Formulas
Cambridge A-Level Chemistry (9701) · Unit 2: Atoms, molecules and stoichiometry · 10 flashcards
Formulas is topic 2.3 in the Cambridge A-Level Chemistry (9701) syllabus , positioned in Unit 2 — Atoms, molecules and stoichiometry , alongside Relative masses of atoms and molecules and The mole and the Avogadro constant. In one line: Aluminum has a charge of +3 (Al³⁺) and oxygen has a charge of -2 (O²⁻). To balance the charges, the formula is Al₂O₃.
Marked as AS Level: examined at AS Level in Paper 1 (Multiple Choice), Paper 2 (AS Structured Questions) and Paper 3 (Advanced Practical Skills). The same content may also be assumed in Paper 4 (A Level Structured Questions).
The deck below contains 10 flashcards — 7 definitions and 3 calculations — covering the precise wording mark schemes reward. Use the 7 definition cards to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.
Write the formula for aluminum oxide
Aluminum has a charge of +3 (Al³⁺) and oxygen has a charge of -2 (O²⁻). To balance the charges, the formula is Al₂O₃.
What the Cambridge 9701 syllabus says
Official 2025-2027 spec · AS LevelThese are the exact learning outcomes Cambridge sets for this topic. The candidate is expected to be able to do each of these on the relevant paper.
- write formulas of ionic compounds from ionic charges and oxidation numbers (shown by a Roman numeral), including: (a) the prediction of ionic charge from the position of an element in the Periodic Table (b) recall of the names and formulas for the following ions: NO3 –, CO3 2–, SO4 2–, OH–, NH4 +, Zn2+, Ag+, HCO3 –, PO4 3–
- define and use the terms empirical and molecular formula
- understand and use the terms anhydrous, hydrated and water of crystallisation
- calculate empirical and molecular formulas, using given data
- www.cambridgeinternational.org/alevel
- perform calculations including use of the mole concept, involving: (a) reacting masses (from formulas and equations) including percentage yield calculations (b) volumes of gases (e.g. in the burning of hydrocarbons) (c) volumes and concentrations of solutions (d) limiting reagent and excess reagent
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 9701 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Tips to avoid common mistakes in Formulas
- › Always check the subscripts in a chemical formula to ensure the correct number of metal ions are used in quantitative calculations.
- › Use the balanced equation to find the exact mole ratio and subtract the mass reacted from the initial mass.
Write the formula for aluminum oxide.
Aluminum has a charge of +3 (Al³⁺) and oxygen has a charge of -2 (O²⁻). To balance the charges, the formula is Al₂O₃.
Write the formula for copper(II) sulfate.
Copper(II) indicates a +2 charge (Cu²⁺). Sulfate is SO₄²⁻. Therefore, the formula is CuSO₄.
Define empirical formula.
The empirical formula is the simplest whole number ratio of atoms of each element in a compound.
Define molecular formula.
The molecular formula is the actual number of atoms of each element in a molecule.
Define anhydrous.
Anhydrous means a substance contains no water molecules. It is the opposite of hydrated.
Define hydrated.
Hydrated means a substance contains water molecules within its crystal structure.
Define water of crystallisation.
Water of crystallisation refers to water molecules that are chemically bonded within the crystal structure of a hydrated compound.
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Calculate its empirical formula.
Assume 100g, so 40g C, 6.7g H, 53.3g O. Convert to moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Divide by smallest (3.33): C = 1, H = 2, O = 1. Empirical formula is CH₂O.
0.100g of magnesium reacts with excess hydrochloric acid to produce 0.00821 mol of hydrogen gas. What is the percentage yield of hydrogen?
Mg + 2HCl → MgCl₂ + H₂. Moles of Mg = 0.100/24.3 = 0.00411 mol. Theoretical yield of H₂ = 0.00411 mol. % yield = (actual yield / theoretical yield) x 100 = (0.00821 / 0.00411) x 100 = 200%. Note, must be less than 100%. Mistake in question, 0.002055 mol of hydrogen gas is correct. Actual yield = 0.002055 % yield = 50%
25.0 cm³ of 0.100 mol dm⁻³ NaOH is neutralised by 20.0 cm³ of HCl. What is the concentration of the HCl?
NaOH + HCl → NaCl + H₂O. Moles of NaOH = (25/1000) * 0.1 = 0.0025 mol. Moles of HCl = moles of NaOH = 0.0025 mol. Concentration of HCl = moles/volume = 0.0025/(20/1000) = 0.125 mol dm⁻³
More Chemistry flashcards
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All Chemistry FlashcardsMore topics in Unit 2 — Atoms, molecules and stoichiometry
Formulas sits alongside these A-Level Chemistry decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Key terms covered in this Formulas deck
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