Atoms, Molecules and Stoichiometry — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Mole calculations with Avogadro’s constant, empirical and molecular formula determination, atomic and charge balancing of chemical equations, limiting reagent and percentage yield calculations, and core concentration units including molarity, ppm, and percent composition.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Atoms, Molecules and Stoichiometry?

Stoichiometry is the quantitative study of the relative amounts of reactants and products involved in chemical reactions, rooted in the conserved behavior of individual atoms and molecules during chemical change. It is the foundational calculation framework for every other unit in the A-Level Chemistry syllabus, from energetics to organic reaction pathways, and accounts for 15-20% of total marks across Papers 1, 2, and 4. Common synonyms include "chemical arithmetic" or "quantitative chemistry", and examiners frequently embed stoichiometry calculations in higher-mark extended response questions to test applied knowledge.

2. Mole concept — Avogadro's number

The mole (symbol: mol) is the SI unit for amount of substance, defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, electrons) as there are atoms in exactly 12 g of pure carbon-12 (${}^{12}\text{C}$). This fixed number is called Avogadro’s constant, denoted $N_A=6.02\times 10^{23}{\text{ mol}}^{-1}$. We use moles because individual atoms are far too small to count directly, so the unit scales subatomic particle counts to measurable mass values for lab use.

Key core formulas for this subtopic: $$n=\frac{m}{M_r}\text{and}N=n\times N_A$$ Where $n$ = amount of substance in mol, $m$ = mass in g, $M_r$ = relative atomic/molecular mass in ${\text{g mol}}^{-1}$, and $N$ = total number of elementary entities.

Worked Example: Calculate the total number of sulfate ions in 14.2 g of sodium sulfate (${\text{Na}}_2{\text{SO}}_4$).

1. Calculate $M_r$ of ${\text{Na}}_2{\text{SO}}_4$: $(2\times 23)+32+(4\times 16)=142{\text{ g mol}}^{-1}$
2. Moles of ${\text{Na}}_2{\text{SO}}_4$: $n=\frac{14.2}{142}=0.10\text{ mol}$
3. Each formula unit of ${\text{Na}}_2{\text{SO}}_4$ contains 1 sulfate ion, so moles of ${\text{SO}}_4^{2-}$ = 0.10 mol
4. Total number of sulfate ions: $N=0.10\times 6.02\times 10^{23}=6.02\times 10^{22}$

Exam tip: Always specify the type of entity you are counting (atoms, molecules, ions) in your working to avoid losing marks for ambiguous answers.

3. Empirical and molecular formulas

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element present in the compound. The molecular formula is the actual number of atoms of each element present in one molecule of the compound, which is always a whole-number multiple of the empirical formula.

Step-by-step calculation method:

1. Record the mass or percentage mass of each element in the compound (assume a 100 g sample for percentage values to simplify calculations)
2. Divide each mass value by the relative atomic mass of the corresponding element to get moles
3. Divide all mole values by the smallest mole value to get a raw ratio
4. Round to the nearest whole number if the ratio is within 0.1 of an integer; if you get values like 1.5, 1.33 or 1.25, multiply all ratios by 2, 3 or 4 respectively to eliminate decimals
5. To find the molecular formula, calculate the multiplier $n=\frac{\text{Molecular mass of compound}}{\text{Empirical formula mass}}$, then multiply all subscripts in the empirical formula by $n$

Worked Example: A compound is 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass, with a relative molecular mass of 46. Find its molecular formula.

1. 100 g sample: C = 52.2 g, H =13.0 g, O=34.8 g
2. Moles: C = $\frac{52.2}{12}=4.35$, H = $\frac{13.0}{1}=13.0$, O = $\frac{34.8}{16}=2.175$
3. Divide by smallest mole value (2.175): C = 2, H=6, O=1 → Empirical formula = ${\text{C}}_2{\text{H}}_6\text{O}$
4. Empirical formula mass = $(2\times 12)+(6\times 1)+16=46$, so $n=\frac{46}{46}=1$
5. Molecular formula = ${\text{C}}_2{\text{H}}_6\text{O}$

4. Balancing chemical equations — atoms and charges

A balanced chemical equation follows the law of conservation of mass, meaning the number of each type of atom is identical on both the reactant (left) and product (right) sides. For ionic equations, the net charge on both sides must also be equal.

