The mole and the Avogadro constant- CIE iGCSE Chemistry Notes - New Syllabus
The mole and the Avogadro constant Notes for iGCSE
Core Syllabus
- State that concentration can be measured in g /dm3 or mol/dm3
Supplement Syllabus
State that the mole, mol, is the unit of amount of substance and that one mole contains $6.02 \times 10^{23}$ particles, e.g. atoms, ions, molecules; this number is the Avogadro constant Use the relationship amount of substance $(\mathrm{mol})=\frac{\text { mass }(\mathrm{g})}{\text { molar mass }(\mathrm{g} / \mathrm{mol})}$ to calculate:
(a) amount of substance
(b) mass
(c) molar mass
(d) relative atomic mass or relative molecular/formula mass
(e) number of particles, using the value of the Avogadro constantUse the molar gas volume, taken as $24 \mathrm{dm}^3$ at room temperature and pressure, r.t.p., in calculations involving gases
Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in $\mathrm{g} / \mathrm{dm}^3$ and $\mathrm{mol} / \mathrm{dm}^3$, including conversion between $\mathrm{cm}^3$ and $\mathrm{dm}^3$
Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution
Calculate empirical formulae and molecular formulae, given appropriate data
Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data
The Mole Concept
The Mole and the Avogadro Constant
One mole is the amount of substance that contains exactly \( 6.02 \times 10^{23} \) particles. These particles can be atoms, molecules, or ions, depending on the substance.
This number is called the Avogadro constant and is used to count particles at the atomic scale.
Examples of what 1 mole contains:
- 1 mole of oxygen molecules (O₂) contains \( 6.02 \times 10^{23} \) O₂ molecules
- 1 mole of sodium atoms (Na) contains \( 6.02 \times 10^{23} \) sodium atoms
- 1 mole of chloride ions (Cl⁻) contains \( 6.02 \times 10^{23} \) Cl⁻ ions
Avogadro Constant:
\( 1 \text{ mol} = 6.02 \times 10^{23} \text{ particles} \)
This constant allows us to link the macroscopic world (grams and litres) to the microscopic world (atoms and molecules).
Example
How many molecules are there in 3 moles of carbon dioxide (CO₂)?
▶️Answer/Explanation
Number of molecules = \( 3 \times 6.02 \times 10^{23} = 1.806 \times 10^{24} \) molecules
Example
How many sodium atoms are present in 0.2 mol of sodium?
▶️Show Answer
Number of atoms = \( 0.2 \times 6.02 \times 10^{23} = 1.204 \times 10^{23} \) atoms
Using the Relationship: Moles, Mass, Molar Mass and Number of Particles
Using the Relationship: Moles, Mass, Molar Mass and Number of Particles
\( \text{Moles (mol)} = \dfrac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \)
This formula links:
- Amount of substance in moles (mol)
- Mass of the substance in grams (g)
- Molar mass in g/mol – same as the relative atomic mass (Ar) or relative molecular/formula mass (Mr)
(a) Calculating Amount of Substance (moles)
To find the number of moles of a substance:
\( \text{Moles} = \dfrac{\text{Mass}}{\text{Molar Mass}} \)
Where molar mass is taken from the periodic table as the relative atomic mass (Ar) or the sum of Ar values for compounds.
(b) Calculating Mass (g)
To find mass when moles and molar mass are known:
\( \text{Mass} = \text{Moles} \times \text{Molar Mass} \)
(c) Calculating Molar Mass (g/mol)
To find molar mass when mass and moles are known:
\( \text{Molar Mass} = \dfrac{\text{Mass}}{\text{Moles}} \)
(d) Calculating Relative Atomic or Molecular/Formula Mass
- Relative atomic mass (Ar) is the average mass of an atom of the element compared to 1/12 the mass of a carbon-12 atom.
- Relative molecular mass (Mr) is the sum of relative atomic masses in a molecule.
- Relative formula mass is used for ionic compounds, found similarly by adding Ar values of the ions.
