Chemistry (9701) Study Notes: Chapter 2.3 Formulas
Welcome to the Formulas chapter! If you sometimes find writing chemical formulas confusing, you are definitely not alone. This chapter is the absolute backbone of Chemistry. Everything we do—from calculating moles and balancing equations to predicting reaction products—depends on getting the formula right. Think of formulas as the chemical language; once you master the grammar here, you can speak Chemistry fluently!
In these notes, we will break down how to confidently write formulas for both simple and complex ions, understand the difference between empirical and molecular structures, and master writing balanced chemical equations.
Section 1: Writing Formulas for Ionic Compounds (The Charge Balancing Game)
Ionic compounds are formed from the electrostatic attraction between positive ions (cations) and negative ions (anions). The golden rule is simple: the total positive charge must exactly cancel out the total negative charge.
1.1 Determining Ionic Charges
The charge on a simple ion is determined by its position in the Periodic Table:
- Group 1: Form \(+1\) ions (e.g., \(\text{Na}^+\))
- Group 2: Form \(+2\) ions (e.g., \(\text{Mg}^{2+}\))
- Group 13 (Aluminium): Forms \(+3\) ions (e.g., \(\text{Al}^{3+}\))
- Group 16 (Non-metals): Form \(-2\) ions (e.g., \(\text{O}^{2-}\), \(\text{S}^{2-}\))
- Group 17 (Halogens): Form \(-1\) ions (e.g., \(\text{Cl}^-\), \(\text{Br}^-\))
- Transition Metals: Often have variable oxidation numbers (charges), which are usually indicated by a Roman numeral (e.g., Iron(III) means \(\text{Fe}^{3+}\)).
Analogy: The Dating Profile
Imagine ions are dating. A guy with a \(+2\) charge needs a partner (or partners) whose total charge is \(-2\) so they become chemically neutral (a stable relationship!).
Example: Magnesium Oxide
Magnesium (\(\text{Mg}\)) is Group 2, so it forms \(\text{Mg}^{2+}\). Oxygen (\(\text{O}\)) is Group 16, so it forms \(\text{O}^{2-}\).
\(+2\) and \(-2\) cancel out perfectly. The formula is \(\text{MgO}\).
Example: Magnesium Chloride
Magnesium is \(\text{Mg}^{2+}\). Chlorine (\(\text{Cl}\)) is Group 17, so it forms \(\text{Cl}^-\).
You need two \(\text{Cl}^-\) ions to balance the single \(\text{Mg}^{2+}\).
Formula: \(\text{MgCl}_2\).
1.2 Recalling Key Complex Ions (Syllabus Checklist)
You must recall the names and formulas for several key polyatomic ions. These ions stick together and carry their charge as a unit.
Important Polyatomic Ions to Memorise:
- Nitrate: \(\text{NO}_3^-\)
- Carbonate: \(\text{CO}_3^{2-}\)
- Sulfate: \(\text{SO}_4^{2-}\)
- Hydroxide: \(\text{OH}^-\)
- Ammonium: \(\text{NH}_4^+\) (This is the only common positive polyatomic ion you need)
- Phosphate: \(\text{PO}_4^{3-}\)
- Hydrogen Carbonate (Bicarbonate): \(\text{HCO}_3^-\)
You also need to recall the charges for these common metal ions:
- Zinc: \(\text{Zn}^{2+}\)
- Silver: \(\text{Ag}^+\)
Trick for Writing Formulas with Polyatomic Ions
If you need more than one polyatomic ion to balance the charge, you must use brackets around the ion.
Example: Calcium Hydroxide
Calcium is \(\text{Ca}^{2+}\). Hydroxide is \(\text{OH}^-\).
We need two \(\text{OH}^-\) groups to balance the \(\text{Ca}^{2+}\).
Formula: \(\text{Ca}(\text{OH})_2\). (If you wrote \(\text{CaO}_2\text{H}_2\), you would be wrong!)
1. Identify the charge on the cation and the anion.
2. Swap the numbers (ignoring the signs) to get the subscripts.
3. If the numbers are the same, they cancel out (e.g., \(\text{Ca}^{2+}\) and \(\text{CO}_3^{2-} \rightarrow \text{CaCO}_3\)).
4. Use brackets if you need more than one polyatomic ion.
Section 2: Empirical and Molecular Formulas
When we talk about formulas, we often use two distinct types: the Empirical Formula and the Molecular Formula. These terms are essential for stoichiometry calculations.
2.1 Defining the Terms (Syllabus 2.3.3)
- Empirical Formula (EF):
The formula showing the simplest whole number ratio of atoms of each element present in a compound. - Molecular Formula (MF):
The formula showing the actual number of atoms of each element present in a molecule.
Analogy: Blueprints vs. Finished House
The Empirical Formula is like the simplest architectural blueprint showing that for every one bedroom, there is one bathroom (1:1 ratio). The Molecular Formula is the actual house built, revealing there are two bedrooms and two bathrooms (\(\text{B}_2\text{Ba}_2\)).