Balancing steps:

1. Write the unbalanced equation with correct, fixed chemical formulas for all reactants and products
2. Count the number of each atom on both sides, starting with the least common element (leave H and O for last)
3. Add whole-number stoichiometric coefficients in front of formulas to balance atom counts
4. For ionic equations, cancel spectator ions (ions that appear unchanged on both sides) and verify net charge is equal on both sides

Worked Example 1 (covalent reaction): Balance the combustion of ethanol: ${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

1. Balance C: 2 on left, add coefficient 2 to ${\text{CO}}_2$: ${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to 2{\text{CO}}_2+{\text{H}}_2\text{O}$
2. Balance H: 6 on left, add coefficient 3 to ${\text{H}}_2\text{O}$: ${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}$
3. Balance O: Right side has $(2\times 2)+(3\times 1)=7$ O, left side has 1 O in ethanol, so add coefficient 3 to ${\text{O}}_2$: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}$
4. Verify all atoms are balanced.

Worked Example 2 (ionic reaction): Balance the reaction of magnesium with hydrochloric acid: $\text{Mg(s)}+{\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)\to {\text{Mg}}^{2+}(aq)+{\text{Cl}}^{-}(aq)+{\text{H}}_2(g)$

1. Balance H: 1 on left, add coefficient 2 to ${\text{H}}^{+}$: $\text{Mg(s)}+2{\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)\to {\text{Mg}}^{2+}(aq)+{\text{Cl}}^{-}(aq)+{\text{H}}_2(g)$
2. Balance charge: Left net charge = +2, right net charge = +2
3. Cancel spectator ${\text{Cl}}^{-}$ ions to get final ionic equation: $\text{Mg(s)}+2{\text{H}}^{+}(aq)\to {\text{Mg}}^{2+}(aq)+{\text{H}}_2(g)$

Exam tip: Never change the subscripts in chemical formulas to balance equations, only adjust coefficients. This is one of the most common mark-losing mistakes in stoichiometry questions.

5. Limiting reagent and percentage yield

The limiting reagent is the reactant that is completely consumed first in a reaction, limiting the maximum amount of product that can be formed. All other reactants are present in excess.

• Theoretical yield: The maximum mass of product that can be formed from the limiting reagent, assuming 100% reaction completion and no product loss
• Actual yield: The real mass of product obtained from an experiment, always lower than theoretical yield due to side reactions, incomplete reaction, or product loss during separation/purification
• Percentage yield: The ratio of actual to theoretical yield, expressed as a percentage: $$\text{Percentage yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\%$$

Worked Example: 12.0 g of magnesium reacts with 12.0 g of oxygen to form magnesium oxide. Calculate the limiting reagent, theoretical yield of MgO, and percentage yield if 17.0 g of MgO is obtained. Balanced equation: $2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$

1. Calculate moles of reactants: $n(\text{Mg})=\frac{12.0}{24}=0.50\text{ mol}$, $n({\text{O}}_2)=\frac{12.0}{32}=0.375\text{ mol}$
2. Mole ratio Mg:O₂ = 2:1. 0.50 mol Mg requires 0.25 mol O₂, and we have 0.375 mol O₂, so Mg is the limiting reagent
3. Theoretical yield: 2 mol Mg forms 2 mol MgO, so 0.50 mol Mg forms 0.50 mol MgO. $M_r(\text{MgO})=40{\text{ g mol}}^{-1}$, so theoretical yield = $0.50\times 40=20.0\text{ g}$
4. Percentage yield = $\frac{17.0}{20.0}\times 100=85.0\%$

6. Concentration — molarity, ppm, percent

Concentration measures the amount of solute dissolved in a fixed volume or mass of solution, and is expressed using three common units in A-Level Chemistry exams:

1. Molarity (molar concentration, $c$): Moles of solute per cubic decimetre (${\text{dm}}^3$) of solution, units ${\text{mol dm}}^{-3}$. Formula: $$c=\frac{n}{V}$$ Where $n$ = moles of solute, $V$ = volume of solution in ${\text{dm}}^3$ (note: $1{\text{ dm}}^3=1\text{ L}=1000{\text{ cm}}^3$)
2. Parts per million (ppm): Used for very dilute solutions (e.g. contaminants in water). 1 ppm = 1 mg of solute per kg of solution, or $1{\text{ mg dm}}^{-3}$ for aqueous solutions (since density of water ≈ $1{\text{ kg dm}}^{-3}$). Formula: $$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$$
3. Percentage concentration: Two standard forms:

• Mass percent (% m/m): $\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$
• Volume percent (% v/v): $\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100\%$

Worked Example: 1.06 g of sodium carbonate (${\text{Na}}_2{\text{CO}}_3$) is dissolved in water to make 500 cm³ of solution. Calculate its molar concentration.

