Home Notes Papers

The mole and the Avogadro constant

Paper 1Paper 2Paper 3Paper 4Paper 5Paper 6

This topic is examined in Paper 1, Paper 2, Paper 3, Paper 4, Paper 5, and Paper 6.

The Mole and the Avogadro Constant
Learning Objective 2: State that the mole, mol, is the unit of amount of substance and that one mole contains 6.02 \times 10^{23} particles.

In chemistry, we deal with atoms and molecules which are incredibly small. To make calculations manageable, we group them into a specific number called one mole.

Definition: One mole of any substance contains exactly the Avogadro constant (N_A) of particles.

N_A = 6.02 \times 10^{23}

This means:

  • 1 mole of Carbon atoms = 6.02 \times 10^{23} atoms
  • 1 mole of Water molecules (H_2O) = 6.02 \times 10^{23} molecules
  • 1 mole of Sodium ions (Na^+) = 6.02 \times 10^{23} ions

Why is this useful?
It allows us to count particles by weighing them. If we know the mass of one mole (the molar mass), we can calculate how many moles are in any sample, and subsequently how many particles it contains.

Building on previous concepts:
Recall that Relative Atomic Mass (A_r) is the weighted average mass of an atom compared to \frac{1}{12}th the mass of a Carbon-12 atom. The Molar Mass (M) is simply the A_r (or M_r) expressed in grams per mole (g/mol). For example, if A_r of Carbon is 12, then 1 mole of Carbon atoms has a mass of 12 g.

Amount of Substance
Learning Objective 3: Use the relationship:

\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

Where:

  • amount of substance is measured in moles (mol)
  • mass is measured in grams (g)
  • molar mass is the mass of one mole of the substance, measured in g/mol. It is numerically equal to the relative atomic/molecular mass (A_r or M_r).

This formula can be rearranged to find:

  1. Mass: \text{mass} = \text{amount} \times \text{molar mass}
  2. Molar Mass: \text{molar mass} = \frac{\text{mass}}{\text{amount}}

Calculating Number of Particles:
To find the number of particles (N) from moles:
N = \text{amount (mol)} \times 6.02 \times 10^{23}

Calculating Moles and Particles
Example 1: Calculate moles from mass
Calculate the number of moles in 48.0 g of Magnesium (Mg). (A_r of Mg = 24)

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{48.0}{24} = 2.0 \text{ mol}

Example 2: Calculate particles from moles
How many atoms are in 0.50 mol of Argon (Ar)? (N_A = 6.02 \times 10^{23})

\text{number of atoms} = 0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} \text{ atoms}

Example 3: Calculating Molar Mass from Moles and Mass
A sample of 0.25 mol of a substance has a mass of 20.0 g. What is its molar mass?

\text{molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{20.0}{0.25} = 80.0 \text{ g/mol}

⚠︎ Confusing Mass and Moles in Stoichiometry
The Error: Students often try to use the mass ratio directly from the balanced equation without converting to moles first. For example, in 2Mg + O_2 \rightarrow 2MgO, a student might think '2g of Mg reacts with 1g of O_2'.

The Correct Understanding: The coefficients in a balanced equation represent moles, not grams. You must convert given masses to moles, use the mole ratio from the equation, and then convert back to mass if required.

Why this matters: Different elements have different atomic masses. 1 mole of Magnesium (24g) is not the same mass as 1 mole of Oxygen atoms (16g), even though they react in a 1:1 ratio in some contexts.

Using Molar Gas Volume at r.t.p.
Learning Objective 4: Use the molar gas volume, taken as 24 dm³ at room temperature and pressure (r.t.p.), in calculations involving gases.

When to use this phrase/concept:
Use this when a question involves the volume of a gas at r.t.p. and asks for moles or mass, or vice versa.

The Formula:
\text{moles of gas} = \frac{\text{volume of gas (dm}^3)}{24}
Or:
\text{volume of gas (dm}^3) = \text{moles of gas} \times 24

Examiner Acceptance:
Examiners accept calculations that explicitly state the use of 24 dm³/mol as the molar volume at r.t.p. Ensure you convert cm³ to dm³ by dividing by 1000 before using this formula.

Example Usage:
'Calculate the volume of 0.5 mol of hydrogen gas at r.t.p.'
Answer: 0.5 \times 24 = 12 \text{ dm}^3.

Connection to other topics:
This relates to Gas Laws. Note that this value (24 dm³) is specific to r.t.p. If the question specifies different temperature or pressure, you cannot use this constant.

Concentration of Solutions
Learning Objective 1 & 5: State that concentration can be measured in g/dm³ or mol/dm³.

