Atoms, molecules and stoichiometry - Chemistry (9701) - Cambridge A-Level

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AS Level Physical Chemistry: Chapter 2 – Atoms, Molecules and Stoichiometry

Welcome to the most important chapter in Chemistry! Stoichiometry (don't worry, you just need to learn how to calculate it, not pronounce it perfectly!) is the backbone of all chemical calculations.
This is where we learn to count atoms and molecules, measure them, and predict exactly how much product we can make from a given amount of reactant. If you can master this chapter, you can handle almost any calculation question in AS Chemistry!

2.1 Relative Masses of Atoms and Molecules

Atoms are tiny, so we use special scales called relative masses to compare them. These masses are all measured against a universal standard: the $\text{Carbon-12}$ isotope.

Key Definitions Based on Carbon-12

Unified Atomic Mass Unit (u.a.m.u.): Defined as one twelfth of the mass of a carbon-12 atom. $1\text{ u} \approx 1.66 \times 10^{-27}\text{ kg}$.
Relative Isotopic Mass: The mass of a specific isotope of an element compared to 1/12th the mass of a $\text{Carbon-12}$ atom.
Relative Atomic Mass ($A_r$): The weighted average mass of an atom of an element compared to 1/12th the mass of a $\text{Carbon-12}$ atom.
Why "weighted average"? Because elements exist as a mixture of isotopes (e.g., $\text{Cl-35}$ and $\text{Cl-37}$). The $A_r$ accounts for the natural abundance of each isotope.
Relative Molecular Mass ($M_r$): The sum of the $A_r$ values of all atoms shown in the molecular formula (used for simple covalent molecules like $\text{CO}_2$ or $\text{H}_2\text{O}$).
Relative Formula Mass ($M_r$): The sum of the $A_r$ values of all atoms shown in the formula (used for ionic compounds like $\text{NaCl}$ or giant structures, where molecules don't technically exist).

Quick Review Box: The Relative Rule

Relative masses are ratios, so they have no units. They are a comparison to the $\text{C-12}$ standard.

Key Takeaway 2.1

Relative masses allow us to use simple, unitless numbers ($A_r$, $M_r$) to compare the heavy masses of atoms and compounds, using $\text{Carbon-12}$ as the baseline.

2.2 The Mole and the Avogadro Constant

You can't count atoms one by one. Imagine trying to count grains of sand in a bucket! Instead, chemists use a gigantic counting unit called the mole.

Defining the Mole

The mole ($n$) is the amount of substance that contains the same number of particles (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of $\text{Carbon-12}$.

This number is called the Avogadro constant ($L$), and its value is massive: $$L = 6.02 \times 10^{23} \text{ particles/mol}$$

Analogy: Just as a "dozen" always means 12, a "mole" always means $6.02 \times 10^{23}$ particles.

Using Moles to Find Mass

The brilliance of the mole concept is the relationship between the relative mass ($M_r$) and the molar mass.

If the $M_r$ of a substance is $X$, then its Molar Mass (mass of 1 mole) is $X$ grams.

The $M_r$ of $\text{H}_2\text{O}$ is 18.0. The molar mass is 18.0 g mol$\text{mol}^{-1}$.

The fundamental calculation linking mass ($m$), molar mass ($M$), and moles ($n$) is: $$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \quad \text{ or } \quad n = \frac{m}{M}$$

Memory Aid: The Triangle of Mass

Draw a triangle with $m$ (mass) at the top, and $n$ (moles) and $M$ (molar mass) at the bottom. Cover the quantity you want to find.

Key Takeaway 2.2

The mole links the microscopic world (atoms and molecules) to the macroscopic world (grams and litres) using the Avogadro constant.

2.3 Chemical Accounting: Formulas and Equations

To perform any calculation, we first need to write the chemical language correctly: the formulas and equations.

Writing Chemical Formulas

You must be able to write formulas, especially for ionic compounds, by balancing the charges.

