The Mole and The Avogadro Constant: Chemistry's Essential Counting Tool
Welcome to perhaps the most important foundational chapter in AS Chemistry! The concepts of the mole and the Avogadro constant are absolutely central to quantitative chemistry—the part of chemistry that deals with numbers and amounts.
Why is this necessary? Atoms and molecules are tiny! If you had a glass of water, it contains more molecules than the number of grains of sand on all the beaches in the world. We need a way to count these minuscule particles in amounts large enough to work with in the lab.
The mole is that bridge! It lets us connect the mass we measure on a balance (the macroscopic world) to the number of particles (the microscopic world). Don't worry if this seems tricky at first—we will break it down using everyday analogies.
1. Building the Foundation: Defining Relative Masses (Syllabus 2.1)
Before we can define the mole, we need to understand how chemists measure the mass of individual atoms and molecules using a relative scale.
1.1 The Standard Reference: The Unified Atomic Mass Unit
Since atoms are too small to weigh directly, we compare them to a standard. The current standard is the carbon-12 atom.
- Unified atomic mass unit (u): This is defined as exactly one twelfth of the mass of a single atom of carbon-12.
Think of the unified atomic mass unit (u) as the standard "weight" used across the atomic world.
1.2 Key Definitions of Relative Mass
These terms are often confused, but they are all relative measurements (meaning they have no units). They are based on comparison with the mass of the carbon-12 atom.
1. Relative Atomic Mass (\(A_r\)):
The weighted average mass of naturally occurring atoms of an element compared to the unified atomic mass unit.
The \(A_r\) is usually not an integer (a whole number) because it accounts for the abundance of all the different isotopes of that element (e.g., Chlorine has an \(A_r\) of 35.5).
2. Relative Isotopic Mass:
The mass of a specific isotope (e.g., Chlorine-35) compared to the unified atomic mass unit.
This value is close to the nucleon number (mass number).
3. Relative Molecular Mass (\(M_r\)):
The mass of a molecule (e.g., \(\text{H}_2\text{O}\)) compared to the unified atomic mass unit.
You calculate this by summing the relative atomic masses (\(A_r\)) of all the atoms in the molecule. For \(\text{H}_2\text{O}\): \( (2 \times 1.0) + 16.0 = 18.0 \).
4. Relative Formula Mass (\(M_r\)):
Used for compounds that don't exist as discrete molecules, typically ionic compounds (e.g., \(\text{NaCl}\)). It is the sum of the relative atomic masses of the ions in the formula unit.
Note: The symbol \(M_r\) is used for both Relative Molecular Mass and Relative Formula Mass in calculations.Key Takeaway (Section 1)
All relative masses (\(A_r\), \(M_r\)) are unitless comparison values based on the mass of \(\frac{1}{12}\) of a carbon-12 atom. They represent the "weight" of the particle.
2. The Mole and The Avogadro Constant (Syllabus 2.2)
Now we can define the unit that links mass to the number of particles.
2.1 Defining the Avogadro Constant (\(L\))
The Avogadro constant (\(L\)) is simply a huge number that chemists use to count particles.
- \(L = \mathbf{6.02 \times 10^{23} \text{ particles per mole}}\)
Did you know? This number is so large that if you stacked \(6.02 \times 10^{23}\) pennies, they would cover the entire surface of the Earth to a depth of over 300 meters!
2.2 Defining the Mole (\(n\))
The mole (symbol \(\mathbf{n}\)) is the amount of substance that contains as many elementary particles (atoms, molecules, ions, or electrons) as there are atoms in exactly 12 g of carbon-12.
A simpler way to state this definition for AS Level (Syllabus 2.2):
Analogy: The Baker's Dozen
A "dozen" is 12. A baker knows that 12 eggs have a certain total mass. The mole is just the "chemist's dozen," but instead of 12, it's \(6.02 \times 10^{23}\).
Crucial Link: Molar Mass (\(M\))
When we take the unitless Relative Formula Mass (\(M_r\)) and add units of grams per mole (\(\text{g mol}^{-1}\)), we get the Molar Mass (\(M\)).
- \(M_r\) (unitless) = Molar Mass (\(\text{g mol}^{-1}\))
- Example: The \(M_r\) of \(\text{H}_2\text{O}\) is 18.0. The Molar Mass is \(18.0 \text{ g mol}^{-1}\). This means 1 mole of water has a mass of 18.0 g.
Key Takeaway (Section 2)
The mole is a counting unit (\(6.02 \times 10^{23}\)). The mass of one mole of a substance is its Molar Mass (\(M\)), which is numerically equal to its Relative Molecular/Formula Mass (\(M_r\)).
3. Mole Calculations: The Three Pathways (Syllabus 2.4)
There are three main ways to calculate the number of moles (\(n\)) depending on the state of the substance: mass (for solids/liquids), volume and concentration (for solutions), and volume, pressure, and temperature (for gases).
3.1 Pathway 1: Moles and Mass (Solids and Pure Substances)
The most fundamental relationship links moles, mass, and molar mass.
Formula:
$$ \text{Number of moles } (n) = \frac{\text{Mass } (m)}{\text{Molar Mass } (M)} $$
or
$$ n = \frac{m}{M} $$
Example Calculation (Mass)
Calculate the moles of \(\text{NaCl}\) present in 58.5 g of salt. (\(A_r\): Na=23.0, Cl=35.5)
- Determine the Molar Mass (\(M\)) of \(\text{NaCl}\): \( 23.0 + 35.5 = 58.5 \text{ g mol}^{-1} \).
- Apply the formula: \( n = 58.5 \text{ g} / 58.5 \text{ g mol}^{-1} = \mathbf{1.00 \text{ mol}} \).
3.2 Pathway 2: Moles and Concentration (Solutions)
When dealing with aqueous solutions, we use concentration (\(c\)) and volume (\(V\)). Concentration is usually measured in \(\text{mol dm}^{-3}\) (moles per cubic decimeter).
Formula:
$$ \text{Number of moles } (n) = \text{Concentration } (c) \times \text{Volume } (V) $$
or
$$ n = c \times V $$
Units Reminder: The most common mistake here is using the wrong volume units!
- Concentration (\(c\)) must be in \(\mathbf{mol \text{ dm}^{-3}}\).
- Volume (\(V\)) must be in \(\mathbf{dm^3}\).
- If given volume in \(\text{cm}^3\), convert it: \(\mathbf{\text{Volume in dm}^3 = \text{Volume in cm}^3 / 1000}\).
Example Calculation (Solutions)
Calculate the moles of \(\text{HCl}\) in 50.0 \(\text{cm}^3\) of a \(0.100 \text{ mol dm}^{-3}\) solution.
- Convert volume to \(\text{dm}^3\): \( V = 50.0 / 1000 = 0.0500 \text{ dm}^3 \).
- Apply the formula: \( n = 0.100 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = \mathbf{0.00500 \text{ mol}} \).
3.3 Pathway 3: Moles and Gas Volume (Ideal Gas Equation)
For gases, the relationship between moles, pressure, volume, and temperature is defined by the Ideal Gas Equation (assuming the gas behaves ideally—an assumption discussed further in Topic 4).
Formula:
$$ pV = nRT $$
Where:
- \(p\) = Pressure (must be in Pascals, Pa)
- \(V\) = Volume (must be in \(\mathbf{m^3}\))
- \(n\) = Moles (mol)
- \(R\) = Molar gas constant (\(8.31 \text{ J K}^{-1} \text{ mol}^{-1}\))
- \(T\) = Temperature (must be in Kelvin, K)
Unit Conversions are VITAL!
- \(\text{Temperature: } T(\text{K}) = t(^{\circ}\text{C}) + 273\)
- \(\text{Volume: } 1 \text{ m}^3 = 1000 \text{ dm}^3 = 1,000,000 \text{ cm}^3\)
- \(\text{Pressure: } 101 \text{ kPa} = 101,000 \text{ Pa}\)
Example Calculation (Gases)
Calculate the moles of gas occupying \(1.25 \text{ dm}^3\) at \(300 \text{ K}\) and \(100 \text{ kPa}\). (\(R=8.31\))
- Convert Pressure: \( p = 100 \text{ kPa} = 100,000 \text{ Pa} \).
- Convert Volume: \( V = 1.25 \text{ dm}^3 = 1.25 / 1000 = 0.00125 \text{ m}^3 \).
- Rearrange the formula for \(n\): $$ n = \frac{pV}{RT} $$
- Calculate: \( n = (100000 \times 0.00125) / (8.31 \times 300) = 125 / 2493 \approx \mathbf{0.0501 \text{ mol}} \).
Quick Review: The Three Mole Formulas
- Mass: \( n = \frac{m}{M} \) (Units: \(m\) in g, \(M\) in \(\text{g mol}^{-1}\))
- Solution: \( n = c \times V \) (Units: \(V\) in \(\text{dm}^3\), \(c\) in \(\text{mol dm}^{-3}\))
- Gas: \( pV = nRT \) (Units: \(p\) in Pa, \(V\) in \(\text{m}^3\), \(T\) in K)
4. Calculating Formulas (Syllabus 2.3)
The mole concept allows us to determine the exact ratios of atoms in a compound, which helps us find its chemical formula.
4.1 Defining Empirical and Molecular Formulas
Empirical Formula:
The simplest whole-number ratio of the atoms of each element in a compound.
Example: Ethyne (\(\text{C}_2\text{H}_2\)) has the empirical formula \(\text{CH}\).
Molecular Formula:
The actual number of atoms of each element in one molecule of the compound.
Example: The molecular formula of ethene is \(\text{C}_2\text{H}_4\).4.2 Step-by-Step Calculation of Empirical Formula
To find the empirical formula from elemental percentage composition or mass data, follow these steps:
- Mass/Percentage: Write down the mass (in grams) or percentage of each element. (If percentages are given, assume a 100 g sample.)
- Moles: Convert each mass/percentage into moles by dividing by the relative atomic mass (\(A_r\)). $$ n = m / A_r $$
- Ratio: Divide all the calculated mole values by the smallest number of moles calculated in Step 2. This gives a mole ratio.
- Whole Numbers: If the ratios are not whole numbers (e.g., 1.5 or 2.33), multiply all ratios by the smallest integer needed to turn them all into whole numbers.
Example Calculation (Empirical Formula)
A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.
| Element | Mass/100g | 2. Moles ($n = m/A_r$) | 3. Simplest Ratio (Divide by smallest \(n\)) | 4. Whole Number Ratio |
|---|---|---|---|---|
| C | 40.0 g | \(40.0 / 12.0 = 3.33\) | \(3.33 / 3.33 = 1\) | 1 |
| H | 6.7 g | \(6.7 / 1.0 = 6.7\) | \(6.7 / 3.33 = 2.01\) | 2 |
| O | 53.3 g | \(53.3 / 16.0 = 3.33\) | \(3.33 / 3.33 = 1\) | 1 |
The empirical formula is therefore \(\mathbf{\text{CH}_2\text{O}}\).
4.3 From Empirical to Molecular Formula
To find the molecular formula, you need the empirical formula and the relative molecular mass (\(M_r\)) of the compound (usually provided by mass spectrometry data).
- Calculate the mass of the empirical formula (Empirical Mass). (For \(\text{CH}_2\text{O}\), Empirical Mass = 12.0 + 2(1.0) + 16.0 = 30.0).
- Determine the ratio between the given \(M_r\) and the Empirical Mass: $$ \text{Factor} = \frac{\text{Given } M_r}{\text{Empirical Mass}} $$
- Multiply the subscripts in the empirical formula by this factor. If the given \(M_r\) was 180.0, the factor is \(180.0 / 30.0 = 6\). Molecular formula = \( (\text{CH}_2\text{O})_6 = \mathbf{\text{C}_6\text{H}_{12}\text{O}_6} \).
5. Stoichiometry, Limiting Reagents, and Yield (Syllabus 2.4)
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It relies entirely on mole ratios derived from balanced equations.
5.1 Using Stoichiometric Ratios
A balanced chemical equation tells you the ratio of moles reacting.
Example: The combustion of methane: $$ \text{CH}_4 (g) + 2\text{O}_2 (g) \longrightarrow \text{CO}_2 (g) + 2\text{H}_2\text{O} (l) $$
This equation means: 1 mole of \(\text{CH}_4\) reacts with 2 moles of \(\text{O}_2\) to produce 1 mole of \(\text{CO}_2\) and 2 moles of \(\text{H}_2\text{O}\).
The core rule for any reacting mass/volume problem is: Convert to moles first!
- Find the moles of the substance you know.
- Use the mole ratio (from the balanced equation) to find the moles of the substance you want to find.
- Convert the new moles back into mass, volume, or concentration as required by the question.
5.2 Limiting Reagents and Excess Reagents
In real laboratory reactions, we rarely mix reactants in perfect stoichiometric ratios. One reactant usually runs out first—this is the limiting reagent.
- Limiting Reagent: The reactant that is completely consumed and therefore determines the maximum amount of product that can be formed.
- Excess Reagent: The reactant present in an amount greater than required to react with all the limiting reagent.
Analogy: Making Sandwiches
If you need 2 slices of bread and 1 slice of cheese to make a sandwich, and you have 10 slices of bread and 3 slices of cheese.
How to find the limiting reagent:
- Calculate the moles of both reactants available.
- Choose one reactant (A) and use the mole ratio from the balanced equation to calculate the moles of the other reactant (B) that would be needed.
-
Compare the required moles of B with the actual moles of B available.
- If (Actual B) < (Required B), then B is the limiting reagent.
- If (Actual B) > (Required B), then A is the limiting reagent.
5.3 Percentage Yield Calculations
The amount of product calculated using stoichiometry is the maximum theoretical amount that could be produced. In reality, due to side reactions, incomplete reactions, or loss during separation, the amount obtained is usually less.
Percentage Yield compares the actual amount obtained in the lab (the actual yield) to the maximum possible amount (the theoretical yield).
Formula:
$$ \text{Percentage Yield} = \frac{\text{Actual Mass (or Moles)}}{\text{Theoretical Mass (or Moles)}} \times 100\% $$
Key Takeaway (Section 5)
Stoichiometry uses the mole ratios from balanced equations. Always identify the limiting reagent first, as it dictates the theoretical yield. Percentage yield measures efficiency.