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 |
State Symbols in Chemical Equations
Missing or incorrect state symbols ((s), (l), (g), (aq)) routinely cost marks on IB exams. Always include them unless a question explicitly states otherwise, especially when writing thermochemical or net ionic equations.
Core Stoichiometry Equations
The Three Key Mole Formulas
- Solids (Mass & Molar Mass): \( n = \dfrac{m}{M} \)
- Solutions (Concentration & Volume): \( n = cV \) (where \(V\) is in \(\text{dm}^3\))
- Gases at STP (Molar Gas Volume): \( n = \dfrac{V}{V_{\text{m}}} = \dfrac{V}{22.7\text{ dm}^3\text{mol}^{-1}} \)
The Stoichiometry Bridge
Convert → Moles of known → Mole ratio → Moles of unknown → Convert
Using Exact Data Booklet Relative Atomic Masses (\(A_{\text{r}}\))
Always use the atomic masses from the IB Data Booklet to two decimal places (e.g. \(\text{C} = 12.01\), \(\text{O} = 16.00\), \(\text{Ca} = 40.08\)). Rounding atomic masses to whole numbers during intermediate calculation steps introduces rounding penalties.
Worked Example: Mass Stoichiometry
Reacting Mass Stoichiometry Calculation
Problem: What mass of \(\text{CO}_2\) forms when \(10.0\text{ g}\) of \(\text{CaCO}_3\) thermally decomposes?
\(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)
Step 1: Calculate moles of \(\text{CaCO}_3\)
\(M(\text{CaCO}_3) = 40.08 + 12.01 + 3(16.00) = 100.09\text{ g mol}^{-1}\)
\(n(\text{CaCO}_3) = \dfrac{10.0}{100.09} = 0.0999\text{ mol}\)
Step 2: Mole ratio
\(\text{CaCO}_3 : \text{CO}_2 = 1 : 1 \implies n(\text{CO}_2) = 0.0999\text{ mol}\)
Step 3: Convert moles to mass of \(\text{CO}_2\)
\(M(\text{CO}_2) = 12.01 + 2(16.00) = 44.01\text{ g mol}^{-1}\)
\(m(\text{CO}_2) = 0.0999 \times 44.01 = \mathbf{4.40\text{ g}}\)
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:
Molar Gas Volume at STP
\( V = n \times V_{\text{m}} = n \times 22.7\text{ dm}^3 \)
At Standard Temperature and Pressure (STP: \(273.15\text{ K}\), \(100\text{ kPa}\)).
Calculating Gas Volume at STP
Problem: Calculate the volume occupied by \(0.500\text{ mol}\) of an ideal gas at STP.
Calculation:
\(V = n \times 22.7 = 0.500 \times 22.7 = \mathbf{11.35\text{ dm}^3} \quad (11.4\text{ dm}^3 \text{ to 3 s.f.})\)
The Ideal Gas Equation
When conditions differ from STP, use the ideal gas equation:
The Ideal Gas Equation
\( PV = nRT \)
\(P\) in \(\text{Pa}\) (or \(\text{N m}^{-2}\)), \(V\) in \(\text{m}^3\), \(n\) in \(\text{mol}\), \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\), \(T\) in \(\text{K}\).
| 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) |
The Individual Gas Laws and Combined Gas Law
- Boyle\'s Law: \( P \propto \dfrac{1}{V} \) or \( P_1 V_1 = P_2 V_2 \) (at constant \(T, n\))
- Charles\'s Law: \( V \propto T \) or \( \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \) (at constant \(P, n\))
- Gay-Lussac\'s Law: \( P \propto T \) or \( \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \) (at constant \(V, n\))
- Combined Gas Law: \( \dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2} \) (for a fixed amount of gas \(n\))
Solution Concentration
Solution Concentration (\(c\))
Concentration (\(c\)): The amount of solute (in \(\text{mol}\)) dissolved per unit volume of solution (in \(\text{dm}^3\)). The SI-derived unit is \(\text{mol dm}^{-3}\) (often abbreviated as \(\text{M}\)).
Concentration and Molarity Formula
\( c = \dfrac{n}{V} \quad \text{or} \quad n = cV \)
Always convert solution volumes in \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by 1000 (\(V_{\text{dm}^3} = \frac{V_{\text{cm}^3}}{1000}\)).
Standard Solution Preparation Calculation
Problem: What mass of \(\text{NaOH}\) is needed to prepare \(250.0\text{ cm}^3\) of a \(0.100\text{ mol dm}^{-3}\) solution?
Step 1: Convert volume to \(\text{dm}^3\)
\(V = \dfrac{250.0}{1000} = 0.2500\text{ dm}^3\)
Step 2: Calculate moles of \(\text{NaOH}\)
\(n = cV = 0.100 \times 0.2500 = 0.0250\text{ mol}\)
Step 3: Convert moles to mass
\(M(\text{NaOH}) = 22.99 + 16.00 + 1.01 = 40.00\text{ g mol}^{-1}\)
\(m = n \times M = 0.0250 \times 40.00 = \mathbf{1.00\text{ g}}\)
Connecting Reacting Mass to Gas Volume
In the \(\text{CaCO}_3\) thermal decomposition calculation above, \(0.0999\text{ mol}\) of \(\text{CO}_2\) was produced. At STP, this gas occupies:
\(V = 0.0999\text{ mol} \times 22.7\text{ dm}^3\text{mol}^{-1} = \mathbf{2.27\text{ dm}^3}\) (or \(2270\text{ cm}^3\)).
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