Predicting Ionic Charge: Metals form positive ions (cations) based on their group number (e.g., Group 1 elements form +1 ions). Non-metals form negative ions (anions) based on the number of electrons needed to achieve a noble gas configuration (e.g., Group 17 forms -1 ions).
Recall Common Polyatomic Ions: Don't worry if this seems tricky at first, you will learn these through practice!
Anions: $\text{Nitrate}$ ($\text{NO}_3^-$), $\text{Carbonate}$ ($\text{CO}_3^{2-}$), $\text{Sulfate}$ ($\text{SO}_4^{2-}$), $\text{Hydroxide}$ ($\text{OH}^-$), $\text{Hydrogen carbonate}$ ($\text{HCO}_3^-$), $\text{Phosphate}$ ($\text{PO}_4^{3-}$).
Cations: $\text{Ammonium}$ ($\text{NH}_4^+$), $\text{Zinc}$ ($\text{Zn}^{2+}$), $\text{Silver}$ ($\text{Ag}^{+}$).

Example: Aluminium sulfate. Aluminium is $\text{Al}^{3+}$, Sulfate is $\text{SO}_4^{2-}$. To balance the charges (total charge must be zero): 2 x (+3) + 3 x (-2) = 0. Formula: $\text{Al}_2(\text{SO}_4)_3$.

Formulas: Empirical vs. Molecular

Empirical Formula: The simplest whole number ratio of atoms of each element in a compound. (The minimal recipe.)
Molecular Formula: The actual number of atoms of each element in one molecule of the compound. (The full recipe.)

Did you know? Ethene ($\text{C}_2\text{H}_4$) and cyclobutane ($\text{C}_4\text{H}_8$) have different molecular formulas but share the same empirical formula: $\text{CH}_2$.

Calculating Empirical and Molecular Formulas (Step-by-Step)

Percentage to Mass: Assume you have 100 g of the substance, converting percentages directly into mass in grams.
Mass to Moles: Divide the mass of each element by its $A_r$ to find the number of moles ($n = m/M$).
Ratio (Moles to Simplest Ratio): Divide all the mole values by the smallest mole value calculated in step 2. This gives the subscripts for the empirical formula.
Molecular Formula (if required): Calculate the mass of the empirical unit. Divide the given $M_r$ of the compound by the mass of the empirical unit. This gives you the multiplication factor.
$$ \text{Factor} = \frac{\text{Relative Molecular Mass}}{\text{Empirical Formula Mass}} $$

Chemical Equations and State Symbols

Equations must be balanced to obey the law of conservation of mass (atoms are neither created nor destroyed).

You must use appropriate state symbols:

(s) for solid
(l) for liquid
(g) for gas
(aq) for aqueous solution (dissolved in water)

Ionic Equations (The Real Action)

Ionic equations show only the particles that are directly involved in the reaction.

Spectator Ions are ions that remain unchanged in the reaction mixture (they "watch" but don't participate) and must be omitted from the final ionic equation.

Example: $\text{AgNO}_3(\text{aq}) + \text{NaCl}(\text{aq}) \rightarrow \text{AgCl}(\text{s}) + \text{NaNO}_3(\text{aq})$
Full ionic: $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq}) + \text{Na}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s}) + \text{Na}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$
Ionic (excluding $\text{Na}^+$ and $\text{NO}_3^-$ spectators): $$\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$$

Hydrated, Anhydrous, and Water of Crystallisation

Anhydrous: A substance containing no water (e.g., anhydrous $\text{CuSO}_4$ is white).
Hydrated: A substance that contains water of crystallisation chemically bound within its crystal structure (e.g., hydrated $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ is blue).
Water of Crystallisation: The specific number of water molecules bound per formula unit of a salt.

Key Takeaway 2.3

Formulas define composition (empirical is ratio, molecular is actual count). Equations represent reactions, must be balanced, and state symbols and ionic equations clarify the participating species.

2.4 Reacting Masses and Volumes (Stoichiometric Calculations)

This section is about applying the mole concept to real chemical processes to calculate masses, volumes, and concentrations. All calculations rely entirely on the mole ratio from the balanced equation.

