Moles.
The one word that makes Grade 12 Physical Science students break out in a cold sweat. But here is what nobody tells you. Stoichiometry is not a thinking topic. It is a recipe. You follow the same steps every single time. The formula is always the same. The method is always the same. If you can follow a recipe, you can do mole calculations.
This topic weaves through almost every chemistry question in Paper 2. Acids and bases? Moles. Equilibrium? Moles. Electrochemistry? Moles. Get this right and you unlock marks across multiple topics.
In This Post You Will Learn
✓ What a mole actually is and why chemists use it
✓ The three formulas you need for every stoichiometry question
✓ How to use a balanced equation to convert between substances
✓ Step-by-step methods for solution, gas, and mass calculations
✓ How to handle limiting reagent problems
✓ How the NSC exam tests this topic and where to find the marks
What is a Mole?
A mole is just a counting word. Like "dozen" means 12, mole means 6.02 x 10²³.
That is Avogadro's number. It is the number of particles (atoms, molecules, ions) in one mole of any substance.
You do not need to understand why it is that specific number. You just need to know that 1 mole = 6.02 x 10²³ particles.
Why do we use moles? Because atoms are tiny. You cannot count them one by one. Moles let us work with manageable numbers.
The Three Formulas That Run Everything
Every stoichiometry question uses one or more of these three formulas.
| Formula | When to Use It |
|---------------------|---------------------------------------|
| n = m / M | When you have mass (grams) |
| n = c x V | When you have a solution (liquid) |
| n = V / Vm | When you have a gas at STP |Where:
n = number of moles
m = mass in grams (g)
M = molar mass in g/mol (from the periodic table)
c = concentration in mol/dm³ (also called mol/L)
V = volume in dm³ (for solutions) or dm³ (for gases)
Vm = molar volume of a gas at STP = 22.4 dm³/mol
The most important thing: Always find moles first. No matter what the question asks, your first step is to calculate n. Everything else follows from there.
Step-by-Step Method for Any Stoichiometry Problem
Follow this method every single time. Do not skip steps.
Step 1: Write the balanced equation
If the equation is not balanced, balance it. If it is given, check it.
Step 2: Calculate moles of the substance you know
Use whichever formula matches your given information (mass, concentration, or gas volume).
Step 3: Use the mole ratio from the balanced equation
This is where the balanced equation comes in. The coefficients tell you the ratio.
Step 4: Calculate what the question asks for
Convert moles back to mass, volume, or concentration as needed.
Worked Example: Mass to Mass Calculation
Question: How many grams of water are produced when 4 g of hydrogen reacts completely with excess oxygen?
Balanced equation: 2H₂ + O₂ → 2H₂O
Step 1: Equation is balanced. ✓
Step 2: Find moles of H₂.
M(H₂) = 2 x 1 = 2 g/mol
n(H₂) = m/M = 4/2 = 2 mol
Step 3: Use the mole ratio.
From the equation: 2 mol H₂ produces 2 mol H₂O
Ratio is 2:2 which simplifies to 1:1
So n(H₂O) = 2 mol
Step 4: Convert to mass.
M(H₂O) = 2(1) + 16 = 18 g/mol
m(H₂O) = n x M = 2 x 18 = 36 g
Answer: 36 g of water is produced.
Worked Example: Solution (Concentration) Calculation
Question: 25 cm³ of a 0.1 mol/dm³ HCl solution reacts with excess NaOH. Calculate the moles of HCl.
Watch the units. Volume must be in dm³, not cm³.
25 cm³ = 25/1000 = 0.025 dm³
n(HCl) = c x V = 0.1 x 0.025 = 0.0025 mol
Unit conversion you must know:
1 dm³ = 1000 cm³ = 1 L
To convert cm³ to dm³: divide by 1000
Students lose marks here all the time. Always check your units before calculating.
Worked Example: Gas Volume at STP
Question: What volume of CO₂ gas (at STP) is produced when 10 g of CaCO₃ decomposes?
CaCO₃ → CaO + CO₂
Step 1: Balanced. ✓ (all coefficients are 1)
Step 2: Moles of CaCO₃.
