Lesson 1 of 3
Particles to formulae
Separate mass ratios from atom ratios.
A-Level 8873, revised syllabus (2026-2027)
I can define the four relative masses.Syllabus
Syllabus 8873, 5(a). Define the four relative masses.
I can define the mole through the Avogadro constant.Syllabus
Syllabus 8873, 5(b). Define the mole through the Avogadro constant.
I can calculate isotope-weighted relative atomic mass.Syllabus
Syllabus 8873, 5(c). Calculate isotope-weighted relative atomic mass.
I can distinguish empirical and molecular formulae.Syllabus
Syllabus 8873, 5(d). Distinguish empirical and molecular formulae.
I can calculate formulae from analytical evidence.Syllabus
Syllabus 8873, 5(e). Calculate formulae from analytical evidence.
Keep relative mass, molar mass and amount distinct
Relative quantities have no units; molar mass converts grams into moles.
Relative masses compare a mass with one twelfth of the mass of a carbon-12 atom. They are ratios, so they have no unit. Relative isotopic mass refers to one isotope; relative atomic mass, Ar, is the weighted mean for the atoms in a specified sample of an element.
| Term | What is compared with one twelfth of a carbon-12 atom |
|---|---|
| Relative isotopic mass | Mass of one atom of the stated isotope |
| Relative atomic mass, Ar | Weighted mean mass of atoms of the element |
| Relative molecular mass, Mr | Mass of one molecule; numerically the sum of its constituent Ar values |
| Relative formula mass, Mr | Mass for one formula unit; the sum of Ar values in that formula, used for ionic compounds |
One mole contains exactly 6.02214076 × 1023 specified entities. The Avogadro constant NA is that number per mole, with unit mol-1. Specify whether you mean atoms, molecules, ions, electrons or formula units. One mole of MgCl2 formula units contains one mole Mg2+ ions and two moles Cl- ions.
Use n = m/M for mass m in g and molar mass M in g mol-1; n is in mol. M has the same numerical value as the corresponding Ar or Mr, but different units. For particles, N = nNA.
Worked exampleWeight the isotope data
A sample contains 75.0% isotope mass 35.0 and 25.0% isotope mass 37.0. Calculate its relative atomic mass using the supplied rounded isotopic masses.
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- Convert percentages into fractions: 0.750 and 0.250.
- Weighted mean = (0.750 × 35.0) + (0.250 × 37.0).
- The result is 35.5; it lies between the two isotope masses and nearer the more abundant one.
Ar = 35.5, with no unit. An unweighted average of 36.0 would incorrectly treat the isotope populations as equal.
Worked exampleFrom a weighed sample to atoms
How many atoms are in 0.486 g Mg, using Ar(Mg) = 24.3?
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- n = 0.486 g / (24.3 g mol-1) = 0.0200 mol.
- N = 0.0200 mol × 6.02214076 × 1023 mol-1.
1.20 × 1022 Mg atoms, to three significant figures.
Check your understanding
Which statement about Mr and molar mass for water is correct?
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Mr is 18.0 with no unit; molar mass is 18.0 g mol-1
Relative masses are ratios to one twelfth of a carbon-12 atom, so they have no unit. Molar mass has the same number but carries g mol-1, which converts grams into moles.
Turn mass evidence into an atom ratio
An empirical formula is the simplest ratio; a molecular formula counts atoms in one molecule.
An empirical formula gives the simplest whole-number ratio of elements. A molecular formula gives the actual number of each atom in one molecule. Ethene has molecular formula C2H4 and empirical formula CH2. Ionic substances such as MgO have formula units rather than discrete molecules.
- Choose a basis
For percentages, imagine 100 g. Each percentage then becomes a mass in grams.
- Convert each element to moles
Divide its mass by its atomic molar mass.
- Divide by the smallest
Obtain the simplest mole ratio, not a mass ratio.
- Reach whole numbers carefully
A reliable ratio such as 1:1.5 must be doubled, not rounded to 1:2.
Worked exampleA percentage formula with a fractional ratio
An oxide contains 69.9% Fe and 30.1% O. Use Ar(Fe) = 55.8 and Ar(O) = 16.0.
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- For 100 g, Fe amount = 69.9/55.8 = 1.253 mol; O amount = 30.1/16.0 = 1.881 mol.
- Divide both by 1.253: approximately 1 : 1.50.
- Multiply both values by two: 2 : 3.
The empirical formula is Fe2O3. The small departures from an exact ratio reflect the rounded input data.
Worked exampleUse combustion and mass by difference
Complete combustion of 0.900 g of a compound containing only C, H and O gives 1.320 g CO2 and 0.540 g H2O. Its Mr is 90.0. Find both formulae; use C = 12.0, H = 1.00, O = 16.0.
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- CO2 amount = 1.320/44.0 = 0.0300 mol, so C amount = 0.0300 mol and C mass = 0.360 g.
- H2O amount = 0.540/18.0 = 0.0300 mol. There are two H atoms per water molecule, so H amount = 0.0600 mol and H mass = 0.0600 g.
- Original O mass = 0.900 - 0.360 - 0.0600 = 0.480 g; O amount = 0.480/16.0 = 0.0300 mol.
- C:H:O = 0.0300:0.0600:0.0300 = 1:2:1, so the empirical formula is CH2O.
- Empirical formula mass = 30.0. The multiplier 90.0/30.0 = 3.
Empirical formula CH2O; molecular formula C3H6O3.
Watch out for this
A student says: "To find the oxygen in the sample, count all the oxygen atoms in the CO2 and water it produced." What is wrong with this?
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Oxygen gas is supplied during combustion. Find original oxygen by mass difference only when the sample is known to contain just the stated elements. The method also assumes complete combustion and quantitative collection of CO2 and water.
Check your understanding
Does the empirical formula CH2 identify ethene?
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No. C2H4, C3H6 and others share CH2; Mr is needed to decide.
An empirical formula is the simplest ratio of atoms. Many molecules have the 1 : 2 ratio, so Mr or other evidence is needed.