Formulae and Amounts of a Substance
Welcome to one of the most fundamental toolkits in chemistry! If you have ever followed a recipe to bake a cake, you already understand the main philosophy of this chapter: measuring the exact quantities of starting ingredients to produce the perfect amount of product. In chemistry, we call this stoichiometry and mole calculations.
Don't worry if maths in chemistry has felt overwhelming before. We will break every single calculation down into simple, manageable steps with clear rules and memory tricks.
1. Counting Atoms: The Mole and Relative Masses
What is a Mole?
Atoms and molecules are impossibly tiny. If you tried to weigh out a single atom of carbon, your balance would need to read around \(0.00000000000000000000002\text{ g}\)! Because individual particles are so small, chemists group them into giant bundles called moles.
Analogy: Just like the word "dozen" always means \(12\) items (whether eggs, doughnuts, or cars), the word mole (symbol: \(\text{mol}\)) always means \(6.02 \times 10^{23}\) particles.
This huge number, \(6.02 \times 10^{23}\text{ mol}^{-1}\), is known as the Avogadro Constant (\(L\) or \(N_A\)).
Did you know? If you had one mole of marbles, they would cover the entire surface of the Earth to a depth of several miles!
Key Definitions of Mass
Because atoms have tiny masses, we measure them relative to a standard: the carbon-12 isotope (\(^{12}\text{C}\)).
Relative Atomic Mass (\(A_r\)): The weighted mean mass of an atom of an element compared to \(\frac{1}{12}\)th of the mass of an atom of carbon-12.
Relative Isotopic Mass: The mass of an atom of an isotope compared to \(\frac{1}{12}\)th of the mass of an atom of carbon-12.
Relative Molecular Mass (\(M_r\)): The weighted mean mass of a molecule compared to \(\frac{1}{12}\)th of the mass of an atom of carbon-12 (used for simple covalent molecules like \(\text{H}_2\text{O}\) or \(\text{CO}_2\)).
Relative Formula Mass (\(M_r\)): The weighted mean mass of a formula unit compared to \(\frac{1}{12}\)th of the mass of an atom of carbon-12 (used for giant structures and ionic compounds like \(\text{NaCl}\)).
Calculating \(M_r\): Add up the \(A_r\) values of all the atoms in the formula.
Example: For sulfuric acid, \(\text{H}_2\text{SO}_4\):
\(M_r = (2 \times 1.0) + (1 \times 32.1) + (4 \times 16.0) = 98.1\)
The Fundamental Mass Formula
To convert between the mass of a solid and the number of moles:
\(\text{Number of moles } (n) = \frac{\text{Mass in grams } (m)}{\text{Molar mass } (M_r)}\)
Quick Memory Triangle: Put \(m\) on top, and \(n \times M_r\) on the bottom.
Key Takeaway: One mole contains \(6.02 \times 10^{23}\) particles and has a mass equal to the relative formula mass (\(M_r\)) in grams.
2. Empirical and Molecular Formulae
What is the Difference?
Empirical Formula: The simplest whole number ratio of atoms of each element present in a compound.
Molecular Formula: The actual number of atoms of each element present in one molecule of a compound.
Example: Glucose has the molecular formula \(\text{C}_6\text{H}_{12}\text{O}_6\). Dividing all numbers by \(6\) gives its empirical formula: \(\text{CH}_2\text{O}\).
Step-by-Step Method: Finding the Empirical Formula
When given experimental data (percentages or masses in grams):
Step 1: Write down the mass or percentage of each element.
Step 2: Divide each mass by its relative atomic mass (\(A_r\)) to find the moles of each element.
Step 3: Divide all mole values by the smallest mole value obtained in Step 2.
Step 4: If you get whole numbers, write down the formula. If you get fractions (like \(.5\) or \(.33\)), multiply all numbers by an integer (e.g., multiply by \(2\) for halves, or by \(3\) for thirds) to achieve whole numbers.
Hydrated Salts and Water of Crystallisation
Some crystalline ionic compounds contain water molecules chemically bound inside their crystal lattice. This is called water of crystallisation (e.g., \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)).
When heated, the water is driven off, leaving behind the anhydrous salt:
\(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\text{ (s)} \rightarrow \text{CuSO}_4\text{ (s)} + x\text{H}_2\text{O}\text{ (g)}\)
To calculate the value of \(x\):
1. Calculate the mass of anhydrous salt and the mass of water lost.
2. Convert both masses into moles (\(n = \frac{m}{M_r}\)).
3. Divide both by the moles of the anhydrous salt to get the mole ratio \(1 : x\).
Key Takeaway: Empirical formula gives the simplest ratio; molecular formula gives the real-world count. For hydrated salts, treat the anhydrous salt and \(\text{H}_2\text{O}\) as two separate units to find their ratio.
3. Working with Solutions and Titrations
Solution Concentrations
Concentration tells us how much solute is dissolved in a given volume of solution.
Units used in chemistry:
- Molar concentration: \(\text{mol dm}^{-3}\)
- Mass concentration: \(\text{g dm}^{-3}\)
Important Volume Conversion: Chemistry volumes are often measured in \(\text{cm}^3\), but concentration uses \(\text{dm}^3\).
\(1\text{ dm}^3 = 1000\text{ cm}^3\)
To convert \(\text{cm}^3\) to \(\text{dm}^3\), divide by \(1000\).
Core Formulas for Solutions
\(n = c \times V\text{ (when } V \text{ is in } \text{dm}^3\text{)}\)
\(n = \frac{c \times V}{1000}\text{ (when } V \text{ is in } \text{cm}^3\text{)}\)
To convert from \(\text{mol dm}^{-3}\) to \(\text{g dm}^{-3}\):
\(\text{Concentration in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times M_r\)
Preparing a Standard Solution
A standard solution is a solution of accurately known concentration.
