C3 Stoichiometry: The Recipe for Chemistry
Welcome to Stoichiometry! This might sound complicated, but it is simply the study of the numerical relationship between reactants and products in a chemical reaction. Think of it as following a precise recipe. If you want to bake 10 biscuits, you know exactly how much flour, sugar, and butter you need. Stoichiometry lets chemists do the same thing for reactions!
We’ll cover three main areas: formulas and equations, calculating relative masses, and the powerhouse unit of amount—the mole.
1. Writing and Interpreting Chemical Formulas (C3.1)
1.1 Formulas and Molecular Composition
A chemical formula tells you exactly which elements and how many atoms of each are present in a compound.
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Molecular Formula: This shows the actual number and type of atoms in one molecule.
Example: Glucose is \(\text{C}_6\text{H}_{12}\text{O}_6\) (6 Carbon, 12 Hydrogen, 6 Oxygen atoms). - Elements and Compounds: Remember to state the formulas of common elements and compounds (like \(\text{H}_2\text{O}\), \(\text{O}_2\), \(\text{CO}_2\)).
Quick Tip for Formulas: Using Diagrams (Core)
You may be asked to deduce the formula of a simple molecular compound from a picture or model. Simply count the atoms! If a diagram shows one Carbon atom bonded to four Chlorine atoms, the formula is \(\text{CCl}_4\).
Deducing Ionic Formulas (Supplement)
Ionic compounds are made up of positive ions (cations) and negative ions (anions). The compound must be electrically neutral overall (charges balance to zero).
Step-by-Step Example: Finding the formula for Aluminum Oxide
- Aluminum forms a \(+3\) ion: \(\text{Al}^{3+}\)
- Oxygen forms a \(-2\) ion: \(\text{O}^{2-}\)
- To balance the total charge: We need 2 aluminum ions (\(2 \times +3 = +6\)) and 3 oxygen ions (\(3 \times -2 = -6\)).
- The resulting neutral formula is: \(\text{Al}_2\text{O}_3\).
1.2 Constructing and Balancing Equations
Chemical equations show the reactants (starting materials) turning into products (final substances).
Word Equations (Core)
These are the simplest way to show a reaction:
Example: Methane + Oxygen \(\rightarrow\) Carbon dioxide + Water
Symbol Equations and Balancing (Core)
A symbol equation uses chemical formulas. It must be balanced, meaning the number of atoms of each element must be the same on both the reactant side (left) and the product side (right).
Analogy: The Conservation Law Balancing equations respects the law of conservation of mass: atoms are never created or destroyed in a chemical reaction, only rearranged.
Step-by-Step Balancing Example:
Reaction: \(\text{Fe} + \text{Cl}_2 \rightarrow \text{FeCl}_3\)
- Count atoms: Fe (1 L, 1 R), Cl (2 L, 3 R). Chlorine is unbalanced.
- Find the lowest common multiple for Cl (2 and 3 is 6).
- Place coefficients: \(2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3\)
- Check atoms: Fe (2 L, 2 R), Cl (6 L, 6 R). It's balanced!
State Symbols (Core and Supplement)
We use state symbols to show the physical state of each substance in the reaction:
- \((s)\): Solid
- \((l)\): Liquid
- \((g)\): Gas
- \((aq)\): Aqueous solution (dissolved in water)
Balanced Example with State Symbols:
\(2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)\)
Ionic Equations (Supplement)
Ionic equations show only the particles (ions or molecules) that actually participate in the reaction. Ions that stay unchanged throughout the reaction are called spectator ions and are left out.
Example: Precipitation reaction between Silver Nitrate and Sodium Chloride:
- Full balanced equation: \(\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)\)
- Separate into ions: \(\text{Ag}^+(aq) + \text{NO}_3^-(aq) + \text{Na}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) + \text{Na}^+(aq) + \text{NO}_3^-(aq)\)
- Remove spectator ions (\(\text{Na}^+\) and \(\text{NO}_3^-\)).
- Ionic Equation: \(\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\)
Key Takeaway: Formulas are how we name substances; balanced equations (with state symbols) are how we describe chemical change numerically.
2. Relative Masses (C3.2)
We need a standardized way to compare the masses of atoms and molecules.
2.1 Relative Atomic Mass (\(A_r\)) (Core)
The Relative Atomic Mass (\(A_r\)) is the average mass of the isotopes of an element compared to 1/12th of the mass of one atom of Carbon-12.
- It is an average because most elements have isotopes (atoms with different numbers of neutrons).
- You generally find the $A_r$ value on the Periodic Table (usually the larger number).
2.2 Relative Molecular Mass (\(M_r\)) and Relative Formula Mass (Core)
The Relative Molecular Mass (\(M_r\)) (or Relative Formula Mass for ionic compounds) is the sum of the relative atomic masses ($A_r$) of all the atoms in the formula.
Calculating \(M_r\) Step-by-Step Example:
Calculate the \(M_r\) of \(\text{Water}\) (\(\text{H}_2\text{O}\)). (Given: \(\text{H}=1\), \(\text{O}=16\))
- Hydrogen: \(2 \times 1 = 2\)
- Oxygen: \(1 \times 16 = 16\)
- \(M_r(\text{H}_2\text{O}) = 2 + 16 = 18\)
Did you know? We use Relative Formula Mass (\(M_r\)) for ionic compounds like \(\text{NaCl}\) because they don't form distinct molecules, but rather continuous giant lattices.