Definition: Concentration is the amount of solute dissolved in a specific volume of solvent/solution.

There are two common units:

  1. g/dm³: Mass of solute (g) per cubic decimetre (dm³) of solution.
  2. mol/dm³ (also written as M): Moles of solute per cubic decimetre (dm³) of solution.

Conversion between cm³ and dm³:
1 \text{ dm}^3 = 1000 \text{ cm}^3
Therefore:
\text{Volume in dm}^3 = \frac{\text{Volume in cm}^3}{1000}

Formulas:
\text{Concentration (mol/dm}^3) = \frac{\text{moles of solute}}{\text{volume of solution (dm}^3)}
\text{Concentration (g/dm}^3) = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3)}

Calculating Concentration and Titration Data

Example 1: Calculating Concentration in mol/dm³
Calculate the concentration of a solution containing 0.05 moles of NaOH in 250 cm³ of solution.

  1. Convert volume to dm³: 250 \text{ cm}^3 = 0.250 \text{ dm}^3
  2. Apply formula:
    \text{Concentration} = \frac{0.05}{0.250} = 0.20 \text{ mol/dm}^3

Example 2: Using Titration Data (Learning Objective 6)
In a titration, 25.0 cm³ of NaOH solution reacts completely with 20.0 cm³ of 0.10 mol/dm³ HCl.
Equation: NaOH + HCl \rightarrow NaCl + H_2O (1:1 ratio)

  1. Calculate moles of HCl:
    \text{moles HCl} = \frac{20.0}{1000} \times 0.10 = 0.0020 \text{ mol}
  2. Use mole ratio (1:1): Moles of NaOH = 0.0020 mol
  3. Calculate concentration of NaOH:
    \text{Conc NaOH} = \frac{0.0020}{25.0/1000} = \frac{0.0020}{0.025} = 0.080 \text{ mol/dm}^3
⚠︎ Forgetting to Convert cm³ to dm³ in Concentration
The Error: Dividing moles by the volume in cm³ directly (e.g., 0.05 / 250) instead of converting to dm³ first.

The Correct Understanding: The unit mol/dm³ means 'moles per thousand cubic centimetres'. If you use cm³, your answer will be 1000 times too small.

Examiner Tip: Always write down the conversion step explicitly: '25 \text{ cm}^3 = 0.025 \text{ dm}^3'. This shows the examiner you understand the units and helps avoid calculation errors.

Empirical and Molecular Formulae

Learning Objective 7: Calculate empirical formulae and molecular formulae.

Empirical Formula: The simplest whole-number ratio of atoms in a compound.
Molecular Formula: The actual number of atoms of each element in a molecule.

Steps to find Empirical Formula from Mass/Percentage Data:

  1. Write the mass (or percentage) of each element below its symbol.
  2. Divide each mass by the relative atomic mass (A_r) to get moles.
  3. Divide all mole values by the smallest mole value obtained in step 2.
  4. If necessary, multiply by a common factor to get whole numbers.

Finding Molecular Formula:

  1. Calculate the molar mass of the empirical formula.
  2. Divide the given actual molar mass by the empirical formula mass.
  3. Multiply the subscripts in the empirical formula by this integer.
Calculating Empirical Formula

Example: A compound contains 80% Carbon and 20% Hydrogen by mass. Find the empirical formula. (A_r: C=12, H=1)

  1. Masses: C = 80g, H = 20g (assuming 100g total)
  2. Moles:
    • C: 80 / 12 = 6.67
    • H: 20 / 1 = 20
  3. Divide by smallest (6.67):
    • C: 6.67 / 6.67 = 1
    • H: 20 / 6.67 \approx 3
  4. Ratio is 1:3.
    Empirical Formula: CH_3
Percentage Yield, Purity, and Composition
Learning Objective 8: Calculate percentage yield, percentage composition by mass, and percentage purity.

1. Percentage Yield:
Measures the efficiency of a reaction. Real reactions rarely produce 100% yield due to side reactions, incomplete reactions, or loss of product during purification.
\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100%

  • Actual Yield: Mass obtained experimentally.
  • Theoretical Yield: Mass calculated from stoichiometry (assuming perfect reaction).

2. Percentage Purity:
Measures how much of an impure sample is the desired substance.
\text{Percentage Purity} = \left( \frac{\text{Mass of Pure Substance}}{\text{Mass of Impure Sample}} \right) \times 100%

3. Percentage Composition by Mass:
The percentage by mass of an element in a compound.
\text{Percentage Composition} = \left( \frac{\text{Total Mass of Element in Formula}}{\text{Molar Mass of Compound}} \right) \times 100%

Calculating Percentage Yield and Purity

Example 1: Percentage Yield
Calcium carbonate decomposes: CaCO_3 \rightarrow CaO + CO_2.
If 100g of CaCO_3 (M_r=100) is heated and produces 56g of CaO (M_r=56), what is the percentage yield?

