Welcome to Quantitative Chemistry
Welcome to Quantitative Chemistry! The word quantitative simply means measuring quantities—in other words, doing chemical calculations. Chemistry is not just about what happens when substances react; it is also about how much of each substance is involved.
Don't worry if maths in science feels a little scary at first. We will break every single calculation down into simple, repeatable steps. By mastering these methods, you can secure valuable marks in your CCEA GCSE Double Award Science Unit C2 exam!
1. Hydrated Salts & Water of Crystallisation
Key Definitions to Learn
- Hydrated substance: A solid chemical compound that contains water molecules chemically bound inside its crystal structure.
- Anhydrous substance: A substance that does not contain any water of crystallisation (often left behind after heating).
- Water of crystallisation: Water chemically bonded into the crystal lattice of a salt. We write this using a dot formula, such as \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\) or \(\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}\).
- Heating to constant mass: Repeatedly heating a sample, allowing it to cool in a desiccator, and reweighing until consecutive balance readings are identical (\(\pm 0.01\text{ g}\)). This proves that all water has been driven off.
Understanding the Dot Formula
Take hydrated copper(II) sulfate: \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\).
The dot does not mean multiply the masses together mathematically in the regular arithmetic way! It means that for every 1 unit of \(\text{CuSO}_4\), there are 5 molecules of \(\text{H}_2\text{O}\) trapped inside the crystal lattice.
Calculating Relative Formula Mass (\(M_r\)) of a Hydrated Salt
To find the total \(M_r\):
\(\text{Total } M_r = M_r(\text{anhydrous salt}) + [x \times M_r(\text{H}_2\text{O})]\)
Remember: \(M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18\).
Worked Example: Find the \(M_r\) of \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\).
(Given: \(A_r(\text{Cu}) = 64\), \(A_r(\text{S}) = 32\), \(A_r(\text{O}) = 16\), \(A_r(\text{H}) = 1\))
- Calculate \(M_r(\text{CuSO}_4) = 64 + 32 + (4 \times 16) = 64 + 32 + 64 = 160\)
- Calculate mass of water: \(5 \times M_r(\text{H}_2\text{O}) = 5 \times 18 = 90\)
- Add them together: \(\text{Total } M_r = 160 + 90 = 250\)
Finding the Degree of Hydration (\(x\)) from Experimental Data
When you heat a hydrated salt, the water evaporates into the air, leaving behind the anhydrous salt residue.
Step-by-Step Method:
- Find mass of water lost:
\(\text{Mass of water lost} = \text{Mass of hydrated salt} - \text{Mass of anhydrous salt remaining}\) - Find moles of anhydrous salt:
\(\text{Moles of anhydrous salt} = \frac{\text{Mass of anhydrous residue}}{M_r(\text{anhydrous salt})}\) - Find moles of water:
\(\text{Moles of water} = \frac{\text{Mass of water lost}}{18}\) - Find the whole number ratio (\(x\)):
\(x = \frac{\text{Moles of water}}{\text{Moles of anhydrous salt}}\)
Quick Review Box:
Common Trap: In the exam, remember to explain heating to constant mass properly: heat, cool in a desiccator, re-weigh, and repeat until the mass stops changing!
Key Takeaway: Water of crystallisation is chemically bound in crystals. Heating drives it off, and comparing the moles of anhydrous salt to the moles of water gives the formula ratio \(x\).
2. Empirical and Molecular Formulae
Definitions
- 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 in one molecule of a compound.
Analogy: Think of the molecular formula as a full recipe (e.g., \(\text{C}_6\text{H}_{12}\text{O}_6\) for glucose), while the empirical formula is the simplified ratio (\(\text{CH}_2\text{O}\)).
Key Takeaway: The empirical formula shows the lowest whole-number ratio; the molecular formula shows the real count of atoms in each molecule.
3. Solution Concentrations and Volumes
Units of Volume
In chemistry, solution volume is standardly measured in cubic decimetres (\(\text{dm}^3\)), where \(1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ litre}\).
Golden Rule: Always convert \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by 1000!