2.3 Core Calculation: Reacting Masses in Simple Proportions
Before introducing the mole concept (which is often Supplement content), Core calculations focus on simple scaling based on known mass data.
Example: If $10.0 \text{g}$ of Magnesium reacts completely with acid to produce $0.8 \text{g}$ of Hydrogen gas. How much Hydrogen gas would be produced if $25.0 \text{g}$ of Magnesium was used?
This is a simple ratio calculation:
$$\frac{\text{Mass of Mg (New)}}{\text{Mass of Mg (Old)}} = \frac{\text{Mass of H}_2\text{ (New)}}{\text{Mass of H}_2\text{ (Old)}}$$
$$\frac{25.0 \text{g}}{10.0 \text{g}} = \frac{X \text{g}}{0.8 \text{g}}$$
$$X = 2.5 \times 0.8 = 2.0 \text{g}$$
Key Takeaway: \(A_r\) is for single atoms (averaged), and \(M_r\) is the sum of these masses for a compound. Core calculations use ratios of masses.
3. The Mole and Calculations (C3.3)
The concept of the mole is crucial for linking the microscopic world of atoms to the macroscopic measurements we make in the lab (like grams or cubic decimeters).
3.1 The Mole and Avogadro Constant (Supplement)
The mole (\(\text{mol}\)) is the unit of amount of substance.
Analogy: The Chemist’s Dozen Just like a dozen is always 12 items, a mole is always a specific number of particles.
One mole of any substance contains \((6.02 \times 10^{23})\) particles (atoms, ions, or molecules). This massive number is called the Avogadro constant.
The mass of one mole of a substance is called the Molar Mass. Numerically, the molar mass (in \(\text{g/mol}\)) is equal to the Relative Molecular Mass (\(M_r\)) or Relative Atomic Mass (\(A_r\)).
Example: If \(M_r(\text{H}_2\text{O}) = 18\), then the Molar Mass of water is \(18 \text{ g/mol}\).
3.2 The Mole Calculation Formula (Supplement)
The most important relationship in stoichiometry links mass and moles:
$$\text{Amount of substance (mol)} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}$$
You can rearrange this equation to find mass or molar mass, too.
Memory Aid: The Mole Triangle
Imagine a triangle with "Mass (g)" at the top, and "Mol" and "Molar Mass" at the bottom. Cover the quantity you want to find!
3.3 Calculations Involving Solutions (Core)
When substances are dissolved in water, we measure their amount using concentration.
Concentration is measured in grams per cubic decimeter (\(\text{g/dm}^3\)), which tells you how many grams of solute are dissolved in 1 \(\text{dm}^3\) (or $1000 \text{ cm}^3$) of solution.
3.4 Molar Gas Volume (Supplement)
For gases, Avogadro's Law states that equal volumes of all gases (at the same temperature and pressure) contain equal numbers of molecules.
At room temperature and pressure (r.t.p.), one mole of any gas occupies a volume of \(24 \text{ dm}^3\).
Step-by-Step Gas Volume Example: If you produce $0.5 \text{ mol}$ of \(\text{CO}_2\) gas at r.t.p., what volume does it occupy?
$$\text{Volume (dm}^3) = \text{Moles} \times 24 \text{ dm}^3/\text{mol}$$ $$\text{Volume} = 0.5 \text{ mol} \times 24 \text{ dm}^3/\text{mol} = 12 \text{ dm}^3$$
Conversion Tip: $1 \text{ dm}^3 = 1000 \text{ cm}^3$. Be careful with units in your exam!
3.5 Stoichiometric Calculations (Supplement)
These complex calculations use the mole concept and balanced equations to find out exactly how much product you get from specific amounts of reactants (or vice versa).
The Key Link: Mole Ratio
The coefficients (the large numbers) in a balanced equation represent the ratio of moles of reactants and products.
Example: \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\) The mole ratio of \(\text{H}_2\) : \(\text{O}_2\) : \(\text{H}_2\text{O}\) is $2 : 1 : 2$.
General Procedure for Mass/Volume Calculations:
- Write and Balance the equation.
- Convert the given mass (or volume) of the known substance into moles.
- Use the Mole Ratio (from the balanced equation) to find the moles of the unknown substance.
- Convert the moles of the unknown substance back into mass (or volume).
Don't worry if this seems tricky at first! Practice is key. Always use the balanced equation to find the mole ratio—it's the heart of stoichiometry.
3.6 Limiting Reactants (Supplement)
In a real-life reaction, we rarely use exactly the right ratio of reactants. One reactant runs out before the other—this is the limiting reactant.
- The limiting reactant determines the maximum amount of product that can be formed.
- The other reactant is in excess.
Analogy: Making Sandwiches If you have 10 slices of bread (5 pairs) and 7 slices of cheese, cheese is the limiting reactant. You can only make 7 cheese sandwiches, even though you have leftover bread.
To identify the limiting reactant, convert both reactants to moles and compare the mole ratio needed (from the equation) versus the mole ratio available.
Quick Review: Stoichiometry Checklist
Formulas (C3.1): Know how to count atoms and balance charges (ionic).
Masses (C3.2): Understand $A_r$ and $M_r$. Be ready for simple scaling calculations (Core).
Moles (C3.3): The link between mass and amount. \(\text{mol} = \text{mass} / \text{Molar Mass}\).
Gases (C3.3): 1 mol of gas is \(24 \text{ dm}^3\) at r.t.p.
Calculations (C3.3): Always go via the mole ratio from the balanced equation!