Quantitative chemistry

Quantitative Chemistry Study Notes (CCEA GCSE)

Welcome to Quantitative Chemistry! Many students find the math-heavy side of chemistry intimidating, but don't worry if this seems tricky at first. Think of quantitative chemistry as simply following a recipe in baking: if you know the right ingredients and the steps to measure them, you can solve any calculation with confidence!

1. Relative Formula Mass (\(M_r\)) and Relative Atomic Mass (\(A_r\))

Before we can count or weigh particles, we need to know how heavy atoms and molecules are compared to each other.

Relative Atomic Mass (\(A_r\)): The average mass of an atom of an element compared to \(\frac{1}{12}\)th the mass of a carbon-12 atom. You can find this on your Periodic Table as the larger number beside each element symbol.

Relative Formula Mass (\(M_r\)): The sum of the relative atomic masses of all the atoms present in a chemical formula.

How to Calculate \(M_r\) Step-by-Step:

1. List each element present in the compound.
2. Count the number of atoms of each element (watch the small subscript numbers and brackets!).
3. Multiply the count by the element's \(A_r\).
4. Add all the totals together.

Worked Example: Find the \(M_r\) of calcium hydroxide, \(\text{Ca(OH)}_2\).
Given \(A_r\): \(\text{Ca} = 40\), \(\text{O} = 16\), \(\text{H} = 1\)
- Calcium (\(\text{Ca}\)): \(1 \times 40 = 40\)
- Oxygen (\(\text{O}\)): \(2 \times 16 = 32\)
- Hydrogen (\(\text{H}\)): \(2 \times 1 = 2\)
- Total \(M_r = 40 + 32 + 2 = 74\)

Key Takeaway: Always look inside brackets! The subscript outside a bracket multiplies everything inside (e.g., in \(\text{Ca(OH)}_2\), there are \(2\) oxygens and \(2\) hydrogens).

2. The Mole Concept

Atoms and molecules are too tiny to weigh individually. Instead, chemists bundle them into large groups called moles.

Analogy: Just like a "dozen" always means \(12\) items (whether eggs or donuts), a mole always contains \(6.02 \times 10^{23}\) particles. This huge number is called Avogadro's constant (\(L\)).

Did you know? One mole of any substance has a mass in grams exactly equal to its relative formula mass (\(M_r\)). For example, \(1\text{ mole}\) of water (\(\text{H}_2\text{O}\), \(M_r = 18\)) weighs exactly \(18\text{ g}\)!

The Golden Mole Formula:

\(\text{Moles} = \frac{\text{Mass (in g)}}{M_r}\)

You can rearrange this triangle in three ways:
- \(\text{Mass} = \text{Moles} \times M_r\)
- \(\text{Moles} = \frac{\text{Mass}}{M_r}\)
- \(M_r = \frac{\text{Mass}}{\text{Moles}}\)

Worked Example: How many moles are in \(22\text{ g}\) of carbon dioxide (\(\text{CO}_2\))?
- \(M_r \text{ of } \text{CO}_2 = 12 + (2 \times 16) = 44\)
- \(\text{Moles} = \frac{22}{44} = 0.5\text{ mol}\)

Key Takeaway: Mass must always be in grams (g). If a question gives you kilograms (\(\text{kg}\)), multiply by \(1000\) first!

3. Percentage Composition by Mass

This calculation tells you what percentage of a compound's total mass comes from a specific element.

Formula:
\(\% \text{ of an Element} = \frac{(\text{Number of atoms of the element} \times A_r)}{M_r \text{ of compound}} \times 100\)

Worked Example: Calculate the percentage by mass of nitrogen in ammonium nitrate, \(\text{NH}_4\text{NO}_3\).
Given \(A_r\): \(\text{N} = 14\), \(\text{H} = 1\), \(\text{O} = 16\)
- Total \(M_r = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80\)
- Mass of nitrogen present \(= 2 \times 14 = 28\)
- \(\% \text{ Nitrogen} = \frac{28}{80} \times 100 = 35\%\)

Key Takeaway: Make sure you count all atoms of that element across the whole formula (in \(\text{NH}_4\text{NO}_3\), nitrogen appears twice!).

4. Empirical and Molecular Formulae

Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.
Molecular Formula: The actual number of atoms of each element in one molecule of a compound.

Finding the Empirical Formula:

Use the simple 4-step table method (M-A-D-R: Mass \(\rightarrow\) \(A_r\) \(\rightarrow\) Divide \(\rightarrow\) Ratio):

1. Mass / Percentage: Write down the mass or \(\%\) for each element.
2. Divide by \(A_r\): Calculate the moles of each element (\(\frac{\text{mass}}{A_r}\)).
3. Divide by Smallest: Divide all mole numbers by the smallest mole value obtained.
4. Ratio: Write the whole-number ratio.

Worked Example: A compound contains \(4.8\text{ g}\) of magnesium and \(3.2\text{ g}\) of oxygen. Find its empirical formula.
- Step 1: \(\text{Mg} = 4.8\text{ g}\), \(\text{O} = 3.2\text{ g}\)
- Step 2 (Moles): \(\text{Mg} = \frac{4.8}{24} = 0.2\text{ mol}\); \(\text{O} = \frac{3.2}{16} = 0.2\text{ mol}\)
- Step 3 (Divide by smallest): \(\text{Mg} = \frac{0.2}{0.2} = 1\); \(\text{O} = \frac{0.2}{0.2} = 1\)
- Step 4: Ratio is \(1 : 1\), so the empirical formula is \(\text{MgO}\).

Finding Molecular Formula from Empirical Formula:

If you know the empirical formula and the true \(M_r\):
\(\text{Scaling Factor} = \frac{\text{Actual } M_r}{\text{Empirical formula mass}}\)
Then multiply each subscript in the empirical formula by this number.

5. Water of Crystallisation

Hydrated salt: A solid crystal that contains water chemically bonded into its crystalline structure (e.g., \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)).
Anhydrous salt: A substance containing no water of crystallisation (e.g., dry \(\text{CuSO}_4\)).
Water of crystallisation: The fixed number of water molecules bound into the crystal structure.

Finding the Value of \(x\) in \(\text{Salt} \cdot x\text{H}_2\text{O}\):

1. Calculate mass of anhydrous salt remaining after heating to constant mass.
2. Calculate mass of water lost: \(\text{Mass of water} = \text{Mass of hydrated salt} - \text{Mass of anhydrous salt}\).
3. Find moles of anhydrous salt (\(\frac{\text{mass}}{M_r \text{ of salt}}\)).
4. Find moles of water lost (\(\frac{\text{mass}}{18}\)).
5. Find the mole ratio: \(x = \frac{\text{moles of water}}{\text{moles of anhydrous salt}}\).

Key Takeaway: "Heating to constant mass" means heating, cooling, weighing, and repeating until the mass no longer changes, ensuring all water has completely evaporated.

6. Reacting Masses (Stoichiometry)

Balanced chemical equations tell us the exact ratio of moles reacting together.

The 3-Step Reacting Mass Method:

1. Moles of known: Find the moles of the substance whose mass you are given (\(\text{Moles} = \frac{\text{Mass}}{M_r}\)).
2. Mole Ratio: Use the big numbers in the balanced equation to find the moles of the unknown substance.
3. Mass of unknown: Convert moles back into mass (\(\text{Mass} = \text{Moles} \times M_r\)).

Worked Example: What mass of magnesium oxide (\(\text{MgO}\)) is produced when \(12\text{ g}\) of magnesium reacts fully with oxygen?
Equation: \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\)
- Step 1 (Moles of \(\text{Mg}\)): \(\text{Moles} = \frac{12}{24} = 0.5\text{ mol}\)
- Step 2 (Mole Ratio): Ratio of \(\text{Mg} : \text{MgO}\) is \(2 : 2\) (or \(1 : 1\)). Therefore, moles of \(\text{MgO} = 0.5\text{ mol}\).
- Step 3 (Mass of \(\text{MgO}\)): \(M_r \text{ of } \text{MgO} = 24 + 16 = 40\).
\(\text{Mass} = 0.5 \times 40 = 20\text{ g}\).

Limiting Reactants: The reactant that is completely used up first is the limiting reactant. It determines the maximum amount of product formed. The other reactant is in excess.

7. Percentage Yield and Atom Economy

Percentage Yield:

Theoretical yield is the maximum mass predicted by calculation. Actual yield is the real mass collected in an experiment.

\(\% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)

Why is percentage yield rarely \(100\%\)?
- The reaction may be reversible and not reach completion.
- Some product is lost during practical steps (e.g., sticking to glassware, left on filter paper).
- Unexpected side reactions occurred forming unwanted products.

Percentage Atom Economy:

Atom economy measures how many of the reactant atoms end up as useful desired products.

\(\% \text{ Atom Economy} = \frac{\text{Mass (or } M_r\text{) of Desired Product}}{\text{Total Mass (or } M_r\text{) of All Products}} \times 100\)

Why is high atom economy important in industry?
- Minimises chemical waste.
- Preserves raw materials.
- Makes production more cost-effective and environmentally friendly (green chemistry).

8. Gas Volumes

At room temperature and pressure (RTP: \(20^\circ\text{C}\) and \(1\text{ atmosphere}\)), one mole of any gas occupies a volume of \(24\text{ dm}^3\) (or \(24\,000\text{ cm}^3\)). This is known as the molar gas volume.

Gas Volume Formulas:

- \(\text{Moles} = \frac{\text{Volume in dm}^3}{24}\)
- \(\text{Moles} = \frac{\text{Volume in cm}^3}{24\,000}\)

Unit Conversion Reminder:
\(1\text{ dm}^3 = 1000\text{ cm}^3\)
- To convert \(\text{cm}^3 \rightarrow \text{dm}^3\), divide by \(1000\).
- To convert \(\text{dm}^3 \rightarrow \text{cm}^3\), multiply by \(1000\).

Worked Example: What volume does \(0.25\text{ moles}\) of oxygen gas occupy at RTP in \(\text{cm}^3\)?
\(\text{Volume} = 0.25 \times 24\,000 = 6000\text{ cm}^3\)

9. Solution Concentrations and Titrations

Concentration tells us how much solute is dissolved in a certain volume of solution.

Key Concentration Formulas:

1. \(\text{Moles} = \text{Concentration (mol/dm}^3\text{)} \times \text{Volume (dm}^3\text{)}\)
2. \(\text{Concentration (mol/dm}^3\text{)} = \frac{\text{Moles}}{\text{Volume (dm}^3\text{)}}\)
3. \(\text{Concentration (g/dm}^3\text{)} = \text{Concentration (mol/dm}^3\text{)} \times M_r\)

Titration Calculations Step-by-Step:

1. Calculate moles of standard solution whose concentration and volume are both known: \(\text{Moles} = \text{Concentration} \times \text{Volume (dm}^3\text{)}\).
2. Use the balanced equation to deduce the mole ratio between the acid and alkali.
3. Determine the unknown concentration: \(\text{Concentration} = \frac{\text{Moles}}{\text{Volume (dm}^3\text{)}}\).

Worked Example: In a titration, \(25.0\text{ cm}^3\) of \(0.10\text{ mol/dm}^3\) \(\text{NaOH}\) is neutralised by \(20.0\text{ cm}^3\) of \(\text{HCl}\). Calculate the concentration of the \(\text{HCl}\).
Equation: \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)
- Step 1 (Moles of \(\text{NaOH}\)):
Volume in \(\text{dm}^3 = \frac{25.0}{1000} = 0.025\text{ dm}^3\)
\(\text{Moles of NaOH} = 0.10 \times 0.025 = 0.0025\text{ mol}\)
- Step 2 (Mole Ratio): Ratio of \(\text{NaOH} : \text{HCl}\) is \(1 : 1\). Therefore, moles of \(\text{HCl} = 0.0025\text{ mol}\).
- Step 3 (Concentration of \(\text{HCl}\)):
Volume of \(\text{HCl} = \frac{20.0}{1000} = 0.020\text{ dm}^3\)
\(\text{Concentration} = \frac{0.0025}{0.020} = 0.125\text{ mol/dm}^3\)

Key Takeaway: Always ensure titration volumes in \(\text{cm}^3\) are divided by \(1000\) to turn them into \(\text{dm}^3\) before using concentration formulas!

Quick Formula Summary Box

- \(\text{Moles} = \frac{\text{Mass}}{M_r}\)
- \(\text{Moles of gas} = \frac{\text{Volume in dm}^3}{24}\)
- \(\text{Moles in solution} = \text{Concentration (mol/dm}^3\text{)} \times \text{Volume (dm}^3\text{)}\)
- \(\text{Concentration (g/dm}^3\text{)} = \text{Concentration (mol/dm}^3\text{)} \times M_r\)
- \(\% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)
- \(\% \text{ Atom Economy} = \frac{M_r \text{ of desired product}}{\text{Total } M_r \text{ of all products}} \times 100\)