Ch 3 · Formulae · AS & A Level Chemistry 9701
Topic 3 of 4 in Atoms, molecules and stoichiometry — AS & A Level Chemistry 9701 — 55 lessons.
Introduction
Every quantitative calculation in chemistry starts from a correctly written formula or a correctly balanced equation. This lesson covers writing ionic formulae from charges, recalling common polyatomic ions, constructing balanced (including ionic) equations, and calculating empirical and molecular formulae, together with the language of hydrated compounds.
Key ideas — Predicting ionic charge and writing ionic formulae
An element's typical ionic charge can be predicted from its position in the Periodic Table: Group 1 metals form 1+ ions, Group 2 metals form 2+ ions, Group 13 (e.g. aluminium) forms 3+ ions, Group 16 non-metals form 2- ions, and Group 17 non-metals (halogens) form 1- ions, since each element gains or loses the number of electrons needed to achieve a full outer shell (noble gas configuration). To write the formula of an ionic compound, the charges of the cation and anion must balance to give an overall neutral formula unit: the number of each ion is chosen so that total positive charge equals total negative charge. For example, calcium (mathrmCa^2+) with chloride (Cl^-) requires two chloride ions per calcium ion, giving CaCl_2; aluminium (mathrmAl^3+) with oxide (mathrmO^2-) requires the lowest common multiple of 3 and 2, giving Al_2O_3.
🧑🏫 Teacher's explanation
Think of ionic compounds like a puzzle where the pieces must fit perfectly so there are no 'loose ends.' In chemistry, 'loose ends' are unbalanced electrical charges. Every atom wants to reach a stable, happy state, which usually means having a full outer shell of electrons, much like how you might want a full set of matching tools in a kit.
📖 Definitions
🔍 Worked example — Easy
Predict the formula for the ionic compound formed between Magnesium (Mg^2+) and Nitrogen (N^3-).
🔍 Worked example — Standard
Write the formula for the compound formed between Aluminium (Al^3+) and Oxygen (O^2-).
🔍 Worked example — Exam-style
A compound is formed from Sodium (Na^+) and the polyatomic ion Sulfate (SO_4^2-). Write its formula.
⚠️ Common errors
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🎯 Exam question
Key ideas — Common polyatomic ions
Certain groups of covalently bonded atoms carry an overall charge and must be recalled with their correct formula and charge: nitrate NO_3^-, sulfate mathrmSO_4^2-, carbonate mathrmCO_3^2-, hydroxide OH^-, ammonium NH_4^+, and hydrogencarbonate HCO_3^-. When more than one polyatomic ion is needed in a formula, the ion (not just one atom of it) is enclosed in brackets before the subscript, e.g. calcium nitrate is Ca(NO_3)_2, not mathrmCaNO_32 or CaN_2O_6 written without brackets.
🧑🏫 Teacher's explanation
Think of polyatomic ions as 'super-atoms'. Instead of a single atom acting as a building block, a group of atoms is covalently bonded together so tightly that they act as a single unit with a single charge. Imagine a Lego brick that has been glued to three smaller bricks; you can't pull them apart easily, so you treat the whole assembly as one single piece when building something larger.
📖 Definitions
🔍 Worked example — Easy
Write the formula for Magnesium hydroxide.
🔍 Worked example — Standard
Write the formula for Ammonium sulfate.
🔍 Worked example — Exam-style
A compound contains Ca^2+ ions and CO_3^2- ions. Write its formula and determine the total charge of the formula unit.
⚠️ Common errors
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🎯 Exam question
(a) Identify the metal M. [1] (b) Write the balanced chemical equation for the reaction between M(NO_3)_2 and NaOH (aq) to produce M(OH)_2 (s) and NaNO_3 (aq). [3] [4 marks]
Key ideas — Balanced equations, including ionic equations
A balanced equation has equal numbers of atoms of each element, and equal total charge, on both sides, with state symbols (s), (l), (g), (aq) given for every species. An ionic equation shows only the species that actually change during the reaction, omitting spectator ions, ions present in solution that do not take part in the reaction and appear unchanged on both sides. For example, the full equation AgNO_3(aq) + NaCl(aq) arrow AgCl(s) + NaNO_3(aq) has Na^+(aq) and NO_3^-(aq) as spectator ions, so the ionic equation is Ag^+(aq) + Cl^-(aq) arrow AgCl(s) An ionic equation must be balanced for both atoms and charge: here, charge on the left (+1 - 1 = 0) equals charge on the right (0, since AgCl is a neutral solid).
