Number of Particles} = Moles x N ,A

  • Example 1: How many atoms in 0.250 moles of copper?

  • Calculation: [ \text{Number of atoms} = 0.250 \times (6.022 \times 10^{23}) = 1.555 \times 10^{23} ]

  • Example 2: How many molecules in 1.3 moles of caffeine?

  • Calculation: [ \text{Number of molecules} = 1.3 \times (6.022 \times 10^{23}) = 7.829 \times 10^{23} ]

Mass and Moles

• The mass of 1 mole of a substance is numerically equal to its relative atomic or molecular mass (g).

• Examples:

  • 1 mole C = 12.0 g

  • 1 mole Fe = 55.8 g

  • 1 mole Ethanol (C(2)H(6)O) = 46 g

Calculating Mass from Nuclear Mass

Example: Carbon (C)

1. Nuclear Composition:

   • 6 protons, 6 neutrons.

2. Total Proton Mass (TPM):

   • Mass of 1 proton: (1.67262 \times 10^{-27}) kg

   • Total Proton Mass: [ \text{TPM} = 6 \times (1.67262 \times 10^{-27}) = 1.00537 \times 10^{-26} \text{ kg} ]

3. Total Neutron Mass (TNM):

   • Mass of 1 neutron: (1.67492 \times 10^{-27}) kg

   • Total Neutron Mass: [ \text{TNM} = 6 \times (1.67492 \times 10^{-27}) = 1.00495 \times 10^{-26} \text{ kg} ]

4. Nuclear Mass:

   • Total Nuclear Mass: [ \text{Nuclear Mass} = ext{TPM} + ext{TNM} = 1.00537 \times 10^{-26} + 1.00495 \times 10^{-26} = 2.01032 \times 10^{-26} \text{ kg} ]

5. Mass of 1 mole of Carbon:

   • [ 1 ext{ mole C} = (2.01032 \times 10^{-26}) \times (6.022 \times 10^{23}) \approx 0.01210 \text{ kg} = 12.1 ext{ g} ]

Example: Iron (Fe)

1. Nuclear Composition:

   • 26 protons, 30 neutrons.

2. Total Proton Mass (TPM):

   • [ ext{TPM} = 26 \times (1.67262 \times 10^{-27}) = 4.3488 \times 10^{-26} \text{ kg} ]

3. Total Neutron Mass (TNM):

   • [ ext{TNM} = 30 \times (1.67492 \times 10^{-27}) = 5.0248 \times 10^{-26} \text{ kg} ]

4. Nuclear Mass:

   • [ ext{Nuclear Mass} = 4.3488 \times 10^{-26} + 5.0248 \times 10^{-26} = 9.37356 \times 10^{-26} \text{ kg} ]

5. Mass of 1 mole of Iron:

   • [ 1 ext{ mole Fe} = (9.37356 \times 10^{-26}) \times (6.022 \times 10^{23}) = 0.056 \text{ kg} = 56 ext{ g} ]

Calculating Moles from Mass

• Given the mass of a substance and its atomic/molecular mass, moles can be calculated using:

  • Formula: [ n = \frac{m}{M_r} ]

• Rearranged for mass: [ m = n \times M_r ]

• Rearranged for molecular mass: [ M_r = \frac{m}{n} ]

Example Calculations:

1. Aluminium Oxide (Al(2)O(3))

   • Given: 2.6 g

   • ( M_r = (2 \times 27.0) + (3 \times 16.0) = 102 g/mol )

   • Moles: [ n = \frac{2.6}{102} = 0.025 \text{ mol} ]

2. Glucose (C(6)H({12})O(_6))

   • Given: 12.0 g

   • ( M_r = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 180 g/mol )

   • Moles: [ n = \frac{12.0}{180} = 0.067 \text{ mol} ]

Mole Calculations in Solutions

• Solutions are defined by concentration (moles per volume, typically in dm³):

  • Formula: [ n = C \times V ]

• Important conversions to remember:

  • If ( V ) is in cm³, convert it to dm³: ( V \text{ (dm³)} = \frac{V \text{ (cm³)}}{1000} )

Example: Sodium Chloride Solution

1. Given Concentration: 1.0 mol dm⁻³, Volume: 50 cm³

   • Moles Calculation: [ n = C \times V / 1000 = 1.0 \times 50 / 1000 = 0.050 \text{ mol} ]

Ideal Gas Equations and Moles

• The ideal gas equation relates pressure, volume, number of moles, and temperature:

  • --> [ pV = nRT ]

  • Variables:

  • ( p ): Pressure (Pascals)

  • ( V ): Volume (m³)

  • ( n ): Moles (mol)

  • ( R ): Gas constant (8.314 J K⁻¹ mol⁻¹)

  • ( T ): Temperature (K)

  • Note: Convert Celsius to Kelvin: ( K = \text{oC} + 273 )

Balanced Chemical Equations

• Ensure the same number of each type of atom appears on both sides of the equation to maintain mass balance.

Example Balancing the Reaction:

• Given the equation: [ H2(g) + O2(g) \rightarrow H_2O(l) ]

1. Balance oxygens: [ H2(g) + 2H2O(l) ]

2. Balance hydrogens: [ 2H2(g) + O2(g) \rightarrow 2H_2O(l) ]

Ionic Equations

• Ionic equations depict reactions and showcase the particles involved:

• Example:

  • Reaction: [ HCl(aq) + NaOH(l) \rightarrow NaCl(aq) + H_2O(l) ]

  • Complete ionic equation: [ H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l) ]

  • Condensed ionic equation: [ Cl^-(aq) + Na^+(aq) \rightarrow NaCl(aq) ]

Empirical and Molecular Formulas

• Empirical Formula: Simplest whole number ratio of atoms in a compound.

• Molecular Formula: Actual number of atoms of each element in a molecule.

• Example:

  • Glucose: Molecular formula: C(6)H({12})O(6), Empirical formula: CH(2)O.

Calculating Empirical Formula from Elemental Composition

1. Divide % by relative atomic mass.

2. Normalize ratios to simplest whole number form.

3. Construct formula.

Molecular Formula Calculations

1. Determine empirical formula using previously described method.

2. Divide molar mass of compound by empirical formula molar mass to find multiplier.

3. Construct molecular formula accordingly.

Percentage Yield and Atom Economy

• Percentage Yield:

  • Formula: [ \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 ]

• Atom Economy: Measures how efficiently reactants are converted into products.

  • Formula: [ \text{Percentage Atom Economy} = \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Products}} \times 100 ]

Example for Atom Economy Calculation

• Reaction: [ C6H{12}O6 \rightarrow 2CH3CH2OH + 2CO2 ]

• Calculation: [ \text{Atom Economy} = \frac{46}{46 + 44} \times 100 \approx 51.1\% ]