M(CaCO₃) = 40 + 12 + 3(16) = 100 g/mol
n(CaCO₃) = 10/100 = 0.1 mol
Step 3: Mole ratio. 1:1 for CaCO₃ to CO₂.
n(CO₂) = 0.1 mol
Step 4: Volume at STP.
V = n x Vm = 0.1 x 22.4 = 2.24 dm³
For full live lessons on stoichiometry and all chemistry topics, see our Grade 12 Physical Science tuition page.
Limiting Reagent Problems
Sometimes both reactants are given and one runs out first. The one that runs out is the limiting reagent. It determines how much product is formed.
How to Find the Limiting Reagent
Step 1: Calculate moles of both reactants.
Step 2: Divide each by its coefficient in the balanced equation.
Step 3: The one with the smaller value is the limiting reagent.
Example: 5 g of Fe reacts with 5 g of O₂.
4Fe + 3O₂ → 2Fe₂O₃
n(Fe) = 5/56 = 0.0893 mol
n(O₂) = 5/32 = 0.1563 mol
Divide by coefficients:
Fe: 0.0893/4 = 0.0223
O₂: 0.1563/3 = 0.0521
Fe has the smaller value. Fe is the limiting reagent.
Use n(Fe) to calculate the amount of product formed.
If you want to see how mole calculations connect to equilibrium, read our post on Chemical Equilibrium Grade 12 - Le Chatelier's Principle Made Simple.
Molar Mass from the Periodic Table
You must be able to read the periodic table to find molar mass. The molar mass of an element is its relative atomic mass in g/mol.
| Element | Symbol | Molar Mass (g/mol) |
|---------|--------|-------------------|
| Hydrogen| H | 1 |
| Carbon | C | 12 |
| Nitrogen| N | 14 |
| Oxygen | O | 16 |
| Sodium | Na | 23 |
| Chlorine| Cl | 35.5 |
| Calcium | Ca | 40 |
| Iron | Fe | 56 |
| Copper | Cu | 63.5 |For compounds, add up the molar masses of each atom.
M(NaCl) = 23 + 35.5 = 58.5 g/mol
M(H₂SO₄) = 2(1) + 32 + 4(16) = 98 g/mol
M(Ca(OH)₂) = 40 + 2(16 + 1) = 74 g/mol
Common Mistakes Students Make
- Not converting cm³ to dm³
The formulas use dm³. If the question gives volume in cm³, you must divide by 1000 before substituting. Getting this wrong makes your answer 1000 times too big or too small.
- Using the wrong molar mass
Students forget to multiply by the number of atoms. H₂O is not 1 + 16 = 17. It is 2(1) + 16 = 18. Check the subscripts.
- Ignoring the mole ratio
The balanced equation tells you the ratio. If 2 mol of A produces 3 mol of B, you cannot assume a 1:1 ratio. Read the coefficients.
- Forgetting to balance the equation
If the equation is not balanced, your mole ratios will be wrong and every calculation after that is wrong too. Always check.
- Not showing units in working
Write n = m/M = 10/100 = 0.1 mol. Not just "0.1". The examiners want to see units. It also helps you catch errors because if the units do not make sense, your formula is wrong.
How This Topic Appears in the NSC Exam
Stoichiometry appears in Paper 2 of the Grade 12 Physical Science NSC exam.
It does not have its own standalone section, but mole calculations appear inside questions on acids and bases, equilibrium, electrochemistry, and organic chemistry. Across all these questions, stoichiometry is worth roughly 15 to 20 marks in total.
The most common format is a titration calculation in the acids and bases question. You are given the concentration and volume of one solution, and asked to find the concentration of the other. This uses n = cV and the mole ratio.
The DBE also tests stoichiometry in equilibrium (ICE table calculations) and electrochemistry (Faraday's Law calculations involving moles of electrons).
In the 2023 NSC exam, stoichiometry appeared in the acids and bases question (titration calculation), in the equilibrium question (calculating Kc from given moles), and in the electrochemistry question (calculating mass deposited).
Key point: You cannot escape mole calculations. They are woven into every chemistry topic. Master n = m/M, n = cV, and n = V/Vm, and you unlock marks across the entire chemistry section of Paper 2.
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