🧑🏫 Teacher's explanation
Think of a chemical equation like a recipe for a sandwich. If the recipe says you need 2 slices of bread and 1 slice of cheese to make 1 sandwich, you can't just say 'bread + cheese arrow sandwich' and expect it to work; you need the right quantities to ensure nothing is wasted and the result is consistent. In chemistry, atoms are like the ingredients. A balanced equation ensures that no atoms 'disappear' or 'magically appear' during the reaction.
📖 Definitions
🔍 Worked example — Easy
Write the full balanced equation for the reaction between aqueous silver nitrate and aqueous sodium chloride, including state symbols.
🔍 Worked example — Standard
Write the net ionic equation for the reaction: AgNO_3(aq) + NaCl_(aq) arrow AgCl_(s) + NaNO_3(aq)
🔍 Worked example — Exam-style
Write the balanced ionic equation for the reaction between aqueous barium chloride and aqueous sodium sulfate: BaCl_2(aq) + Na_2 extSO_4(aq) arrow BaSO_4(s) + 2NaCl_(aq)
⚠️ Common errors
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🎯 Exam question
Key ideas — Empirical and molecular formula
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in one molecule of the compound; it is always a whole-number multiple of the empirical formula. For example, glucose has molecular formula mathrmC_6H_12O_6 and empirical formula CH_2O, since dividing each subscript by their highest common factor, 6, gives the simplest ratio 1:2:1. For ionic and giant covalent compounds, the empirical formula is also the formula normally written and used in calculations, since such compounds do not exist as discrete molecules.
🧑🏫 Teacher's explanation
Imagine you are making a fruit smoothie. If your recipe calls for 2 bananas and 4 strawberries, the 'molecular formula' is the actual amount in your blender: 2 ext bananas + 4 ext strawberries. However, the 'empirical formula' is the simplest ratio of the recipe: 1 ext banana : 2 ext strawberries. You can scale the recipe up (making a giant batch) or down, but the ratio stays the same.
📖 Definitions
🔍 Worked example — Easy
A compound contains 40.0\% carbon, 6.7\% hydrogen, and 53.3\% oxygen by mass. Determine its empirical formula.
🔍 Worked example — Standard
A compound has an empirical formula of CH_2 and a molar mass of 56 g mol^-1. Determine its molecular formula.
🔍 Worked example — Exam-style
A sample of a hydrocarbon contains 85.7\% C and 14.3\% H. If the relative molecular mass (M_r) is 56, find the molecular formula.
⚠️ Common errors
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🎯 Exam question
Key ideas — Anhydrous, hydrated compounds and water of crystallisation
A hydrated compound contains a fixed number of water molecules incorporated within its crystal lattice structure, called water of crystallisation, shown in the formula after a dot, e.g. CuSO_4 · 5H_2O (hydrated copper(II) sulfate, or copper(II) sulfate pentahydrate). An anhydrous compound contains no water of crystallisation; heating a hydrated salt drives off its water of crystallisation, leaving the anhydrous form, often with a distinct colour change, e.g. blue hydrated CuSO_4 · 5H_2O turns white anhydrous CuSO_4 on heating.
🧑🏫 Teacher's explanation
Imagine a crystal lattice is like a neatly organized apartment complex. The 'main residents' are the ions (like Cu^2+ and SO_4^2-), which form the solid structure. However, sometimes 'guest' molecules—specifically water molecules—get trapped in the gaps between these apartments to help stabilize the structure. These guests are the 'water of crystallisation'.
📖 Definitions
🔍 Worked example — Exam-style
A 5.00 g sample of a hydrated salt contains 1.20 g of water. If the anhydrous salt is CaCl_2, calculate the formula of the hydrated salt. (Use A_r: Ca=40, Cl=35.5, O=16, H=1)
⚠️ Common errors
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🎯 Exam question
Worked example
Question. A hydrated salt has the formula MgSO_4 · xH_2O. When 5.00 g of this hydrated salt is heated to constant mass, 2.44 g of anhydrous MgSO_4 remains. (M_r: MgSO_4 = 120; H_2O = 18.0.) Determine the value of x. [4]
🔍 Worked example — Standard
A sample of hydrated copper(II) sulfate, CuSO_4 · xH_2 extO, has a mass of 5.00 g. After heating, the anhydrous CuSO_4 mass is 3.20 g. Calculate x. (M_r: CuSO_4 = 160; H_2 extO = 18.0).
🔍 Worked example — Exam-style
A hydrated salt MgSO_4 · xH_2 extO has a mass of 5.00 g. When heated to constant mass, 2.44 g of anhydrous MgSO_4 remains. Determine x. (M_r: MgSO_4 = 120; H_2 extO = 18.0).
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Common mistakes
Practice
Answers