  1. Moles of CaCO_3: 100 / 100 = 1.0 \text{ mol}
  2. Theoretical moles of CaO: 1:1 ratio, so 1.0 mol.
  3. Theoretical mass of CaO: 1.0 \times 56 = 56 \text{ g}.
  4. Actual yield is given as 56g.
    \text{Yield} = \left( \frac{56}{56} \right) \times 100% = 100%

Example 2: Percentage Purity
A 5.0g sample of impure calcium carbonate reacts with acid to produce 2.2g of CO_2 (M_r=44). What is the purity?

  1. Moles of CO_2: 2.2 / 44 = 0.05 \text{ mol}
  2. From equation CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2, ratio is 1:1.
  3. Moles of pure CaCO_3 reacted: 0.05 mol.
  4. Mass of pure CaCO_3: 0.05 \times 100 = 5.0 \text{ g}.
  5. Purity:
    \text{Purity} = \left( \frac{5.0}{5.0} \right) \times 100% = 100%
    (Note: If the actual sample mass was different, e.g., 6.0g impure sample yielding 5.0g pure, purity would be 5/6 \times 100 = 83.3%)
Identifying Limiting Reactants
Learning Objective 5: Calculate stoichiometric reacting masses and limiting reactants.

When to use this concept:
Use this when the question gives amounts for both reactants. You must determine which one runs out first (the limiting reactant) because it determines the maximum amount of product formed.

Examiner Acceptance:
Examiners look for clear working showing:

  1. Calculation of moles for both reactants.
  2. Comparison of mole ratios based on the balanced equation.
  3. A statement identifying the limiting reactant.

Example Usage:
'Calculate the mass of MgCl_2 formed from 48.0g Mg and excess HCl.'
Here, Mg is the limiting reactant because HCl is in excess. You calculate moles of Mg, use the 1:1 ratio to find moles of MgCl_2, then convert to mass.

Common Pitfall:
Do not assume the reactant with the smaller mass is the limiting reactant. Always compare moles, not mass.

Past Paper Style Questions
Q:
Limestone (CaCO_3) is used to remove SO_2 from flue gases. Equation: CaCO_3 + SO_2 \rightarrow CaSO_3 + CO_2. What is the smallest mass of CaCO_3 required to remove 1 tonne (1000 kg) of SO_2? (A_r: Ca=40, C=12, O=16, S=32)
A:
  1. Moles of SO_2: Mass = 1,000,000 g. M_r(SO_2) = 32 + (2 \times 16) = 64.
    Moles = 1,000,000 / 64 = 15,625 \text{ mol}.
  2. Mole ratio is 1:1. So moles of CaCO_3 needed = 15,625 mol.
  3. Mass of CaCO_3: M_r(CaCO_3) = 40 + 12 + (3 \times 16) = 100.
    Mass = 15,625 \times 100 = 1,562,500 \text{ g} or 1.56 tonnes.
Q:
Calculate the volume of hydrogen gas produced at r.t.p. when 0.5 mol of magnesium reacts with excess hydrochloric acid. (Mg + 2HCl \rightarrow MgCl_2 + H_2)
A:
  1. Mole ratio Mg : H_2 is 1:1.
  2. Moles of H_2 produced = 0.5 mol.
  3. Volume at r.t.p. = 0.5 \times 24 \text{ dm}^3 = \mathbf{12 \text{ dm}^3}.
Q:
A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. Its relative molecular mass is 60. Determine its empirical and molecular formulae. (A_r: C=12, H=1, O=16)
A:

Empirical Formula:

  1. Moles: C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.
  2. Divide by smallest (3.33): C=1, H=2, O=1.
    Empirical Formula: CH_2O.

Molecular Formula:

  1. Mass of empirical formula (CH_2O) = 12 + 2 + 16 = 30.
  2. Ratio = 60 / 30 = 2.
  3. Molecular Formula: C_2H_4O_2.
Q:
In an experiment, 5.0g of calcium carbonate was heated. The theoretical yield of calcium oxide was calculated to be 2.8g. The actual yield obtained was 2.5g. Calculate the percentage yield.
A:
\text{Percentage Yield} = \left( \frac{\text{Actual}}{\text{Theoretical}} \right) \times 100%
\text{Yield} = \left( \frac{2.5}{2.8} \right) \times 100% = \mathbf{89.3%}
Beta v0.7.8 Free while we're in beta — it transitions to paid post launch. Thank you for supporting us at this stage!