1. Core Mole Relationships
Quantitative chemistry is built upon the concept of the mole. A mole is defined as the amount of substance containing exactly \(6.022 \times 10^{23}\) elementary entities (the Avogadro constant, \(L\)).
The weighted average mass of a molecule relative to \(1/12\text{th}\) of the mass of an atom of carbon-12. Relative formula mass is used for compounds with giant lattices (such as ionic solids).
Solid Mass and Moles
Core Formula
\[ n = \frac{m}{M_r} \]
Where \(n\) is amount of substance in moles (\(\text{mol}\)), \(m\) is mass in grams (\(\text{g}\)), and \(M_r\) is relative molecular or formula mass in \(\text{g mol}^{-1}\).
Solution Concentrations
For solutions, concentration represents the amount of solute dissolved in a specified volume of solvent:
Concentration Formula
\[ n = c \times V \]
Where \(c\) is concentration in \(\text{mol dm}^{-3}\), and \(V\) is volume in \(\text{dm}^3\). To convert volume from \(\text{cm}^3\) to \(\text{dm}^3\), divide by \(1000\):
\[ V(\text{dm}^3) = \frac{V(\text{cm}^3)}{1000} \]
To convert molar concentration (\(\text{mol dm}^{-3}\)) to mass concentration (\(\text{g dm}^{-3}\)):
\[ \text{Concentration in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times M_r \]
2. The Ideal Gas Equation
At low pressure and high temperature, real gas molecules occupy negligible volume compared to the container and experience negligible intermolecular forces. Under these conditions, gases obey the ideal gas equation:
Ideal Gas Law
\[ pV = nRT \]
Strict SI Unit Requirements
| Variable | Physical Quantity | Strict SI Unit | Standard Conversion Rule |
|---|---|---|---|
| p | Pressure | Pascals (\(\text{Pa}\)) | \(1\text{ kPa} = 1000\text{ Pa}\); \(1\text{ MPa} = 1\times 10^6\text{ Pa}\) |
| V | Volume | Cubic metres (\(\text{m}^3\)) | \(1\text{ dm}^3 = 1\times 10^{-3}\text{ m}^3\); \(1\text{ cm}^3 = 1\times 10^{-6}\text{ m}^3\) |
| n | Amount | Moles (\(\text{mol}\)) | \(n = m / M_r\) |
| R | Gas Constant | \(\text{J K}^{-1}\text{mol}^{-1}\) | \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\) (Provided on data sheet) |
| T | Temperature | Kelvin (\(\text{K}\)) | \(T(\text{K}) = \theta(^\circ\text{C}) + 273.15\) |
Step 1: Convert all measurements to standard SI units:
- \(p = 101.3\text{ kPa} = 101300\text{ Pa}\)
- \(V = 78.5\text{ cm}^3 = 78.5 \times 10^{-6}\text{ m}^3\)
- \(T = 98.0 + 273.15 = 371.15\text{ K}\)
- \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\)
Step 2: Rearrange ideal gas equation for moles (\(n\)):
\[ n = \frac{pV}{RT} = \frac{101300 \times 78.5 \times 10^{-6}}{8.314 \times 371.15} = \frac{7.95205}{3085.741} = 0.00257703\text{ mol} \]
Step 3: Calculate relative molecular mass (\(M_r\)):
\[ M_r = \frac{m}{n} = \frac{0.284}{0.00257703} = 110.204\text{ g mol}^{-1} \]
Reported to 3 significant figures: \(M_r = 110\).
3. Empirical and Molecular Formulas
The simplest whole-number ratio of atoms of each element present in a compound.
The actual number of atoms of each element present in one molecule of a compound.
The molecular formula is always an integer multiple (\(k\)) of the empirical formula:
\[ k = \frac{M_r\text{ of compound}}{\text{Empirical formula mass}} \]
Step 1: Calculate masses of carbon and hydrogen:
- Mass of C: \(3.96 \times (12.0 / 44.0) = 1.08\text{ g}\)
- Mass of H: \(2.16 \times (2.0 / 18.0) = 0.24\text{ g}\)
- Mass of O by difference: \(1.80 - (1.08 + 0.24) = 0.48\text{ g}\)
Step 2: Find mole ratio:
- Moles of C: \(1.08 / 12.0 = 0.090\text{ mol}\)
- Moles of H: \(0.24 / 1.0 = 0.240\text{ mol}\)
- Moles of O: \(0.48 / 16.0 = 0.030\text{ mol}\)
Step 3: Divide by smallest value (0.030):
- C: \(0.090 / 0.030 = 3\)
- H: \(0.240 / 0.030 = 8\)
- O: \(0.030 / 0.030 = 1\)
Empirical formula = \(\text{C}_3\text{H}_8\text{O}\).
