Stoichiometry Grade 12 -- Mole Calculations Made Easy

You open the Physical Science Paper 2 and you see it. A chemical equation with numbers. Moles this. Concentration that. Limiting reagent. Something about a gas collected over water.

Your heart sinks a little.

Stoichiometry is the topic that breaks more Grade 12 students than almost any other on Paper 2. It is not that it is inherently difficult. It is that it is taught badly in most classrooms and most students never get a clear method they can repeat under pressure. They understand it in class. They watch the teacher do it on the board. Then they sit alone with a past paper and it all falls apart.

That is what this post fixes.

By the end of this you will have a repeatable method for every stoichiometry question that shows up in the NSC exam. Real numbers. Real working. Nothing vague.

In This Post You Will Learn:

• What the mole actually means and why it matters for every calculation
• The five formulas you must know and when to use each one
• How to balance equations correctly before you start calculating
• A step-by-step method for limiting reagent questions that always works
• How to handle gas volume calculations including the 22,4 dm3 rule
• The empirical and molecular formula method that never fails
• Common mistakes that cost students full marks on calculation questions

What Is Stoichiometry Actually?

Let us start with the basics because the basics are where most students lose the thread.

Stoichiometry is the mathematics of chemical reactions. When chemicals react, they do so in fixed numerical proportions. The numbers in front of the chemical formula in a balanced equation tell you exactly how many moles of each substance react and how many moles of each product form.

That word mole keeps showing up. Let us make sure you actually understand it.

One mole of any substance contains Avogadro's number of particles. That number is 6,02 times 10 to the power of 23. That is a huge number. You will never calculate with it directly. What you will do is use it to convert between mass, number of particles, volume of gas, and concentration.

The mole is simply a counting unit. Like a dozen means 12, a ream means 500, a mole means 6,02 x 10^23. The difference is that a mole is used for atoms, molecules and ions.

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The Five Formulas You Must Know

Every stoichiometry question in the NSC exam is built on one or more of these five formulas. Learn them now. Know which one to use in which situation. That is 80% of the battle won.

Write these down now. Not on a phone. On paper. By hand. You need to be able to write all five from memory before you sit the exam.

What Each Symbol Means

Before you can use any formula you need to know what the symbols stand for. This trips students up more than you would think.

n is the number of moles. Always.

m is the mass in grams. Not kilograms. Grams.

M is the molar mass in grams per mole. You find this by adding up the atomic masses from the periodic table. The one on your data sheet at the back of the exam paper.

N is the number of particles. Atoms, molecules, ions. Whatever the question gives you.

NA is Avogadro's constant. 6,02 x 10^23. It is on your data sheet. Use it.

c is concentration in mol per dm3. The unit matters. Watch your conversions.

V is volume. It can be in cm3 or dm3. You will need to convert between them.

Vm is the molar volume of gas. 22,4 dm3 per mole at standard temperature and pressure. STP means 0 degrees Celsius and 1 atmosphere or 101,3 kilopascals. The value is on your data sheet.

Step 1 Always: Balance the Equation

Here is something most students do not realise. If the equation is not balanced, every calculation that follows is wrong. Full stop.

Balancing equations is not optional. It is the first step in every stoichiometry question. Always. Even if the question does not explicitly tell you to balance it.

The coefficients in a balanced equation give you the mole ratio. That ratio is the key that unlocks every calculation.

Here is an example.

Hydrogen reacts with oxygen to produce water.

H2 + O2 -- H2O

This is not balanced. One oxygen on the left. Two oxygens on the right. Fix it.

2H2 + O2 -- 2H2O

Now it is balanced. The mole ratio is 2 to 1 to 2. Two moles of hydrogen for every one mole of oxygen. Two moles of water produced.

Balancing Checklist

Before you start any calculation, run through this.

• Count the atoms of each element on the left.
• Count the atoms of each element on the right.
• Are they equal? If not, add coefficients one element at a time.
• Never change a subscript. H2O stays H2O. You change the number in front.
• Check all elements again when you are done.

The Mole Calculation Method: Step by Step

Here is the method Mr Sawaya teaches every student who walks through his door struggling with stoichiometry. It works. Every time. You just have to follow it.

Example: Find the mass of water produced when 4g of hydrogen reacts with excess oxygen.

Step 1: Write the balanced equation.

2H2 + O2 -- 2H2O

Step 2: Extract what you know.

m of H2 = 4g. We need n of H2 first.

Step 3: Calculate moles of what you are given.

Molar mass of H2 = 2 x 1 = 2 g/mol

n = m / M = 4 / 2 = 2 moles of H2

Step 4: Use the mole ratio from the balanced equation.

The ratio is 2H2 to 2H2O. That is 1 to 1.

n of H2O produced = 2 moles

Step 5: Calculate what the question asks for.

m of H2O = n x M = 2 x 18 = 36g

Answer: 36g of water is produced.

That is the method. Every question follows this structure. Given. Convert to moles. Mole ratio. Convert from moles to what is asked. Do not deviate from it.

Limiting Reagent Questions: The Method That Always Works

This is the question type that scares students the most. Two reactants. One runs out first. Which one? How much product? What is left over?

Here is the problem. Most students look at the numbers and guess. They feel overwhelmed by the calculation. They pick one reactant and hope it is right.

It is not hard. Here is the process.

You calculate the moles of each reactant. Then you divide the moles you have by the coefficient in the balanced equation. The one with the smaller answer is the limiting reagent. It runs out first. Everything else is in excess.

Example: 10g of iron reacts with 8g of sulphur to produce iron sulphide. Which is limiting?

First, write the balanced equation.

Fe + S -- FeS

Calculate moles of iron.

Molar mass of Fe = 55,8 g/mol (from periodic table)

n of Fe = 10 / 55,8 = 0,179 moles

Calculate moles of sulphur.

Molar mass of S = 32 g/mol

n of S = 8 / 32 = 0,25 moles

Divide by coefficients. Both coefficients are 1.

Fe gives 0,179. S gives 0,25.

Iron has the smaller value. Iron is the limiting reagent.

Everything else you calculate flows from iron. That is your starting point.

This exact type of question appeared in the 2022 NSC November exam. Question 4 on Paper 2 asked students to calculate the mass of magnesium carbonate that reacted with hydrochloric acid, identify the limiting reagent, and calculate the mass of salt produced. The method is identical every time.

Limiting Reagent Checklist

Run through this for every limiting reagent question.

• Write the balanced equation first.
• Calculate moles of each reactant. Do not skip this step.
• Divide each mole value by its coefficient in the balanced equation.
• The smallest result is the limiting reagent. That is your reference point.
• Calculate moles of product from the limiting reagent using the mole ratio.
• Convert to whatever the question asks for. Mass, volume, concentration.

Gas Volume Calculations: The 22,4 Rule

When a reaction produces a gas or uses a gas, you need the molar gas volume rule.

At standard temperature and pressure -- that is 0 degrees Celsius and 1 atmosphere or 101,3 kilopascals -- one mole of any gas occupies 22,4 dm3. This value is on your data sheet at the back of the exam paper.

The most common question is a gas collected over water. Something reacts and a gas bubbles up and gets collected in a graduated cylinder.

Here is what you need to watch for. The volume you measure is not pure gas. It is gas plus water vapour. You have to subtract the water vapour pressure to get the volume of the actual gas.

The question will usually give you the water vapour pressure. Subtract it from the total pressure first. Then use the combined gas law if temperature is not STP. Or if it is exactly STP conditions, just use Vm = 22,4 dm3.

Example: 0,5 moles of hydrogen gas is collected at STP. What volume does it occupy?

n = 0,5 moles. STP means we can use Vm = 22,4 dm3/mol directly.

V = n x Vm = 0,5 x 22,4 = 11,2 dm3

Simple. When you know n and conditions are STP, multiply by 22,4.

Empirical and Molecular Formula: The Method

This one has three steps and students either get full marks or zero. There is no middle ground. That is because the method is exact. Follow it and you cannot go wrong.

Step 1: Find the mass of each element in the compound. If you are given percentage composition, assume 100g total. That turns percentages into grams directly.

Step 2: Convert each mass to moles by dividing by the molar mass of that element.

Step 3: Divide all the mole values by the smallest mole value. This gives you the simplest whole number ratio.

If the ratios are not whole numbers, multiply by the smallest factor that makes them whole. If you get 0,5 as one of your numbers, multiply everything by 2. If you get 0,33, multiply by 3. If you get 0,25, multiply by 4.

Example: A compound contains 40% carbon, 6,67% hydrogen and 53,33% oxygen. What is the empirical formula?

Assume 100g of compound.

Carbon: 40g. Hydrogen: 6,67g. Oxygen: 53,33g.

Convert to moles.

C: 40 / 12 = 3,33 moles

H: 6,67 / 1 = 6,67 moles

O: 53,33 / 16 = 3,33 moles

Divide by the smallest value (3,33).

C: 1. H: 2. O: 1.

Empirical formula is CH2O.

If the question then asks for the molecular formula and tells you the molar mass is 180 g/mol, you find the molar mass of the empirical formula (12 + 2 + 16 = 30) and divide 180 by 30 = 6. Multiply all subscripts by 6. Molecular formula is C6H12O6.

Concentration and Titration Calculations

The concentration formula n = cV comes up in two main ways in the NSC exam. As a standalone calculation, and as part of a full titration.

For concentration questions, you are usually given two of the three variables and asked to find the third. The unit of volume matters. Concentration c is in mol per dm3. Volume V is often given in cm3. Convert cm3 to dm3 by dividing by 1000.

For titration questions, the formula you need is this one.

na/nb = caVa / cbVb

This means the mole ratio of the two reactants equals the concentration times volume ratio. You use this when one solution is titrated against another and you know three of the four values and need to find the fourth.

Example: 25 cm3 of 0,2 mol/dm3 hydrochloric acid is titrated against sodium hydroxide. The end point is reached when 20 cm3 of sodium hydroxide has been added. What is the concentration of the sodium hydroxide?

The reaction is HCl + NaOH -- NaCl + H2O. The mole ratio is 1 to 1.

Convert volumes to dm3.

Va = 25 / 1000 = 0,025 dm3

Vb = 20 / 1000 = 0,020 dm3

Use the formula for equal coefficients (na/nb = 1).

caVa = cbVb

cb = caVa / Vb = 0,2 x 0,025 / 0,020 = 0,25 mol/dm3

Answer: 0,25 mol/dm3

This type of calculation appeared in the 2019 NSC November exam. Question 7 asked students to calculate the concentration of a hydrochloric acid solution used to titrate sodium hydrogen carbonate, including finding the pH of the acid solution. The method is the same every single time.

Common Mistakes Students Make in Stoichiometry

1. Forgetting to balance the equation before calculating.

Every mark in the mole ratio section assumes the equation is balanced. If you use the wrong ratio, you lose full marks for that part. Balance first. Always.

2. Mixing up units.

Concentration is in mol per dm3. Volume in the formula is in dm3. Most calculation errors come from feeding cm3 into the formula without converting. Convert first. Then calculate.

3. Choosing the wrong formula.

Students look at a gas volume question and reach for n = m/M. Students look at a concentration question and reach for n = V/Vm. Each formula has a specific trigger. Know the trigger for each one. m/M for mass. cV for concentration. V/Vm for gas volume. N/NA for number of particles.

4. Incorrectly identifying the limiting reagent.

Students pick the reactant with the smaller mass. That is wrong. You compare moles divided by coefficients. A little bit of a substance with a low molar mass can actually be more moles than a large amount of something with a high molar mass. Always calculate moles.

5. Not reading the question for what it actually asks.

Some questions ask for mass. Some ask for concentration. Some ask for volume. Students calculate the wrong thing and lose marks for an otherwise correct solution. Underline what the question wants before you start. Write the unit they want next to it.

How This Topic Appears in the NSC Exam

Stoichiometry and mole calculations appear throughout Paper 2 Chemistry. They are not limited to one question. They show up in the multiple choice section, in quantitative analysis questions, in titration questions, and in reaction rate questions that involve calculations. They are woven through the entire paper.

Here is what the DBE examination guidelines say about Paper 2 Chemistry.

The Paper 2 exam has 9 questions. Question 1 is always 10 multiple choice questions worth 20 marks. Questions 2 to 9 are longer structured questions worth the remaining 130 marks. The questions are not fixed to specific topics from one year to the next. The DBE arranges them differently each year. What does not change is that stoichiometry appears in the quantitative analysis questions and it appears every single year.

A stoichiometry calculation involving a reaction between an acid and a metal carbonate or metal appeared in the 2022 NSC November exam. A titration calculation appeared in the 2019 NSC November exam. A stoichiometry question appeared in the multiple choice section of the 2023 NSC November exam. The context changes. The method does not.

The DBE Information Sheet at the back of the exam paper gives you the formulas and constants. You still have to know which one to use and how to apply it. The sheet does not do the thinking for you.

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