Learning objective
Calculate empirical formulae from experimental mass data.
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At a glance
5
Flashcards
7
Questions
Topic
Amount of substance
Subtopic
Empirical and molecular formula
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Short explanation
In the subtopic Empirical and molecular formula, this AQA A-Level Chemistry 7405 learning objective focuses on calculate empirical formulae from experimental mass data. It belongs to Amount of substance, so revision should stay anchored to this exact subtopic rather than drifting into a generic GCSE-level chemistry summary. Approved keywords to use include empirical formula. Percentage composition. means the percentage by mass of each element in a compound, calculated using the formula: (mass of element / molar mass of compound) x 100 Avoid the mistake of students often forget to convert the mass of each element to moles before determining the simplest whole-number ratio; instead, to calculate the empirical formula, first convert the mass of each element to moles using the formula: moles = mass / Ar. For example, if you have 10g of carbon (Ar = 12) and 8g of hydrogen (Ar = 1), the calculations would be: moles of carbon = 10g / 12 = 0.833, moles of hydrogen = 8g / 1 = 8. Then, divide both by the smallest number of moles (0.833) to find the ratio: carbon = 0.833 / 0.833 = 1, hydrogen = 8 / 0.833 = 9.6, which rounds to 10. Therefore, the empirical formula is C1H10 For exam answers, to calculate the empirical formula from experimental mass data, first convert the masses of each element to moles using their relative atomic masses. Then, divide by the smallest number of moles to find the simplest whole-number ratio
Key concepts
Why it matters
This objective helps connect Empirical and molecular formula to exam-style questions, flashcards, and revision notes for Amount of substance.
Common mistakes
1 linked- Common Mistake in Empirical Formula Calculation: To calculate the empirical formula, first convert the mass of each element to moles using the formula: moles = mass / Ar. For example, if you have 10g of carbon (Ar = 12) and 8g of hydrogen (Ar = 1), the calculations would be: moles of carbon = 10g / 12 = 0.833, moles of hydrogen = 8g / 1 = 8. Then, divide both by the smallest number of moles (0.833) to find the ratio: carbon = 0.833 / 0.833 = 1, hydrogen = 8 / 0.833 = 9.6, which rounds to 10. Therefore, the empirical formula is C1H10.
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Revision notestopic notes
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Open revision notesRelated learning objectives
- Define relative atomic mass in relation to carbon-12.
Relative atomic mass and relative molecular mass
- Define relative molecular mass in relation to carbon-12.
Relative atomic mass and relative molecular mass
- Use relative formula mass for ionic compounds.
Relative atomic mass and relative molecular mass
- Calculate relative molecular or formula mass from a chemical formula and relative atomic masses.
Relative atomic mass and relative molecular mass
- Explain the Avogadro constant as the number of particles in one mole.
The mole and the Avogadro constant