Empirical formula mass = \((3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\).
Since empirical mass equals \(M_r\) (60.0), integer multiplier \(k = 1\). The molecular formula is also \(\text{C}_3\text{H}_8\text{O}\).
4. Percentage Yield and Atom Economy
In synthetic chemistry and industrial processes, measuring reaction efficiency requires two complementary parameters:
Percentage Yield
\[ \% \text{ Yield} = \frac{\text{Actual mass of product}}{\text{Theoretical mass calculated}} \times 100 \]
Measures practical efficiency. Why is yield almost always less than 100%?
- Incomplete reaction reaching dynamic equilibrium.
- Unwanted side reactions producing by-products.
- Mechanical losses during filtration, transfer, or recrystallisation.
Atom Economy
\[ \% \text{ Atom Economy} = \frac{M_r\text{ of desired product}}{\sum M_r\text{ of all reactants}} \times 100 \]
Measures green chemistry efficiency. Addition reactions always possess 100% atom economy because all reactant atoms are incorporated into a single product.
5. Vertical Quantitative Integration
The Quantitative Spine of A-Level Chemistry
Stoichiometry is not an isolated topic. The principles mastered in Unit 1 directly feed into every subsequent modular paper:
- Unit CH02 (Energetics): Calculating molar enthalpy change \(\Delta H = -q / (n \times 1000)\) relies on accurate mole determination from solution mass or limiting reactants.
- Unit CH03 (Equilibria & Acids): Setting up ICE tables (Initial, Change, Equilibrium) for \(K_c\) and \(K_p\) requires strict mole conversion before evaluating equilibrium concentrations.
- Unit CH03 (Electrochemistry): Quantitative electrolysis connects charge to moles of electrons via \(Q = I \times t = n \times F\).
- Unit CH05 & Required Practicals (RP1 & RP7): Paper 5 practical evaluation demands flawless titration arithmetic, back-titration deductions, and percentage uncertainty error budgets.
6. Strict Examiner Rounding Rule
Examiners routinely penalise candidates who round numbers during intermediate calculation steps. Intermediate numbers must remain stored in your calculator registers.
Mathematical Demonstration: Why Early Rounding Fails
Consider calculating \(X = (10.0 / 3.0) \times 6.0\):
- Correct unrounded method: \(10.0 / 3.0 = 3.333333...\). Multiply by \(6.0\) gives exactly 20.0.
- Erroneous premature rounding: If rounded to 2 significant figures mid-calculation as \(3.3\): \(3.3 \times 6.0 = \mathbf{19.8}\).
The premature rounding creates an error of 1.0%, causing the final answer to fall completely outside the mark scheme acceptance range!
7. Exam-Style Practice Questions
Equation: \(\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\)
Step 1: Calculate moles of reactants:
- \(n(\text{HCl}) = c \times V = 0.200 \times (50.0 / 1000) = 0.0100\text{ mol}\)
- \(n(\text{Mg}) = m / A_r = 0.150 / 24.3 = 0.0061728\text{ mol}\)
Step 2: Determine limiting reagent:
0.0061728 mol of Mg requires \(2 \times 0.0061728 = 0.01235\text{ mol}\) of HCl. We only have 0.0100 mol of HCl. Therefore, HCl is the limiting reactant.
Step 3: Calculate moles of H2 gas:
From stoichiometry, \(n(\text{H}_2) = n(\text{HCl}) / 2 = 0.0100 / 2 = 0.00500\text{ mol}\).
Step 4: Calculate gas volume via pV = nRT:
\[ V = \frac{nRT}{p} = \frac{0.00500 \times 8.314 \times 295}{100000} = \frac{12.26315}{100000} = 1.2263 \times 10^{-4}\text{ m}^3 \]
Convert to \(\text{cm}^3\): \(1.2263 \times 10^{-4} \times 10^6 = \mathbf{123\text{ cm}^3}\) (to 3 significant figures).