1. $M_r({\text{Na}}_2{\text{CO}}_3)=(2\times 23)+12+(3\times 16)=106{\text{ g mol}}^{-1}$
2. $n({\text{Na}}_2{\text{CO}}_3)=\frac{1.06}{106}=0.010\text{ mol}$
3. Volume = $500{\text{ cm}}^3=0.50{\text{ dm}}^3$
4. $c=\frac{0.010}{0.50}=0.020{\text{ mol dm}}^{-3}$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using the molecular mass of diatomic elements (e.g. $M_r({\text{O}}_2)=32$) when calculating moles of single atoms. Why: Students confuse atomic and molecular mass values. Correct move: Always check if you are counting atoms or molecules; use $A_r$ for single atoms, $M_r$ for molecules.
• Wrong move: Changing subscripts in chemical formulas to balance equations. Why: Students forget formulas represent fixed compound composition. Correct move: Only adjust stoichiometric coefficients in front of formulas, never modify subscripts.
• Wrong move: Using the excess reactant to calculate theoretical yield. Why: Students assume the reactant with larger mass is the limiting reagent. Correct move: Convert all reactant masses to moles, use the balanced equation mole ratio to find which reactant produces less product, and use that for yield calculations.
• Wrong move: Using ${\text{cm}}^3$ directly in molarity calculations. Why: Students forget molarity units are ${\text{mol dm}}^{-3}$. Correct move: Divide ${\text{cm}}^3$ values by 1000 to convert to ${\text{dm}}^3$ before plugging into $c=n/V$.
• Wrong move: Rounding empirical formula ratios too early (e.g. rounding 1.33 to 1 instead of multiplying by 3). Why: Students prioritize whole numbers over accurate ratios. Correct move: Round only if the ratio is within 0.1 of an integer, otherwise multiply by 2, 3 or 4 to eliminate decimals.

8. Practice Questions (A-Level Chemistry Style)

Q1 [3 marks]

A 0.750 g sample of a pure hydrocarbon undergoes complete combustion to produce 2.20 g of ${\text{CO}}_2$ and 1.125 g of ${\text{H}}_2\text{O}$. What is the empirical formula of the hydrocarbon? Solution:

1. Moles of C: $n({\text{CO}}_2)=\frac{2.20}{44}=0.050\text{ mol}$, so $n(\text{C})=0.050\text{ mol}$
2. Moles of H: $n({\text{H}}_2\text{O})=\frac{1.125}{18}=0.0625\text{ mol}$, so $n(\text{H})=2\times 0.0625=0.125\text{ mol}$
3. Ratio C:H = $0.050:0.125=1:2.5$, multiply by 2 to get whole numbers: 2:5. Empirical formula = ${\text{C}}_2{\text{H}}_5$

Q2 [4 marks]

For the reaction $2\text{Al}+3{\text{Cl}}_2\to 2{\text{AlCl}}_3$, 5.4 g of Al is reacted with 21.3 g of ${\text{Cl}}_2$. Calculate the percentage yield if 20.0 g of ${\text{AlCl}}_3$ is produced. Solution:

1. Moles of reactants: $n(\text{Al})=\frac{5.4}{27}=0.20\text{ mol}$, $n({\text{Cl}}_2)=\frac{21.3}{71}=0.30\text{ mol}$
2. Mole ratio Al:Cl₂ = 2:3, so 0.20 mol Al requires 0.30 mol Cl₂, both reactants are fully consumed, no excess.
3. Theoretical yield: 0.20 mol Al forms 0.20 mol ${\text{AlCl}}_3$, $M_r({\text{AlCl}}_3)=27+(3\times 35.5)=133.5{\text{ g mol}}^{-1}$, so theoretical yield = $0.20\times 133.5=26.7\text{ g}$
4. Percentage yield = $\frac{20.0}{26.7}\times 100=74.9\%$

Q3 [2 marks]

A 500 cm³ sample of drinking water contains 0.0015 g of fluoride ions. Calculate the concentration of fluoride ions in ppm, assuming the density of water is $1{\text{ g cm}}^{-3}$. Solution:

1. Mass of solution = $500{\text{ cm}}^3\times 1{\text{ g cm}}^{-3}=500\text{ g}$
2. ppm = $\frac{0.0015}{500}\times 10^6=3\text{ ppm}$

9. Quick Reference Cheatsheet

10. What's Next

Mastery of stoichiometry is non-negotiable for success in the rest of the A-Level Chemistry syllabus. You will use these mole calculation skills to determine enthalpy changes in energetics, calculate equilibrium constants for reversible reactions, measure reaction rates, and predict product yields for organic synthesis pathways. Examiners frequently embed stoichiometry calculations in 5-8 mark extended response questions at A2, so strong foundational skills now will save you significant marks in later papers.

If you struggle with any of the concepts, worked examples, or practice questions in this guide, you can ask Ollie for step-by-step explanations, extra practice problems, or clarification on exam marking conventions at any time by visiting Ollie. You can also access more study guides for other core A-Level Chemistry units, including Atomic Structure and Chemical Bonding, on the homepage to continue your structured A-Level Chemistry preparation.

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