Example: Ethane and Ethene
Ethene: Molecular Formula \(\text{C}_2\text{H}_4\). Ratio is 2:4, which simplifies to 1:2. Empirical Formula is \(\text{CH}_2\).
Benzene: Molecular Formula \(\text{C}_6\text{H}_6\). Ratio is 6:6, which simplifies to 1:1. Empirical Formula is \(\text{CH}\).
Did you know? Ionic compounds, like \(\text{NaCl}\), only have an empirical formula because they don't exist as discrete molecules. The formula \(\text{NaCl}\) simply represents the simplest ratio of ions in the giant lattice structure.
2.2 Calculating Empirical and Molecular Formulas (Syllabus 2.3.5)
This is a common exam calculation. We convert the mass or percentage composition into moles to find the ratio.
Step-by-Step Guide to Calculating the Empirical Formula (EF)
Let's say a compound contains 40.0% Carbon (C), 6.7% Hydrogen (H), and 53.3% Oxygen (O). (Assume 100 g sample, so percentages are masses in grams).
Step 1: Divide Mass by Molar Mass (\(A_r\)) to find Moles.
(Molar masses: C = 12.0, H = 1.0, O = 16.0)
- Carbon: \(\text{Moles} = \frac{40.0}{12.0} = 3.33\)
- Hydrogen: \(\text{Moles} = \frac{6.7}{1.0} = 6.7\)
- Oxygen: \(\text{Moles} = \frac{53.3}{16.0} = 3.33\)
Step 2: Find the Simplest Ratio.
Divide all the mole values by the smallest number of moles calculated (which is 3.33).
- Carbon: \(\frac{3.33}{3.33} = 1.0\)
- Hydrogen: \(\frac{6.7}{3.33} \approx 2.0\)
- Oxygen: \(\frac{3.33}{3.33} = 1.0\)
Step 3: Write the Empirical Formula.
The simplest whole number ratio is C:H:O = 1:2:1.
The Empirical Formula is \(\text{CH}_2\text{O}\).
!! Important Note for Step 2 !!
If your ratio does not result in whole numbers (e.g., you get 1:1.5), you must multiply ALL ratios by a factor (in this case, 2) until they are all whole numbers (1:1.5 becomes 2:3). Never round numbers like 1.5 or 0.33.
Calculating the Molecular Formula (MF)
To find the MF, you need the EF and the Relative Molecular Mass (\(M_r\)) of the compound (given in the question).
Step 1: Calculate the Mass of the Empirical Formula (\(M_{\text{EF}}\)).
For our example EF (\(\text{CH}_2\text{O}\)): \(M_{\text{EF}} = 12.0 + (1.0 \times 2) + 16.0 = 30.0\)
Step 2: Determine the Scaling Factor (\(n\)).
Let's assume the experimental Relative Molecular Mass (\(M_r\)) is 60.0.
Factor \(n = \frac{\text{Relative Molecular Mass} (M_r)}{\text{Empirical Formula Mass} (M_{\text{EF}})}\)
$$n = \frac{60.0}{30.0} = 2$$
Step 3: Multiply the Empirical Formula by the Factor (\(n\)).
$$ \text{MF} = n \times \text{EF} = 2 \times (\text{CH}_2\text{O}) = \text{C}_2\text{H}_4\text{O}_2 $$
Empirical formulas define the ratio; molecular formulas define the reality. The conversion requires the overall molar mass (\(M_r\)).
Section 3: Hydrated Salts and Water of Crystallisation (Syllabus 2.3.4)
Many salts, when crystallised from aqueous solutions, incorporate water molecules into their lattice structure. These are called hydrated salts.
3.1 Key Definitions
- Hydrated Compound/Salt: A salt crystal that contains water molecules chemically bound within its structure.
- Water of Crystallisation: The specific number of water molecules bound to each formula unit of the salt.
- Anhydrous Compound/Salt: The compound remaining after all the water of crystallisation has been removed, usually by heating. (Anhydrous means "without water").
We represent hydrated salts by adding the water molecules to the formula, separated by a dot. For example, hydrated copper(II) sulfate is \(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\), where \(x\) is the number of moles of water of crystallisation.
3.2 Determining the Value of \(x\)
We find \(x\) experimentally by heating the hydrated salt until all the water evaporates, leaving the anhydrous salt.
Step-by-Step Example (Determining \(x\) in \(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\))
Assume you start with 5.00 g of hydrated salt and end with 3.20 g of anhydrous \(\text{CuSO}_4\). (Molar masses: \(\text{CuSO}_4 = 159.6\), \(\text{H}_2\text{O} = 18.0\))
Step 1: Calculate the Mass of Water and Anhydrous Salt.
