Edexcel IGCSE Chemistry · Spec 1.31, 1.33
Empirical and Molecular Formulae from Data
Covers find empirical formulae from experimental masses, Use water-of-crystallisation data and Find molecular formulae from Mr.
Chemistry revision video
Empirical and Molecular Formulae from Data
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Explained
Working a formula out from experimental data
An empirical formula is the simplest whole number ratio of atoms in a compound. Given masses or percentages from an experiment, you can find it, and the method is the same every time.
The four steps
Write down the mass of each element, or its percentage, since percentages can be treated as masses out of 100.
Divide each one by that element's relative atomic mass. This converts a mass into a number of moles, and it is the step that turns grams into atoms.
Divide every answer by the smallest of them. That scales the smallest to 1 and gives the ratio.
Round to whole numbers, or multiply up if you are left with a half or a third. A ratio of 1 to 1.5 becomes 2 to 3 when doubled.
Set the work out in labelled rows rather than as one calculation, because mark schemes award the division by the relative atomic masses and the division by the smallest as separate marks.
Molecular formula from the empirical formula
The molecular formula is a whole number multiple of the empirical formula, and the relative formula mass tells you which multiple.
Find the mass of the empirical formula unit, divide the given relative formula mass by it, and multiply every subscript by the answer.
If the empirical formula is NO2 its mass is 46. Given a relative formula mass of 92, dividing gives 2, so the molecular formula is N2O4.
Water of crystallisation
A hydrated salt contains water locked into its crystal structure, written after a dot, as in CuSO4·5H2O.
Heating drives that water off and leaves the anhydrous salt, so weighing before and after gives you both masses. The mass of water is the initial mass minus the final mass, and nothing else needs measuring.
Then treat it as an empirical formula problem with two components. Divide the anhydrous mass by the relative formula mass of the salt, divide the water mass by 18, and divide both by the smaller. The answer is the number in front of the H2O.
Heat, cool and reweigh until two consecutive masses agree. Constant mass is the evidence that all the water has gone, and a sample still holding water gives a value too high.
What the mark scheme accepts and rejects
An Edexcel International GCSE Chemistry mark scheme sets a three mark empirical formula question from percentages, and prints the working in two rows: each percentage over the relative atomic mass, then each result divided by the smallest, giving the formula C3H7OBr.
Its notes contain two refusals and one concession, and they repay reading together.
Zero marks for an upside down calculation. Dividing the relative atomic mass by the mass, rather than the other way round, produces numbers that still form a ratio and it is the wrong ratio. The mark scheme does not treat it as a slip in one step; the whole question goes.
Zero marks for using any atomic number. It even names the values it is watching for, 6, 8 and 35, which are the atomic numbers of carbon, oxygen and bromine. Reaching for the wrong number on the periodic table is common enough that the mark scheme has anticipated exactly which wrong numbers will appear.
And then the concession: an error carried forward is allowed if an incorrect relative atomic mass was used, provided it is not one of those atomic numbers. So a genuine misreading is forgiven and confusing the two numbers on the periodic table is not, because the second is a mistake about what the numbers mean.
The relative atomic mass is the larger of the two figures in a periodic table box. Circling the right number before starting costs nothing and protects all three marks.
Checking your answer
Add up the relative formula mass of your empirical formula and see whether the proportions make sense against the data you were given. The element with the largest mass in the sample should generally have a large share of the mass in your formula.
Be careful with rounding. Divide to two or three decimal places rather than rounding early, because a value of 1.33 is a third and a value of 1.5 is a half, and rounding either to 1 destroys the ratio.
And check that your subscripts are whole numbers. An empirical formula with a decimal in it has not been finished.
Spec 1.31, 1.33
What you need to know
- Find empirical formulae from experimental masses
- Use water-of-crystallisation data
- Find molecular formulae from Mr
Active recall
Quick check
Answer each question before opening the answer.
Why heat to constant mass?
To check that all removable water has been driven off
How is water mass found?
Initial hydrated mass minus final anhydrous mass
What is the molecular formula if the empirical formula is NO2 and Mr is 92?
N2O4
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