(e) Calculating Number of Particles Using Avogadro Constant
To calculate the number of particles (atoms, ions or molecules) from moles, use:
\( \text{Number of Particles} = \text{Moles} \times 6.02 \times 10^{23} \)
6.02 × 10²³ is the Avogadro constant — the number of particles in 1 mole.
Example
Calculate the mass of 0.25 mol of carbon dioxide (CO2).
▶️Answer/Explanation
Step 1 – Write the formula:
\( \text{mass} = \text{moles} \times \text{molar mass} \)
Step 2 – Calculate molar mass (Mr) of CO₂:
\( \text{Mr of CO}_2 = 12 + (2 \times 16) = 44 \, \text{g/mol} \)
Step 3 – Calculate mass:
\( \text{mass} = 0.25 \times 44 = 11 \, \text{g} \)
Answer: The mass of 0.25 mol of CO₂ is 11 g.
Example
A sample of sulfur trioxide (SO₃) has a mass of 64 g. Calculate:
- The number of moles of SO₃ present
- The number of molecules of SO₃ in the sample
- The total number of atoms in the sample
▶️Answer/Explanation
Step 1 – Calculate Molar Mass (Mr) of SO₃:
\( \text{Mr of SO}_3 = 32 + (3 \times 16) = 80 \, \text{g/mol} \)
Step 2 – Moles of SO₃:
\( \text{Moles} = \dfrac{\text{Mass}}{\text{Molar Mass}} = \dfrac{64}{80} = 0.8 \, \text{mol} \)
Step 3 – Number of Molecules:
Use Avogadro constant: \( 1 \, \text{mol} = 6.02 \times 10^{23} \) particles
\( \text{Molecules of SO}_3 = 0.8 \times 6.02 \times 10^{23} = 4.816 \times 10^{23} \) molecules
Step 4 – Total Number of Atoms:
Each SO₃ molecule contains 1 sulfur + 3 oxygen = 4 atoms
Total atoms = \( 4.816 \times 10^{23} \times 4 = 1.9264 \times 10^{24} \) atoms
Using Moles in Gas and Solution Calculations
Using the Molar Gas Volume in Calculations Involving Gases
At room temperature and pressure (r.t.p.), 1 mole of any gas occupies 24 dm³ (or 24,000 cm³). This is known as the molar volume of a gas.
Formula:
\( \text{Volume of gas (dm}^3\text{)} = \text{Moles of gas} \times 24 \)
\( \text{Moles of gas} = \dfrac{\text{Volume (dm}^3\text{)}}{24} \)
If using cm³: \( \text{Moles} = \dfrac{\text{Volume (cm}^3\text{)}}{24,000} \)
Units:
- 1 dm³ = 1,000 cm³
- Always convert to dm³ if needed: divide cm³ by 1,000
Application:
- Can be used in stoichiometry problems involving gaseous reactants or products
- Used to calculate volume of gas produced or used in a chemical reaction, if moles are known
Example
How many dm³ of hydrogen gas are produced when 2 mol of zinc reacts with hydrochloric acid?
Equation: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
▶️Answer/Explanation
From the equation: 1 mol Zn gives 1 mol H₂
So 2 mol Zn gives 2 mol H₂
Volume = \( 2 \times 24 = 48 \text{ dm}^3 \)
Example
What volume of carbon dioxide gas (in dm³) is produced when 10 g of calcium carbonate reacts with excess hydrochloric acid? (Ar of Ca = 40, C = 12, O = 16)
Equation: \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \)
▶️Answer/Explanation
Mr of CaCO₃ = 40 + 12 + (3×16) = 100 g/mol
Moles of CaCO₃ = \( \dfrac{10}{100} = 0.1 \text{ mol} \)
From the equation: 1 mol CaCO₃ gives 1 mol CO₂
So 0.1 mol gives 0.1 mol CO₂
Volume = \( 0.1 \times 24 = 2.4 \text{ dm}^3 \)
Reacting masses, limiting reactants, gas volumes at r.t.p.