- Mass of anhydrous \(\text{CuSO}_4 = 3.20 \text{ g}\)
- Mass of water removed: \(5.00 \text{ g} - 3.20 \text{ g} = 1.80 \text{ g}\)
Step 2: Convert Mass to Moles.
- Moles of \(\text{CuSO}_4 = \frac{3.20 \text{ g}}{159.6 \text{ g mol}^{-1}} \approx 0.0200 \text{ mol}\)
- Moles of \(\text{H}_2\text{O} = \frac{1.80 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.100 \text{ mol}\)
Step 3: Find the Mole Ratio (\(x\)).
Divide the moles of water by the moles of the anhydrous salt.
$$x = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{CuSO}_4} = \frac{0.100}{0.0200} = 5$$
The formula is \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) (Copper(II) sulfate pentahydrate).
Common Mistake to Avoid
When heating the salt, ensure you heat it gently until the mass is constant. If you heat it too strongly (especially salts like calcium carbonate), you risk decomposing the anhydrous salt itself (e.g., \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)), leading to an artificially high mass loss and an incorrect value for \(x\).
Section 4: Writing and Balancing Chemical Equations
Chemical equations represent chemical reactions. A correct equation must satisfy two criteria: it must be balanced (Law of Conservation of Mass) and it must contain appropriate state symbols.
4.1 Balancing Equations (Syllabus 2.3.2a)
Balancing means adjusting the coefficients (the large numbers in front of the formulas) so that the number of atoms of each element is the same on both the reactant side and the product side.
Example: Burning Propane (\(\text{C}_3\text{H}_8\))
Unbalanced: \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\)
Step 1: Balance C. 3 C on the left, so put 3 in front of \(\text{CO}_2\).
$$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}$$
Step 2: Balance H. 8 H on the left, so put 4 in front of \(\text{H}_2\text{O}\) (\(4 \times 2 = 8\)).
$$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$
Step 3: Balance O. Count O atoms on the product side: (\(3 \times 2\)) + (\(4 \times 1\)) = 10 O atoms. Need 10 O atoms on the left, so put 5 in front of \(\text{O}_2\).
Balanced: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\)
4.2 Using State Symbols (Syllabus 2.3.2b)
State symbols tell us the physical state of the substance under reaction conditions. They are crucial for writing equations, especially in thermodynamics and kinetics.
- \((s)\): Solid (e.g., \(\text{NaCl}(s)\))
- \((l)\): Liquid (e.g., \(\text{H}_2\text{O}(l)\))
- \((g)\): Gas (e.g., \(\text{CO}_2(g)\))
- \((aq)\): Aqueous (dissolved in water) (e.g., \(\text{HCl}(aq)\))
4.3 Writing Ionic Equations (Syllabus 2.3.2a)
Ionic equations focus only on the particles that actually participate in the reaction. This is usually applied to reactions in aqueous solution, like precipitation or neutralisation.
The Spectator Ion Concept
Spectator ions are ions that are present in the solution but do not change state or chemical form during the reaction. They are like the audience watching a sports match—they are there, but they don't play!
Step 1: Write the full balanced equation (with state symbols).
Example: Reaction between aqueous Lead(II) Nitrate and aqueous Potassium Iodide. A yellow precipitate of Lead(II) Iodide forms.
$$\text{Pb}(\text{NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$$
Step 2: Dissociate all aqueous (aq) species into ions.
$$\text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq) + 2\text{K}^+(aq) + 2\text{I}^-(aq) \rightarrow \text{PbI}_2(s) + 2\text{K}^+(aq) + 2\text{NO}_3^-(aq)$$
(Note: solids, liquids, and gases stay together.)
Step 3: Identify and cancel out the spectator ions.
The ions that appear identically on both sides are \(\text{K}^+(aq)\) and \(\text{NO}_3^-(aq)\).
Step 4: Write the net ionic equation.
$$\text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \rightarrow \text{PbI}_2(s)$$
Always balance atoms first. Use state symbols (s, l, g, aq). For ionic equations, only include the reactants that form a precipitate, gas, or liquid product—exclude the spectators!
Summary Table: Formulas and Equations
| Concept | Definition/Goal | Crucial Skill |
| Ionic Formula | Compound where charges balance to zero. | Recalling complex ion charges (e.g., \(\text{SO}_4^{2-}\), \(\text{NH}_4^+\)). |
| Empirical Formula (EF) | Simplest whole number ratio of atoms. | Converting mass/percentage to moles and finding the ratio. |
| Molecular Formula (MF) | Actual number of atoms in a molecule. | Using \(M_r\) to find the scaling factor (\(n\)). |
| Hydrated Salt | Salt containing water of crystallisation (\(\cdot x\text{H}_2\text{O}\)). | Calculating \(x\) by finding the mole ratio of water lost to anhydrous salt remaining. |
| Ionic Equation | Equation showing only species involved in change. | Identifying and removing spectator ions (those in the \((aq)\) state on both sides). |