Empirical and molecular formula — AQA A-Level Chemistry
Test yourself on Empirical and molecular formula with AQA A-Level practice questions.
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Empirical and molecular formula explained
The empirical formula gives the ratio of atoms, reduced as far as it will go, so ethanoic acid and glucose both reduce to CH₂O even though their molecules differ enormously.
Read the full explanation
The standard route from composition is: treat percentages as masses in 100 g, divide each by the relative atomic mass of that element to get a relative amount, then divide every result by the smallest of them. If a value such as 1.5 appears, multiply the whole set by 2 to reach whole numbers, since the ratio must be whole. Where a question names only some elements and says the rest is another, that missing percentage must be found first by subtracting from 100.
Molecular formula is the actual number of atoms of each element in a compound.
The molecular formula states how many atoms of each element are actually present in one molecule. Ethanoic acid is C₂H₄O₂, benzene is C₆H₆, glucose is C₆H₁₂O₆. It carries more information than the empirical formula but less than a structural formula, since it says nothing about how the atoms are joined: C₂H₆O could be ethanol or methoxymethane. Sometimes the molecular and empirical formulae are identical, as in water, methane and ammonia, because the subscripts already have no common factor. Molecular formulae are only meaningful for substances made of discrete molecules; giant ionic and giant covalent substances are described by a formula unit instead.
The relationship between empirical formula and molecular formula.
The molecular formula is always a whole-number multiple of the empirical formula, so molecular = n × empirical where n = relative molecular mass ÷ empirical formula mass. Work out the empirical formula mass by adding the relative atomic masses in the empirical unit, divide the relative molecular mass by it, then multiply every subscript in the empirical formula by that value. For an empirical formula of CH₂O with mass 30.0 and a relative molecular mass of 180.0, n is 6 and the molecular formula is C₆H₁₂O₆. The multiplier must be a whole number: a value such as 2.9 signals an arithmetic error or a rounded relative molecular mass rather than a fractional formula.
Students should be able to: calculate empirical formula from data giving composition by mass or percentage by mass calculate molecular formula from the empirical formula and relative molecular mass.
This is the full two-part calculation. Part one: convert composition to an empirical formula. Find any missing percentage by subtraction, divide each mass or percentage by the relative atomic mass, divide the results by the smallest, and multiply up if a half or a third remains. A sample of 37.08 per cent carbon, 5.15 per cent hydrogen, 24.72 per cent oxygen and the rest sulfur gives relative amounts of 3.09, 5.15, 1.55 and 1.03, which reduce to 3, 5, 1.5, 1 and then double to C₆H₁₀O₃S₂. Part two: divide the relative molecular mass by the empirical formula mass and multiply every subscript by the whole number obtained.
Your focus
- Calculate an empirical formula from percentage composition by mass, including an element given only as the remainder.
- Convert a ratio containing a half or a third into whole numbers by multiplying the whole set.
- Explain why two different compounds can share the same empirical formula.
Show all 12 objectives
- