Quantitative Chemistry and Analysis

What you'll learn

This revision guide covers the quantitative aspects of chemistry that appear throughout your Edexcel GCSE Chemistry examination. You'll master calculations involving relative formula mass, moles, concentration, yields, and titrations. These mathematical skills are essential for securing marks across multiple papers and form the foundation for A-Level Chemistry.

Key terms and definitions

Relative formula mass (Mr) — the sum of the relative atomic masses of all atoms in a formula

Mole — the unit for amount of substance; one mole contains 6.02 × 10²³ particles (Avogadro's constant)

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

Concentration — the amount of solute dissolved in a given volume of solution, measured in g/dm³ or mol/dm³

Percentage yield — the actual mass of product obtained divided by the theoretical maximum mass, expressed as a percentage

Titration — a quantitative technique used to determine the precise volume of one solution that reacts with a known volume of another solution

Limiting reactant — the reactant that is completely used up in a reaction, determining the maximum amount of product formed

Core concepts

Relative formula mass and the mole

The relative formula mass (Mr) is calculated by adding the relative atomic masses (Ar) of all atoms in a chemical formula. Find Ar values on the periodic table.

For example, water (H₂O):

• 2 × H = 2 × 1 = 2
• 1 × O = 1 × 16 = 16
• Mr(H₂O) = 18

The mole links mass to number of particles. One mole of any substance has a mass in grams equal to its relative formula mass.

Key equation: Number of moles = mass (g) ÷ Mr

For ionic compounds, use the same method. Calcium chloride (CaCl₂):

• 1 × Ca = 1 × 40 = 40
• 2 × Cl = 2 × 35.5 = 71
• Mr(CaCl₂) = 111

This means 111 g of calcium chloride contains exactly 1 mole.

Reacting masses and balancing equations

Balanced symbol equations show the molar ratios of reactants and products. Use these ratios to calculate reacting masses.

Method:

1. Write the balanced equation
2. Calculate Mr for relevant substances
3. Use the molar ratio from the equation
4. Scale to the actual mass given

Example: Calculate the mass of magnesium oxide formed when 12 g of magnesium burns completely.

2Mg + O₂ → 2MgO

From the equation: 2 moles of Mg produces 2 moles of MgO (1:1 ratio)

• Mr(Mg) = 24
• Mr(MgO) = 40

If 24 g Mg produces 40 g MgO, then 12 g Mg produces 20 g MgO.

Empirical and molecular formulas

The empirical formula represents the simplest ratio. The molecular formula shows the actual number of atoms in a molecule.

To find empirical formula from mass or percentage data:

1. List the mass (or percentage) of each element
2. Divide each by the Ar of that element to get moles
3. Divide all answers by the smallest number of moles
4. If necessary, multiply to get whole numbers

Example: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

Element	Mass	Ar	Moles	Divide by smallest	Ratio
C	40	12	3.33	3.33 ÷ 3.33	1
H	6.7	1	6.7	6.7 ÷ 3.33	2
O	53.3	16	3.33	3.33 ÷ 3.33	1

Empirical formula = CH₂O

If the Mr of the compound is 180, find the molecular formula:

• Mr of CH₂O = 12 + 2 + 16 = 30
• 180 ÷ 30 = 6
• Molecular formula = C₆H₁₂O₆

Concentration of solutions

Concentration measures how much solute dissolves in a solvent. It can be expressed in grams per cubic decimetre (g/dm³) or moles per cubic decimetre (mol/dm³).

Key equations:

Concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)

Concentration (mol/dm³) = number of moles ÷ volume (dm³)

Remember: 1 dm³ = 1000 cm³

To convert cm³ to dm³: divide by 1000

Example conversions:

• 250 cm³ = 0.25 dm³
• 50 cm³ = 0.05 dm³
• 2000 cm³ = 2 dm³

When diluting solutions, the number of moles of solute remains constant, but concentration changes as volume increases.

Percentage yield and atom economy

In practical chemistry, reactions rarely produce the maximum theoretical amount of product.

Percentage yield = (actual yield ÷ theoretical yield) × 100%

Reasons for yields below 100%:

• Incomplete reactions
• Side reactions producing unwanted products
• Loss during transfer or filtration
• Reversible reactions reaching equilibrium

Atom economy measures the efficiency of reactions in terms of atoms used:

Atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100%

Higher atom economy means more sustainable chemistry with less waste. Addition reactions have 100% atom economy because all reactants form products.

Titration calculations

Titrations determine unknown concentrations by reacting known volumes of solutions. A burette delivers the titrant, and an indicator shows the end point when the reaction is complete.

Standard procedure:

1. Measure a known volume of solution into a conical flask using a pipette
2. Add a suitable indicator (e.g., phenolphthalein or methyl orange)
3. Add titrant from a burette until the indicator changes colour
4. Record the volume added (titre)
5. Repeat until concordant results obtained (within 0.1 cm³)

Calculation method:

1. Calculate moles of the known substance using n = CV (where C is in mol/dm³ and V in dm³)
2. Use the balanced equation to find moles of unknown
3. Calculate concentration or volume required

Concordant titres are those within 0.1 cm³ of each other. Calculate a mean using only concordant values, ignoring rough titres and outliers.

Gas volumes and molar volume

At room temperature and pressure (RTP: 20°C and 1 atmosphere), one mole of any gas occupies approximately 24 dm³. This is the molar volume.

Volume of gas (dm³) = moles of gas × 24

This allows calculation of gas volumes in reactions.

Example: What volume of carbon dioxide forms when 0.5 moles of calcium carbonate reacts with excess acid?

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

From equation: 1 mole CaCO₃ produces 1 mole CO₂

0.5 moles CaCO₃ produces 0.5 moles CO₂

Volume of CO₂ = 0.5 × 24 = 12 dm³

Limiting reactants

When reactants are not in the exact stoichiometric ratio, one runs out first. This limiting reactant determines how much product forms.

To identify the limiting reactant:

1. Calculate moles of each reactant
2. Use the balanced equation to determine the ratio required
3. Compare actual moles to required ratio
4. The reactant in short supply is limiting

The other reactant is in excess. Some remains unreacted after the reaction stops.

Worked examples

Example 1: Mole calculations and reacting masses

Question: Sulfuric acid reacts with sodium hydroxide. Calculate the mass of sodium sulfate formed when 49 g of sulfuric acid reacts with excess sodium hydroxide. (Ar: H = 1, O = 16, S = 32, Na = 23)

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Solution:

Step 1: Calculate Mr of H₂SO₄ Mr = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98

Step 2: Calculate moles of H₂SO₄ Moles = mass ÷ Mr = 49 ÷ 98 = 0.5 moles

Step 3: Use equation ratio From equation: 1 mole H₂SO₄ produces 1 mole Na₂SO₄ Therefore: 0.5 moles H₂SO₄ produces 0.5 moles Na₂SO₄

Step 4: Calculate Mr of Na₂SO₄ Mr = (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142

Step 5: Calculate mass Mass = moles × Mr = 0.5 × 142 = 71 g (3 marks)

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Example 2: Concentration and titration

Question: A student titrated 25.0 cm³ of sulfuric acid with sodium hydroxide of concentration 0.10 mol/dm³. The mean titre was 20.0 cm³. Calculate the concentration of the sulfuric acid in mol/dm³.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Solution:

Step 1: Calculate moles of NaOH Volume = 20.0 cm³ = 20.0 ÷ 1000 = 0.020 dm³ Moles = concentration × volume = 0.10 × 0.020 = 0.002 moles

Step 2: Use equation ratio From equation: 1 mole H₂SO₄ reacts with 2 moles NaOH Moles of H₂SO₄ = 0.002 ÷ 2 = 0.001 moles

Step 3: Calculate concentration of H₂SO₄ Volume of H₂SO₄ = 25.0 cm³ = 0.025 dm³ Concentration = moles ÷ volume = 0.001 ÷ 0.025 = 0.04 mol/dm³ (4 marks)

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Example 3: Percentage yield

Question: A student heated 15.9 g of copper carbonate. The equation for the reaction is:

CuCO₃ → CuO + CO₂

The student obtained 6.4 g of copper oxide. Calculate the percentage yield. (Ar: C = 12, O = 16, Cu = 64)

Solution:

Step 1: Calculate moles of CuCO₃ Mr(CuCO₃) = 64 + 12 + (3 × 16) = 124 Moles = 15.9 ÷ 124 = 0.128 moles (allow 0.13)

Step 2: Find theoretical moles of CuO From equation: 1 mole CuCO₃ produces 1 mole CuO Theoretical moles of CuO = 0.128 moles

Step 3: Calculate theoretical mass of CuO Mr(CuO) = 64 + 16 = 80 Theoretical mass = 0.128 × 80 = 10.24 g (allow 10.4 g)

Step 4: Calculate percentage yield Percentage yield = (6.4 ÷ 10.24) × 100 = 62.5% (allow 61.5% if rounded earlier) (4 marks)

Common mistakes and how to avoid them

• Forgetting to convert cm³ to dm³ — always divide by 1000 when using the equation n = CV. Write the conversion step clearly to avoid errors.

• Not balancing equations before calculating reacting masses — the coefficients in balanced equations give molar ratios. Check your equation is balanced first.

• Confusing empirical and molecular formulas — empirical shows the simplest ratio; molecular shows actual numbers. The molecular formula is always a whole number multiple of the empirical formula.

• Using the wrong units for concentration — g/dm³ uses mass directly; mol/dm³ requires you to calculate moles first using n = m ÷ Mr.

• Including rough titres in mean calculations — only average concordant results. Rough titres help you find the approximate end point but should not be included in calculations.

• Mixing up actual and theoretical yield — actual yield is what you obtain practically; theoretical yield is the maximum calculated from the equation. Percentage yield = (actual ÷ theoretical) × 100.

Exam technique for "Quantitative Chemistry and Analysis"

• Show all working clearly — in calculation questions worth 3+ marks, examiners award method marks even if your final answer is wrong. Write each step on a new line.

• Include units in your final answer — marks are often lost for missing units. Common units: g, mol, dm³, cm³, g/dm³, mol/dm³, %.

• Use the correct number of significant figures — generally give answers to 2 or 3 significant figures unless told otherwise. Match the data given in the question.

• Command words matter — "Calculate" requires numerical working and an answer; "Determine" may allow alternative methods; "Show that" means you must arrive at the given answer by clear steps.

Quick revision summary

Quantitative chemistry uses mathematical relationships to predict and analyse chemical reactions. Master calculating Mr and moles using n = m ÷ Mr. Use balanced equations to find reacting mass ratios. Derive empirical formulas from percentage composition by converting to moles and simplifying ratios. Calculate solution concentration using C = n ÷ V, essential for titration calculations. Evaluate reaction efficiency through percentage yield and atom economy. At RTP, one mole of gas occupies 24 dm³. Always show working, include units, and convert cm³ to dm³ where necessary.