Procedure Steps:
1. Weigh the solid solute accurately using a balance and weighing boat (weigh by difference).
2. Dissolve the solid in a beaker using a small volume of deionised water.
3. Transfer the solution into a volumetric flask using a funnel.
4. Rinse the beaker, stirring rod, and funnel with deionised water, adding all washings into the flask.
5. Fill the flask with deionised water until the bottom of the meniscus touches the graduation mark at eye level.
6. Invert the stoppered flask several times to ensure thorough mixing.
Volumetric Titrations
Titrations are used to find the concentration of an unknown solution by reacting it with a standard solution.
Good Titration Technique:
- Rinse the burette with the titrant and the pipette with the solution it will measure.
- Swirl the conical flask during addition.
- Add titrant dropwise near the end-point until a distinct colour change occurs.
- Concordant titres are results within \(\pm 0.10\text{ cm}^3\) of each other. Only use concordant results to calculate the mean titre!
Key Takeaway: Always convert volumes to \(\text{dm}^3\) (divide \(\text{cm}^3\) by \(1000\)) before multiplying by concentration.
4. Gas Volumes and the Ideal Gas Equation
1. Molar Gas Volume at Room Temperature and Pressure (RTP)
Avogadro’s Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
At room temperature and pressure (\(20^\circ\text{C}\) and \(101\text{ kPa}\)), one mole of any gas occupies approximately \(24.0\text{ dm}^3\) (or \(24000\text{ cm}^3\)).
\(\text{Moles of gas } (n) = \frac{\text{Volume in dm}^3}{24.0} = \frac{\text{Volume in cm}^3}{24000}\)
2. The Ideal Gas Equation
When conditions are not at standard room temperature and pressure, we use the Ideal Gas Equation:
\(pV = nRT\)
Variables and Crucial Units:
- \(p = \text{Pressure in Pascals (Pa)}\) (If given in \(\text{kPa}\), multiply by \(10^3\))
- \(V = \text{Volume in cubic metres (m}^3\text{)}\) (To convert \(\text{cm}^3 \rightarrow \text{m}^3\), multiply by \(10^{-6}\); to convert \(\text{dm}^3 \rightarrow \text{m}^3\), multiply by \(10^{-3}\))
- \(n = \text{Amount of substance in moles (mol)}\)
- \(R = \text{Molar gas constant } = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)
- \(T = \text{Temperature in Kelvin (K)}\) (To convert \(^\circ\text{C} \rightarrow \text{K}\), add \(273.15\), or \(+ 273\))
Common Mistake to Avoid: Forgetting to convert units is the most frequent error! Always double-check that pressure is in \(\text{Pa}\), volume is in \(\text{m}^3\), and temperature is in \(\text{K}\).
Key Takeaway: Use \(V / 24\) at RTP. Use \(pV = nRT\) whenever temperature or pressure vary from standard room conditions, taking care of unit conversions.
5. Reaction Yield and Atom Economy
Percentage Yield
The theoretical yield is the maximum mass of product that could be formed, calculated using the balanced equation. The actual yield is the mass of product actually collected in the experiment.
\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)
Why is yield never \(100\%\) in practice?
- Reaction may be reversible and reach equilibrium.
- Side reactions may produce unexpected by-products.
- Product is lost during purification processes (e.g., sticking to glassware, remaining in solution during filtration).
Percentage Atom Economy
Atom economy is a measure of how efficiently the atoms in your starting materials are converted into the desired useful product.
\(\text{Atom Economy} = \frac{\text{Mass of Desired Product}}{\text{Total Mass of All Products}} \times 100\%\)
(Using balanced equation molar masses: \(\frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\%\))
Did you know? Addition reactions always have an atom economy of \(100\%\) because only one product is formed and no atoms are wasted!
Yield vs Atom Economy:
- High Percentage Yield: You carried out the practical technique effectively without losing material.
- High Atom Economy: The chemical process is fundamentally green and produces minimal waste.
Key Takeaway: Yield measures practical efficiency; atom economy measures theoretical sustainability and green chemistry potential.
6. Experimental Uncertainties (Errors)
Every measuring instrument has an inherent limit of precision. The uncertainty of a digital instrument is usually \(\pm\) the last decimal place, while for an analogue scale it is typically \(\pm\) half the smallest graduation mark.
Calculating Percentage Uncertainty
\(\text{Percentage Uncertainty} = \frac{\text{Total Uncertainty in Measurement}}{\text{Measured Value}} \times 100\%\)
Note for Burettes and Balances: Because a burette reading involves two measurements (initial reading and final reading), the total uncertainty is multiplied by \(2\):
\(\text{Uncertainty} = 2 \times (\pm 0.05\text{ cm}^3) = \pm 0.10\text{ cm}^3\)
How to Minimise Percentage Uncertainty
1. Increase the measured value: Using a larger mass of solid or a larger volume/titre reduces the overall percentage error.
2. Use more precise apparatus: Use a balance measuring to \(3\) decimal places instead of \(2\), or a volumetric pipette instead of a measuring cylinder.
Summary Checklist: The Calculation Toolkit
Before sitting your exam, make sure you can confidently use these core relationships:
1. Solids / Masses: \(n = \frac{m}{M_r}\)
2. Solutions: \(n = \frac{c \times V}{1000}\) (where \(V\) is in \(\text{cm}^3\))
3. Gases at RTP: \(n = \frac{V}{24\text{ dm}^3}\)
4. Gases at Any Condition: \(pV = nRT\)
5. Stoichiometric Ratio: Convert mass/volume \(\rightarrow\) moles \(\rightarrow\) use molar ratio from balanced equation \(\rightarrow\) convert back to required quantity.