Subject: Chemistry | Level: GCSE | Exam Board: WJEC
Mastering chemical formulae and the mole concept is the key to unlocking quantitative chemistry. This topic bridges the atomic world with the macroscopic world, forming the foundation for calculating reacting masses, gas volumes, and concentrations.
Revision Notes & Key Concepts
Revision Podcast Transcript
Welcome to your GCSE Chemistry revision podcast. I'm your tutor, and today we're diving into one of the most important and most frequently examined topics in the entire specification: Chemical Formulae, Equations, and Amount of Substance. Whether you're sitting AQA, Edexcel, or OCR, I promise you this topic will come up — and if you nail it, you're looking at some seriously easy marks. So get comfortable, grab a pen, and let's get into it. Let me start by telling you why this topic matters so much. Everything in quantitative chemistry — from calculating yields in industrial processes to working out how much medicine to prescribe — comes back to the ideas we're covering today. The mole concept is the bridge between the tiny atomic world and the measurable world of grams and litres. Once you get it, chemistry starts to make a whole lot more sense. Right, let's get into the core concepts. First up: chemical symbols and formulae. Every element has a symbol — usually one or two letters, with the first always capitalised. So hydrogen is H, oxygen is O, carbon is C, sodium is Na, chlorine is Cl. You need to know these. The formula of a compound tells you which elements are present and in what ratio. Water is H2O — two hydrogen atoms bonded to one oxygen. Carbon dioxide is CO2 — one carbon, two oxygens. Sodium chloride is NaCl — one sodium, one chloride. The small numbers in a formula are called subscripts, and they tell you how many atoms of each element are in one formula unit. This is really important — you must never change subscripts when balancing equations. I'll come back to that. Now, relative formula mass — or Mr. This is simply the sum of all the relative atomic masses in a formula. You add up the Ar values for every atom present. For water, H2O: that's 2 times 1 for hydrogen, plus 16 for oxygen, giving Mr equals 18. For carbon dioxide, CO2: 12 plus 2 times 16 equals 44. For sulfuric acid, H2SO4: 2 plus 32 plus 4 times 16 equals 98. Examiners will always give you a periodic table, so you don't need to memorise atomic masses — but you do need to be able to use them quickly and accurately. Next: balancing equations. A chemical equation shows reactants on the left and products on the right, separated by an arrow. The law of conservation of mass tells us that atoms cannot be created or destroyed in a chemical reaction — so the number of atoms of each element must be the same on both sides. We balance equations by adjusting coefficients — the big numbers in front of formulae. Never, ever change the subscripts. That would change the substance entirely. Let me walk you through an example. Take the reaction between hydrogen and oxygen to make water. We start with: H2 plus O2 gives H2O. Count the atoms: left side has 2 H and 2 O. Right side has 2 H and 1 O. Oxygen is unbalanced. So we put a 2 in front of H2O: H2 plus O2 gives 2H2O. Now the right has 4 H and 2 O. But now hydrogen is unbalanced — we only have 2 H on the left. So we put a 2 in front of H2: 2H2 plus O2 gives 2H2O. Check: left has 4 H and 2 O, right has 4 H and 2 O. Balanced! And don't forget state symbols: g for gas, l for liquid, s for solid, and aq for aqueous — meaning dissolved in water. So the full equation is: 2H2 gas plus O2 gas gives 2H2O liquid. Now for the star of the show: the mole. This is the concept that trips up so many students, but it's actually quite elegant once you see it. A mole is simply a counting unit — like a dozen, but instead of 12, it's 6.02 times 10 to the power of 23. That number is called the Avogadro constant, symbol N-A. One mole of any substance contains 6.02 times 10 to the 23 particles — whether those are atoms, molecules, or ions. Why that number? Because it's the number of atoms in exactly 12 grams of carbon-12. And this leads us to the key relationship: one mole of any substance has a mass equal to its relative formula mass in grams. So one mole of water, Mr equals 18, has a mass of 18 grams. One mole of carbon dioxide, Mr equals 44, has a mass of 44 grams. This is the foundation of all mole calculations. The formula you need is: moles equals mass divided by Mr. Or rearranged: mass equals moles times Mr. Or: Mr equals mass divided by moles. I like to remember this as a triangle — mass on top, moles and Mr on the bottom. Cover what you want to find, and the triangle tells you the formula. Let's do a quick calculation. How many moles are in 36 grams of water? Mr of water is 18. Moles equals 36 divided by 18 equals 2 moles. Simple. What mass of carbon dioxide contains 0.5 moles? Mass equals 0.5 times 44 equals 22 grams. You're getting it. Now, molar gas volume. At room temperature and pressure — which we abbreviate as rtp — one mole of any gas occupies 24 dm cubed, or 24,000 cm cubed. This is a really useful fact. The formula is: moles equals volume in dm cubed divided by 24. So 48 dm cubed of oxygen contains 48 divided by 24 equals 2 moles of oxygen gas. Now let's talk about using moles in stoichiometric calculations — that's the fancy word for calculations based on balanced equations. The coefficients in a balanced equation tell you the mole ratio of reactants and products. In the equation 2H2 plus O2 gives 2H2O, the ratio is 2 to 1 to 2. So 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water. Here's a typical exam question: what mass of water is produced when 4 grams of hydrogen reacts completely with excess oxygen? Step 1: write the balanced equation — 2H2 plus O2 gives 2H2O. Step 2: find moles of hydrogen — Mr of H2 is 2, so moles equals 4 divided by 2 equals 2 moles. Step 3: use the mole ratio — the ratio of H2 to H2O is 2 to 2, which is 1 to 1, so 2 moles of H2 produces 2 moles of H2O. Step 4: find the mass of water — mass equals 2 times 18 equals 36 grams. That's your answer: 36 grams of water. Finally, let's cover empirical formulae. The empirical formula is the simplest whole number ratio of atoms in a compound. To find it from mass data: divide each mass by its Ar to get moles, then divide all by the smallest number of moles to get the ratio. If the ratios aren't whole numbers, multiply up. For example, if a compound contains 2.4 grams of carbon and 0.4 grams of hydrogen: moles of C equals 2.4 divided by 12 equals 0.2; moles of H equals 0.4 divided by 1 equals 0.4. Divide by smallest, which is 0.2: C equals 1, H equals 2. Empirical formula is CH2. And limiting reactants — this is Higher tier content. The limiting reactant is the one that runs out first and determines how much product is made. To identify it, calculate the moles of each reactant, compare to the mole ratio in the equation, and the one that gives the smaller amount of product is the limiting reactant. The other reactant is in excess. Right, now let's talk exam technique. This is where marks are won and lost. Number one: always show your working. Even if you get the final answer wrong, you can still earn method marks. Examiners are instructed to award marks for correct steps, so never skip steps. Number two: never change subscripts when balancing equations. This is the single most common mistake I see. Changing H2O to H2O2 changes water into hydrogen peroxide — a completely different substance. Only change coefficients. Number three: check your units. If the question gives volume in cm cubed, convert to dm cubed by dividing by 1000 before using the molar volume formula. 1 dm cubed equals 1000 cm cubed. Number four: use the correct number of significant figures. Match your answer to the data given in the question — usually 2 or 3 significant figures. Number five: when a question says show that or verify, you must show all your working clearly. The answer is given to you — the marks are for the method. Number six: state symbols. If the question asks for a full equation with state symbols, you must include them. Forgetting state symbols when they're specifically asked for will cost you marks. Number seven: for limiting reactant questions, always calculate moles of both reactants and compare to the equation ratio. Don't just guess which one runs out. Now for our quick-fire recall quiz. I'll ask a question, pause, then give the answer. Ready? Question 1: What is the Avogadro constant? ... The Avogadro constant is 6.02 times 10 to the 23 — the number of particles in one mole. Question 2: What is the formula for calculating moles from mass? ... Moles equals mass divided by Mr. Question 3: What volume does one mole of gas occupy at room temperature and pressure? ... 24 dm cubed. Question 4: What is the empirical formula of a compound with the molecular formula C6H12O6? ... CH2O — divide all subscripts by 6. Question 5: In the equation N2 plus 3H2 gives 2NH3, what is the mole ratio of hydrogen to ammonia? ... 3 to 2 — three moles of hydrogen produces two moles of ammonia. How did you do? If you got all five, you're in great shape. If not, go back and review those sections. Let me leave you with the key things to remember. One: the mole is a counting unit — 6.02 times 10 to the 23 particles. Two: moles equals mass divided by Mr. Three: one mole of gas at rtp occupies 24 dm cubed. Four: balance equations using coefficients only — never subscripts. Five: state symbols are s, l, g, and aq. Six: the limiting reactant is the one that runs out first. Seven: always show your working in calculations. This topic is genuinely one of the most rewarding in GCSE Chemistry. Once you've mastered the mole, you'll find that so many other topics — electrolysis, titrations, percentage yield — all click into place. Keep practising those calculations, and you'll be picking up marks with confidence in the exam. Thanks for listening. Good luck with your revision — you've got this!