\(\text{Volume in dm}^3 = \frac{\text{Volume in cm}^3}{1000}\)
Core Concentration Formulae
Concentration tells us how much solute is dissolved in a specific volume of solution. It can be expressed in two ways:
- Concentration in \(\text{mol/dm}^3\):
\(\text{Moles} = \text{Concentration } (\text{mol/dm}^3) \times \text{Volume } (\text{dm}^3)\)
\(\text{Concentration } (\text{mol/dm}^3) = \frac{\text{Moles}}{\text{Volume } (\text{dm}^3)}\)
If volume is given in \(\text{cm}^3\):
\(\text{Moles} = \frac{\text{Concentration } (\text{mol/dm}^3) \times \text{Volume } (\text{cm}^3)}{1000}\) - Concentration in \(\text{g/dm}^3\):
\(\text{Concentration } (\text{g/dm}^3) = \frac{\text{Mass of solute } (\text{g})}{\text{Volume } (\text{dm}^3)}\)
Converting Between \(\text{mol/dm}^3\) and \(\text{g/dm}^3\)
You can quickly convert between both concentration units using the compound's relative formula mass (\(M_r\)):
\(\text{Concentration } (\text{g/dm}^3) = \text{Concentration } (\text{mol/dm}^3) \times M_r\)
\(\text{Concentration } (\text{mol/dm}^3) = \frac{\text{Concentration } (\text{g/dm}^3)}{M_r}\)
Worked Example:
A student dissolves \(0.05\text{ moles}\) of sodium hydroxide (\(\text{NaOH}\)) in \(250\text{ cm}^3\) of water. Calculate the concentration in \(\text{mol/dm}^3\) and in \(\text{g/dm}^3\). (\(M_r(\text{NaOH}) = 40\))
- Convert volume: \(250\text{ cm}^3 = \frac{250}{1000} = 0.25\text{ dm}^3\)
- Calculate concentration in \(\text{mol/dm}^3\):
\(\text{Concentration} = \frac{0.05\text{ moles}}{0.25\text{ dm}^3} = 0.20\text{ mol/dm}^3\) - Convert to \(\text{g/dm}^3\):
\(\text{Concentration in g/dm}^3 = 0.20 \times 40 = 8.0\text{ g/dm}^3\)
Key Takeaway: Always check the units for volume in the question. If you see \(\text{cm}^3\), divide by 1000 to get \(\text{dm}^3\) before calculating!
4. Atom Economy and Green Chemistry
What is Atom Economy?
Atom economy is a measure of the proportion of starting reactants that end up as useful, desired products in a balanced chemical equation. It is a key idea in sustainable development and "green chemistry".
Formula for Percentage Atom Economy
\(\text{Percentage Atom Economy} = \frac{\text{Mass (or } M_r\text{) of desired product}}{\text{Total mass (or total } M_r\text{) of all products}} \times 100\)
Note: Due to the conservation of mass, the total mass of all products equals the total mass of all reactants, so you can use either for the denominator!
Why is High Atom Economy Important?
- Reduces waste: Less unwanted by-product is produced and dumped into the environment.
- Saves resources: Maximises the efficiency of precious starting materials.
- Economic benefits: Lowers costs associated with separating, treating, and disposing of waste products.
Did You Know? Reactions with only one single product (such as addition reactions) always have an atom economy of \(100\%\) because every single starting atom ends up in the product!
Important Distinction: Atom Economy vs. Percentage Yield
- Atom economy: A theoretical value calculated from the balanced chemical equation. It tells you how efficient the reaction design is on paper.
- Percentage yield: A practical value calculated from real lab experiments. It compares the actual mass of product obtained to the maximum theoretical mass possible.
Key Takeaway: High atom economy means less chemical waste and greener processes. Remember to include balancing numbers when calculating the total \(M_r\) of products!
Chapter Summary & Exam Checklist
- Water of Crystallisation: Calculate \(M_r\) by adding the anhydrous mass to the mass of water (\(x \times 18\)).
- Constant Mass: Heat, cool in a desiccator, reweigh, and repeat until the mass is identical (\(\pm 0.01\text{ g}\)).
- Solutions: Always divide \(\text{cm}^3\) by 1000 to work in \(\text{dm}^3\).
- Concentration link: \(\text{Concentration (g/dm}^3\text{)} = \text{Concentration (mol/dm}^3\text{)} \times M_r\).
- Atom Economy: Divide the \(M_r\) of the desired product by the total \(M_r\) of all products and multiply by 100.