Balancing Equations and State Symbols
Stoichiometry is the study of quantitative relationships in chemical reactions. The coefficients in a balanced equation give the mole ratios of reactants and products. Every balanced equation must also include state symbols, which are mandatory in all IB exams.
| Symbol | State | Meaning |
|---|---|---|
| (s) | Solid | Fixed shape and volume |
| (l) | Liquid | Pure liquid (not a solution) |
| (g) | Gas | Gaseous state |
| (aq) | Aqueous | Dissolved in water |
Core Stoichiometry Equations
The Three Key Formulas
\( n = \frac{m}{M} \) (mass → moles, where M = molar mass in g mol⁻¹)
\( n = cV \) (solution: concentration in mol dm⁻³ × volume in dm³)
\( n = \frac{V}{V_m} \) (gas at STP: Vm = 22.7 dm³ mol⁻¹)
The Stoichiometry Bridge
Convert → Moles of known → Mole ratio → Moles of unknown → Convert
Worked Example: Mass Stoichiometry
CaCO₃(s) → CaO(s) + CO₂(g)
Step 1: Calculate moles of CaCO₃:
\( n = \frac{m}{M} = \frac{10.0}{100.09} = 0.0999 \text{ mol} \)
Step 2: Use the mole ratio (1:1):
Moles of CO₂ = 0.0999 mol
Step 3: Convert back to mass:
\( m = n \times M = 0.0999 \times 44.01 = \textbf{4.40 g} \)
Molar Volume of Gases at STP
At STP (273.15 K, 100 kPa), one mole of any ideal gas occupies 22.7 dm³. This allows rapid conversion between moles and gas volume:
\( V = n \times V_m = n \times 22.7 \text{ dm}^3 \)
\( V = 0.500 \times 22.7 = 11.35 \text{ dm}^3 \approx 11.4 \text{ dm}^3 \text{ (3 s.f.)} \)
The Ideal Gas Equation
When conditions differ from STP, use the ideal gas equation:
\( PV = nRT \)
| Symbol | Meaning | SI Units |
|---|---|---|
| \( P \) | Pressure | Pa (or kPa if V in dm³) |
| \( V \) | Volume | m³ (or dm³ if P in kPa) |
| \( n \) | Amount of substance | mol |
| \( R \) | Gas constant | 8.314 J K⁻¹ mol⁻¹ |
| \( T \) | Temperature | K (= °C + 273) |
Related Gas Laws
The ideal gas equation combines three historical laws:
- Boyle's Law: \( P \propto \frac{1}{V} \) at constant T (inverse relationship)
- Charles's Law: \( V \propto T \) at constant P (direct relationship)
- Gay-Lussac's Law: \( P \propto T \) at constant V (direct relationship)
For a fixed mass of gas changing conditions: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
Solution Concentration
The amount of solute dissolved per unit volume of solution, measured in mol dm⁻³ (sometimes written as M).
\( c = \frac{n}{V} \quad \text{or} \quad n = cV \)
V must be in dm³ (divide cm³ by 1000)
Step 1: Convert volume:
\( V = \frac{250.0}{1000} = 0.2500 \text{ dm}^3 \)
Step 2: Calculate moles:
\( n = cV = 0.100 \times 0.2500 = 0.0250 \text{ mol} \)
Step 3: Convert to mass:
\( m = n \times M = 0.0250 \times 40.00 = \textbf{1.00 g} \)
Think About It
In the CaCO₃ worked example above, what volume would the CO₂ occupy at STP?
V = n × Vm = 0.0999 × 22.7 = 2.27 dm³