Key Terms & Definitions
- Relative Formula Mass (Mr)
- The sum of the relative atomic masses of the elements as given in the formula for any non-molecular compound.
- Mole
- The measure of the amount of substance. One mole contains exactly 6.02 x 10^23 elementary entities.
- Avogadro Constant
- The number of atoms, molecules or ions in a mole of a given substance (6.02 x 10^23 per mole).
- Limiting Reactant
- The reactant in a chemical reaction that is completely consumed when the reaction goes to completion, thereby limiting the amount of product formed.
- Empirical Formula
- The simplest whole number ratio of atoms of each element present in a compound.
- Conservation of Mass
- The principle stating that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.
Worked Examples
Worked Example
Question: Calculate the mass of magnesium oxide produced when 6.0g of magnesium burns completely in oxygen. (4 marks)
Solution: Step 1: Write the balanced equation. $2Mg + O_2 \rightarrow 2MgO$ Step 2: Calculate the moles of the known substance (Mg). $A_r$ of Mg = 24 Moles of Mg = Mass / $A_r$ = 6.0 / 24 = 0.25 mol Step 3: Use the molar ratio from the balanced equation to find moles of MgO. Ratio of Mg : MgO is 2:2 (which simplifies to 1:1). Therefore, moles of MgO produced = 0.25 mol. Step 4: Calculate the mass of the unknown substance (MgO). $M_r$ of MgO = 24 + 16 = 40 Mass of MgO = Moles $\times M_r$ = 0.25 $\times$ 40 = 10g Final answer: 10g
Worked Example
Question: A compound contains 3.2g of copper, 0.6g of carbon, and 2.4g of oxygen. Calculate its empirical formula. [Ar: Cu=64, C=12, O=16] (4 marks)
Solution: Step 1: Write down the masses of each element. Cu = 3.2g, C = 0.6g, O = 2.4g Step 2: Divide each mass by the respective Relative Atomic Mass ($A_r$) to find moles. Cu: 3.2 / 64 = 0.05 mol C: 0.6 / 12 = 0.05 mol O: 2.4 / 16 = 0.15 mol Step 3: Divide all mole values by the smallest number of moles (0.05) to find the simplest whole number ratio. Cu: 0.05 / 0.05 = 1 C: 0.05 / 0.05 = 1 O: 0.15 / 0.05 = 3 Final answer: $CuCO_3$
Worked Example
Question: In a reaction, 4.8g of magnesium is added to 0.5 moles of sulfuric acid. Determine the limiting reactant and calculate the moles of hydrogen gas produced. $Mg + H_2SO_4 \rightarrow MgSO_4 + H_2$ (4 marks)
Solution: Step 1: Calculate the moles of magnesium. $A_r$ of Mg = 24 Moles of Mg = 4.8 / 24 = 0.20 mol Step 2: Identify the limiting reactant. We have 0.20 mol of Mg and 0.50 mol of $H_2SO_4$. The balanced equation shows a 1:1 reacting ratio. Since 0.20 < 0.50, Magnesium is the limiting reactant (and $H_2SO_4$ is in excess). Step 3: Calculate the moles of hydrogen gas produced. The amount of product depends entirely on the limiting reactant. Ratio of Mg : $H_2$ is 1:1. Therefore, moles of $H_2$ produced = 0.20 mol. Final answer: Limiting reactant is Mg; 0.20 moles of $H_2$ produced.
Practice Questions
Question: Balance the following equation: $Fe_2O_3 + C \rightarrow Fe + CO_2$
Answer:
Question: Calculate the number of molecules in 11g of carbon dioxide ($CO_2$). [Ar: C=12, O=16, Avogadro constant = $6.02 \times 10^{23}$]
Answer:
Question: Sodium reacts with water to produce sodium hydroxide and hydrogen gas: $2Na + 2H_2O \rightarrow 2NaOH + H_2$. Calculate the volume of hydrogen gas produced at rtp when 4.6g of sodium reacts completely. [Ar: Na=23]
Answer:
Question: A student heated 1.20g of magnesium in a crucible. It reacted with oxygen to form 2.00g of magnesium oxide. Use this data to deduce the empirical formula of magnesium oxide. [Ar: Mg=24, O=16]
Answer:
Question: Explain why the mass of the contents of a crucible increases when magnesium is heated in air.
Answer:
