Welcome to the exciting world of stoichiometry! This chapter, "The Mole and the Avogadro Constant," is arguably the single most important concept in IGCSE Chemistry.
Why is it so important? Because chemistry is about mixing substances, and substances are made of incredibly tiny particles (atoms, molecules, ions). We can’t count them directly, so we use the concept of the Mole to count them in bulk. Mastery of the mole allows you to predict exactly how much product you will make in a reaction, how much gas will be produced, and how strong a solution is.
Don't worry if the numbers seem massive at first. We will break down every calculation step-by-step!
The Central Idea: What is The Mole? (Supplement 2)
In everyday life, we use special names for large collections of things: a pair of socks (2), a dozen of eggs (12). In Chemistry, we need a special name for a huge collection of particles. That name is the Mole.
Defining the Amount of Substance
The unit used for the amount of substance is the mole (abbreviated as mol).
A mole is defined as the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12.
The Avogadro Constant
When we say "one mole," we are referring to a very specific, huge number of particles. This number is called the Avogadro Constant (\(N_A\)).
1 mole contains: \[6.02 \times 10^{23} \text{ particles}\]
Particles can be anything: atoms (like in copper metal), ions (like in NaCl), or molecules (like in water, H₂O).
Did you know?
If you had a mole of pennies, and you spread them across the entire surface of the Earth, the layer of pennies would be about 300 meters deep! That’s how huge \(6.02 \times 10^{23}\) is.
Key Takeaway: The mole is simply a counting unit for chemists, just like a dozen is a counting unit for eggs. It always equals \(6.02 \times 10^{23}\) of whatever you are counting.
Molar Mass (Supplement 3c, 3d)
The Molar Mass (\(M\)) is the mass of one mole of a substance.
The units for molar mass are always grams per mole (\(g/mol\)).
The Relationship between Relative Mass and Molar Mass
The fantastic thing about the mole is that the molar mass of an element or compound is numerically equal to its Relative Atomic Mass (\(A_r\)) or Relative Molecular Mass (\(M_r\)).
For example:
- The \(A_r\) of Oxygen (O) is 16.0. Therefore, the molar mass of Oxygen atoms is \(16.0\ g/mol\).
- The \(M_r\) of Water (\(H_2O\)) is \((2 \times 1.0) + 16.0 = 18.0\). Therefore, the molar mass of Water molecules is \(18.0\ g/mol\).
- For ionic compounds, we use Relative Formula Mass, \(M_r\). The \(M_r\) of NaCl is \(23.0 + 35.5 = 58.5\). Therefore, the molar mass of NaCl is \(58.5\ g/mol\).
The Fundamental Mole Calculation (Supplement 3)
The most common calculation you will perform links mass, moles, and molar mass. This relationship is so important, we often use a triangle trick to remember it:
The Mole Triangle:
If you cover up the quantity you want to find, the formula appears.
\[\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]
Or, using symbols: \[\mathbf{n} = \frac{\mathbf{m}}{\mathbf{M}}\]
Where:
- \(n\) = amount of substance (moles)
- \(m\) = mass (grams)
- \(M\) = Molar Mass (\(g/mol\))
Step-by-Step Example: Calculating Moles
Question: How many moles are in 80 g of Sulfur Dioxide (\(SO_2\))?
(Given: \(A_r\) of S = 32, O = 16)
- Find the Molar Mass (\(M\)):
\(M(\text{SO}_2) = 32 + (2 \times 16) = 64\ g/mol\). - Use the Formula:
\(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\) - Calculate:
\(\text{Moles} = \frac{80\ g}{64\ g/mol} = 1.25\ mol\)
Quick Review: To convert between mass and moles, you must use the Molar Mass (\(M_r\)) as the conversion factor.
Calculations Involving Molar Gas Volume (Supplement 4, 5)
Gases are special because, unlike solids and liquids, their volume is determined mostly by temperature and pressure, not the type of particle.
The Molar Gas Volume Concept
At room temperature and pressure (r.t.p.), one mole of any gas occupies a standard volume.
Molar Gas Volume (at r.t.p.) = \(\mathbf{24\ dm^3}\)
Note: This volume (24 dm³) applies to all gases, whether it’s hydrogen (\(H_2\)) or carbon dioxide (\(CO_2\)).
Gas Volume Formula
To find the amount of substance (moles) of a gas: \[\text{amount of substance (mol)} = \frac{\text{Volume of gas (dm}^3\text{)}}{\text{24 dm}^3\text{/mol}}\]
Unit Conversion Warning!
You must use the volume unit \(\mathbf{dm^3}\) (cubic decimeters) for this formula. If the question gives volume in \(\mathbf{cm^3}\) (cubic centimeters), convert first!
\[\mathbf{1\ dm^3 = 1000\ cm^3}\]
To convert \(cm^3\) to \(dm^3\), divide by 1000.
Step-by-Step Example: Calculating Gas Volume
Question: What volume in \(cm^3\) is occupied by 0.5 moles of nitrogen gas (\(N_2\)) at r.t.p.?
- Use the rearranged formula:
\(\text{Volume} = \text{Moles} \times 24\ dm^3\text{/mol}\) - Calculate volume in \(dm^3\):
\(\text{Volume} = 0.5\ mol \times 24\ dm^3\text{/mol} = 12\ dm^3\) - Convert to \(cm^3\):
\(\text{Volume} = 12\ dm^3 \times 1000 = 12,000\ cm^3\)
Key Takeaway: The magic number 24 only works for gases at r.t.p. Always check your units!
Calculations Involving Solutions and Concentration (Core 1, Supplement 5, 6)
When dealing with liquids, especially in experiments like titrations (Supplement 6), we need to know the concentration of the dissolved substance (the solute).
Units of Concentration (Core 1)
Concentration can be measured in two main ways:
- Mass Concentration: \(\mathbf{g/dm^3}\) (grams of solute per cubic decimeter of solution).
- Molar Concentration (Molarity): \(\mathbf{mol/dm^3}\) (moles of solute per cubic decimeter of solution).
Molar Concentration Formula
This formula links moles, concentration, and volume: \[\text{amount of substance (mol)} = \text{Concentration (mol/dm}^3\text{)} \times \text{Volume (dm}^3\text{)}\]
Or, using symbols: \[\mathbf{n} = \mathbf{C} \times \mathbf{V}\]
Crucial Note on Units: Just like with gases, the volume (\(V\)) must be in \(\mathbf{dm^3}\) if the concentration (\(C\)) is in \(\mathbf{mol/dm^3}\).
Converting between Concentration Types
Since Molar Mass (\(M\)) links grams and moles, we can easily switch between the two concentration units:
Titration Calculations (Supplement 6)
Titrations use this formula extensively. You measure the volume of a solution of known concentration (the standard solution) needed to exactly neutralize a solution of unknown concentration.
The key steps in a titration calculation always involve finding the number of moles of the known substance first, then using the mole ratio from the balanced equation to find the moles of the unknown substance.
Common Mistake to Avoid: Never use volumes in \(cm^3\) when calculating moles using molar concentration! Always convert \(cm^3\) to \(dm^3\) by dividing by 1000.
Working with Particles (Supplement 3e)
If a question asks for the actual number of atoms or molecules, you must use the Avogadro Constant (\(N_A\)).
\[\text{Number of particles} = \text{Moles (mol)} \times 6.02 \times 10^{23}\]
Example: Finding the Number of Atoms
Question: How many atoms are in 0.1 moles of Gold (Au)?
\(\text{Number of atoms} = 0.1\ mol \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\) atoms.
Key Takeaway: The mole is the bridge between mass, volume, concentration, and the number of individual particles.
Advanced Stoichiometry and Formulae (Extended Content)
These calculations require you to combine the mole concept with balanced chemical equations to solve practical problems. (Supplement 5, 7, 8)
1. Reacting Masses and Limiting Reactants (Supplement 5)
Stoichiometry uses the ratios in a balanced chemical equation to relate the amounts of reactants and products.
For example, in the reaction \(2H_2 + O_2 \rightarrow 2H_2O\), the ratio is 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water.
The Moles Calculation Route (The Golden Rule)
To find the mass of an unknown substance (B) from the mass of a known substance (A):
Limiting Reactants (Supplement 5)
In most real reactions, one reactant runs out before the other. The reactant that is completely used up is called the limiting reactant.
The amount of product formed is always determined by the limiting reactant. If you run out of flour, you can’t make any more cake, even if you still have eggs!
How to identify the limiting reactant:
- Calculate the moles of ALL reactants provided.
- Use the mole ratio from the equation to see which reactant requires the least amount of the other reactant.
- The one that gives the smaller amount of product is the limiting reactant.
2. Empirical and Molecular Formulae (Supplement 7)
Definitions
The Empirical Formula is the simplest whole number ratio of the atoms (or ions) of the different elements in a compound.
The Molecular Formula is the actual number and type of atoms of each element in one molecule of the compound.
Example: Ethyne has a molecular formula of \(C_2H_2\). Its empirical formula is \(CH\) (simplest ratio 1:1).
Calculating the Empirical Formula (The MOLE method)
Given the mass or percentage composition by mass of a compound, you can find the empirical formula using this routine:
- List Elements: Write down the masses (or percentages) of each element.
- Convert to Moles: Divide each mass (g) by the element's \(A_r\) (Molar Mass). This gives the number of moles of each element.
- Find the Ratio: Divide all the mole values by the smallest mole value calculated in step 2.
- Simplify: If the ratios are whole numbers, you have the empirical formula. If not, multiply everything by a small integer (like 2 or 3) until they are whole numbers.
Finding the Molecular Formula
Once you have the empirical formula, you need the Relative Molecular Mass (\(M_r\)) of the compound to find the molecular formula.
\[\text{Multiplier} = \frac{\text{Actual } M_r \text{ (given)}}{\text{Empirical Formula Mass}}\]
Multiply the subscripts in the empirical formula by this multiplier to get the molecular formula.
3. Percentage Yield and Percentage Purity (Supplement 8)
In the real world, chemical reactions rarely produce 100% of the theoretical product. We use percentages to measure efficiency.
Percentage Yield
The theoretical yield is the mass of product calculated using stoichiometry (assuming 100% reaction and no loss). The actual yield is the mass of product obtained experimentally.
\[\text{Percentage Yield} = \frac{\text{Actual Yield (g)}}{\text{Theoretical Yield (g)}} \times 100\%\]
Reasons why yield is rarely 100%: loss during separation, reversible reactions, side reactions.
Percentage Purity and Composition
Percentage Purity measures how much of a sample is the desired compound, compared to impurities.
\[\text{Percentage Purity} = \frac{\text{Mass of pure substance}}{\text{Mass of impure sample}} \times 100\%\]
Percentage Composition by Mass (usually refers to the percentage of a specific element within a pure compound):
\[\text{\% Composition of Element X} = \frac{\text{Mass of Element X in Formula}}{\text{Total Molar Mass of Compound}} \times 100\%\]
Final Key Takeaway: The mole calculations are the toolkit for Stoichiometry. Always start by converting known quantities (mass, volume) into moles. Moles give you the crucial ratio needed to solve any problem!