Reacting Masses
You can calculate the mass of reactants or products using the mole concept and balanced equations:
- Find moles of a given substance using:
\( \text{mol} = \dfrac{\text{mass (g)}}{\text{Mr}} \) - Use the mole ratio from the balanced equation to find moles of the other substance
- Convert moles to mass using:
\( \text{mass} = \text{mol} \times \text{Mr} \)
Limiting Reactants
The limiting reactant is the one that gets used up first and controls the amount of product formed. Steps:
- Calculate the number of moles of each reactant
- Compare using the mole ratio from the balanced equation
- The one that gives fewer moles of product is the limiting reactant
Example
Calculate the mass of aluminium oxide formed when 5.4 g of aluminium reacts with excess oxygen.
▶️Answer/Explanation
Step 1 – Write balanced equation:
\( 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \)
Step 2 – Calculate moles of Al:
\( \text{Mr of Al} = 27 \)
\( \text{mol of Al} = \dfrac{5.4}{27} = 0.2 \, \text{mol} \)
Step 3 – Use ratio Al:Al₂O₃ = 4:2 → 0.2 mol Al gives:
\( \dfrac{2}{4} \times 0.2 = 0.1 \, \text{mol of Al}_2\text{O}_3 \)
Step 4 – Find mass of Al₂O₃:
\( \text{Mr of Al}_2\text{O}_3 = (2 \times 27) + (3 \times 16) = 102 \)
\( \text{mass} = 0.1 \times 102 = 10.2 \, \text{g} \)
Answer: 10.2 g of aluminium oxide is formed.
Example
20.0 cm³ of hydrogen reacts with 10.0 cm³ of oxygen. What volume of water vapour is produced, and which is the limiting reactant?
▶️Answer/Explanation
Step 1 – Write balanced equation:
\( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)
Step 2 – Ratio H₂:O₂ is 2:1
20.0 cm³ H₂ would need 10.0 cm³ O₂ → both are exact ratio
Step 3 – Volume of H₂O formed:
From ratio 2 mol H₂ gives 2 mol H₂O → same volume
So 20.0 cm³ H₂ gives 20.0 cm³ H₂O
Answer:
Volume of water vapour = 20.0 cm³
No limiting reactant (exact stoichiometric ratio)
Solution Concentration
Concentration is the amount of a substance dissolved in a given volume of solution. It tells us how strong or dilute a solution is.
Common Units of Concentration:
- g/dm³ — grams per cubic decimetre
- mol/dm³ — moles per cubic decimetre
Formulae:
- \( \text{mol} = \text{concentration (mol/dm}^3\text{)} \times \text{volume (dm}^3\text{)} \)
- To convert cm³ to dm³: divide by 1000
Converting between g/dm³ and mol/dm³:
The two units are related by the molar mass of the solute:
\( \text{Concentration (mol/dm}^3) = \frac{\text{Concentration (g/dm}^3)}{\text{Molar mass (g/mol)}} \)
\( \text{Concentration (g/dm}^3) = \text{Concentration (mol/dm}^3) × \text{Molar mass (g/mol)} \)
Example
Calculate the concentration in mol/dm³ of a sodium chloride solution with concentration 58.5 g/dm³
▶️Answer/Explanation
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Concentration in mol/dm³ = \( \frac{58.5}{58.5} = 1.0 \) mol/dm³
Example
A solution of sulfuric acid has a concentration of 0.5 mol/dm³. What is its concentration in g/dm³?
(Relative atomic masses: H = 1, S = 32, O = 16)
▶️Show Answer
Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98 g/mol
Concentration in g/dm³ = 0.5 × 98 = 49 g/dm³
Example
Calculate the concentration in mol/dm³ of a solution containing 1.5 g of NaOH in 250 cm³.
▶️Answer/Explanation
Step 1 – Find molar mass of NaOH:
\( \text{Mr of NaOH} = 23 + 16 + 1 = 40 \)
Step 2 – Moles of NaOH:
\( \text{mol} = \dfrac{1.5}{40} = 0.0375 \, \text{mol} \)
Step 3 – Convert volume to dm³:
\( \dfrac{250}{1000} = 0.25 \, \text{dm}^3 \)
Step 4 – Calculate concentration:
\( \text{concentration} = \dfrac{0.0375}{0.25} = 0.15 \, \text{mol/dm}^3 \)
Answer: Concentration = 0.15 mol/dm³
Using titration data to calculate moles, concentrations, or volumes
Using titration data to calculate moles, concentrations, or volumes
Titration is an experiment used to determine the concentration of a solution by reacting it with a solution of known concentration. It is commonly used in acid-base reactions.
Key formula:
\( \text{mol} = \text{concentration (mol/dm}^3\text{)} \times \text{volume (dm}^3\text{)} \)
Steps in a titration calculation:
- Write the balanced chemical equation
- Convert all volumes from cm³ to dm³ by dividing by 1000
- Calculate the moles of the known solution (volume × concentration)
- Use the mole ratio from the equation to find moles of the unknown
- Use the mole formula to calculate the unknown concentration or volume
Units to remember:
- Volume must be in dm³ for the formula to work. (1 dm³ = 1000 cm³)
- Concentration is usually in mol/dm³ or can be converted to g/dm³ using:
\( \text{g/dm}^3 = \text{mol/dm}^3 \times \text{Mr} \)
Example types:
- Given volume and concentration of one solution and volume of the other – find the unknown concentration
- Calculate volume needed to neutralise a known amount of acid or alkali
Example
25.0 cm³ of sodium hydroxide solution is neutralised by 27.5 cm³ of 0.100 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide solution.
▶️Answer/Explanation
Step 1 – Write the balanced equation:
\( \text{NaOH (aq)} + \text{HCl (aq)} \rightarrow \text{NaCl (aq)} + \text{H}_2\text{O (l)} \)
Step 2 – Convert volumes to dm³:
NaOH: \( \frac{25.0}{1000} = 0.025 \, \text{dm}^3 \)
HCl: \( \frac{27.5}{1000} = 0.0275 \, \text{dm}^3 \)
Step 3 – Calculate moles of HCl:
\( \text{mol} = \text{conc} \times \text{vol} = 0.100 \times 0.0275 = 0.00275 \, \text{mol} \)
Step 4 – Use mole ratio (1:1):
Moles of NaOH = 0.00275 mol
Step 5 – Calculate concentration of NaOH:
\( \text{conc} = \frac{\text{mol}}{\text{vol}} = \frac{0.00275}{0.025} = 0.110 \, \text{mol/dm}^3 \)
Answer: The concentration of NaOH is 0.110 mol/dm³.
Example
25.0 cm³ of sulfuric acid (H2SO4) was diluted to 250 cm³. Then, 20.0 cm³ of this diluted acid required 32.0 cm³ of 0.10 mol/dm³ sodium hydroxide (NaOH) to neutralise it. Calculate the concentration of the original sulfuric acid solution before dilution.
▶️Answer/Explanation
Step 1 – Write the balanced equation:
\( \text{H}_2\text{SO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{H}_2\text{O} (l) \)
Step 2 – Convert volumes to dm³:
Acid (diluted sample): \( \frac{20.0}{1000} = 0.020 \, \text{dm}^3 \)
NaOH: \( \frac{32.0}{1000} = 0.032 \, \text{dm}^3 \)
Step 3 – Calculate moles of NaOH:
\( \text{mol NaOH} = 0.10 \times 0.032 = 0.0032 \, \text{mol} \)
Step 4 – Use mole ratio to find moles of H₂SO₄ (1:2):
\( \text{mol H}_2\text{SO}_4 = \frac{0.0032}{2} = 0.0016 \, \text{mol} \)
Step 5 – Calculate concentration of diluted H₂SO₄:
\( \text{conc} = \frac{0.0016}{0.020} = 0.080 \, \text{mol/dm}^3 \)
Step 6 – Use dilution to find original concentration:
The diluted solution was made by diluting 25.0 cm³ to 250 cm³, a 10× dilution.
\( \text{original concentration} = 0.080 \times 10 = 0.80 \, \text{mol/dm}^3 \)
Answer: The concentration of the original sulfuric acid solution was 0.80 mol/dm³.
Calculations of Empirical Formulae, Molecular Formulae, and Percentage Yield
Empirical and Molecular Formulae
Empirical Formula is the simplest whole number ratio of atoms of each element in a compound.
Molecular Formula is the actual number of atoms of each element in one molecule of the compound.
Steps to calculate the empirical formula from masses or percentages:
- Write down the mass or percentage of each element.
- Divide each by the element’s relative atomic mass (Ar) to get the number of moles.
- Divide all mole values by the smallest mole value to get the simplest whole number ratio.
- Use this ratio to write the empirical formula.
To find the molecular formula:
- Calculate the relative formula mass (Mr) of the empirical formula.
- Divide the given molecular Mr by the empirical Mr.
- Multiply the empirical formula by this value to get the molecular formula.
Note: Always round mole ratios to the nearest whole number, or close fractions like 1.5 can be multiplied up to whole numbers (e.g. ×2 to get 3).
Example
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass (Mr) is 60. Determine its empirical and molecular formulae.
▶️Answer/Explanation
Step 1 – Assume 100 g of compound
So: C = 40.0 g, H = 6.7 g, O = 53.3 g
Step 2 – Divide by relative atomic mass (Ar)
C: \( \frac{40.0}{12} = 3.33 \)
H: \( \frac{6.7}{1} = 6.7 \)
O: \( \frac{53.3}{16} = 3.33 \)
Step 3 – Divide by smallest mole value
C: \( \frac{3.33}{3.33} = 1 \)
H: \( \frac{6.7}{3.33} \approx 2 \)
O: \( \frac{3.33}{3.33} = 1 \)
Empirical formula = CH2O
Step 4 – Calculate Mr of empirical formula
C = 12, H = 1×2 = 2, O = 16 → Mr = 12 + 2 + 16 = 30
Step 5 – Determine multiplier for molecular formula
\( \frac{60}{30} = 2 \)
Molecular formula = C2H4O2
Percentage Yield, Composition by Mass, and Purity
1. Percentage Yield
It compares the actual amount of product obtained to the theoretical (expected) amount.
Formula:
\( \text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \)
Yield is always less than 100% due to loss during reaction, side reactions, or incomplete reactions.
2. Percentage Composition by Mass
This shows how much of each element is present by mass in a compound.
Formula:
\( \text{Percentage by mass of element} = \frac{\text{A}_r \times \text{number of atoms}}{\text{M}_r \text{ of compound}} \times 100 \)
This helps in identifying unknown compounds or checking consistency with a formula.
3. Percentage Purity
Used when a substance contains impurities. It compares the mass of the pure substance to the total mass of the sample.
Formula:
\( \text{Percentage purity} = \frac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100 \)
This is important in manufacturing and titration calculations where purity affects product quality and safety.
Example
Iron(III) oxide (Fe2O3) reacts with carbon monoxide (CO) to form iron and carbon dioxide:
\( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)
In a reaction, 160 g of iron(III) oxide was used. The mass of iron actually obtained was 100 g. The sample of iron obtained was only 80% pure. Calculate:
- The theoretical yield of iron
- The percentage yield
- The percentage composition by mass of iron in iron(III) oxide
- The mass of pure iron in the sample
- The percentage purity of the sample
▶️Answer/Explanation
Step 1 – Find Mr values:
Fe = 56, O = 16
\( M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \)
Step 2 – Theoretical yield of iron:
From the balanced equation: 160 g of Fe2O3 gives 2 mol of Fe → 2 × 56 = 112 g
So theoretical yield = 112 g
Step 3 – Percentage yield:
Actual yield = 100 g
\( \text{Percentage yield} = \frac{100}{112} \times 100 \approx 89.3\% \)
Step 4 – % composition by mass of iron in Fe2O3:
\( \frac{112}{160} \times 100 = 70\% \)
Step 5 – Mass of pure iron in sample:
\( \frac{80}{100} \times 100 = 80 \text{ g} \)
Step 6 – Percentage purity:
\( \text{Percentage purity} = \frac{80}{100} \times 100 = 80\% \)