2.4.1 Reacting Masses and Percentage Yield

The General Mole Calculation Pathway

If you are given Mass A, and want to find Mass B (using the balanced equation $\text{A} \rightarrow \text{B}$):

Mass A to Moles A: $n(\text{A}) = m(\text{A}) / M(\text{A})$.
Moles A to Moles B: Use the mole ratio from the balanced equation. If $\text{A}$ reacts 2:1 with $\text{B}$, then $n(\text{B}) = n(\text{A}) / 2$.
Moles B to Mass B: $m(\text{B}) = n(\text{B}) \times M(\text{B})$.

Percentage Yield

In real life, reactions never achieve 100% efficiency due to side reactions or loss of product.

The Theoretical Yield is the maximum possible mass calculated from stoichiometry.
The Actual Yield is the mass obtained in the experiment.

$$ \text{Percentage Yield} = \frac{\text{Actual Mass Yield}}{\text{Theoretical Mass Yield}} \times 100\% $$

2.4.2 Volumes and Concentrations of Solutions

For solutions, we link moles to volume ($V$) and concentration ($c$) using the relationship:

$$\text{Moles} = \text{Concentration} \times \text{Volume}$$ $$n = c \times V$$

Concentration ($c$): Measured in $\text{mol dm}^{-3}$ (moles per cubic decimetre, or molarity, $\text{M}$).
Volume ($V$): Must be measured in $\text{dm}^3$ (litres) for this formula.
Remember: $1 \text{ dm}^3 = 1000 \text{ cm}^3$. To convert $\text{cm}^3$ to $\text{dm}^3$, divide by 1000.

2.4.3 Volumes of Gases and the Ideal Gas Equation

Gases are special because 1 mole of any ideal gas occupies the same volume under the same conditions (temperature and pressure).

The relationship governing the behaviour of ideal gases is the Ideal Gas Equation: $$pV = nRT$$

$p$: Pressure (in Pascals, Pa)
$V$: Volume (in $\text{m}^3$)
$n$: Amount of substance (in moles, mol)
$R$: Ideal gas constant ($8.31 \text{ J K}^{-1} \text{mol}^{-1}$)
$T$: Temperature (in Kelvin, K)

Crucial Conversion Reminder:
For the ideal gas equation, you MUST use SI units:

Pressure: $\text{kPa}$ must be converted to $\text{Pa}$ ($\times 1000$).
Volume: $\text{dm}^3$ must be converted to $\text{m}^3$ ($\div 1000$).
Temperature: $\text{}^{\circ}\text{C}$ must be converted to $\text{K}$ ($+ 273$).

2.4.4 Limiting and Excess Reagents

When mixing reactants, we rarely use exactly the stoichiometric ratio. One reactant will run out first, stopping the reaction—this is the limiting reagent.

Limiting Reagent: The reactant that is completely consumed and determines the maximum amount of product that can be formed (the theoretical yield).
Excess Reagent: The reactant that is left over after the reaction is complete.

Step-by-Step: Identifying the Limiting Reagent

Convert both reactant masses/volumes into moles.
Find the required ratio: Look at the balanced equation to find the required mole ratio (e.g., $A:B$ is 1:2).
Test the ratios: Pick one reactant (A) and calculate how many moles of the other reactant (B) you would need to react completely.
Compare: If you have more moles of B than required, B is in excess (A is limiting). If you have less moles of B than required, B is limiting.

Common Mistake to Avoid (Significant Figures)

In all calculations (2.4.1-2.4.4), your final answer should reflect the minimum number of significant figures given in the data used in the question. If the data provides 3 significant figures, your answer should generally be given to 3 or 4 significant figures, unless specified otherwise.

Key Takeaway 2.4

Stoichiometry is the tool kit for quantitative chemistry. Always start by finding the number of moles ($n = m/M$ or $n = cV$ or $n = pV/RT$) and use the mole ratio from the balanced equation to move between reactants and products.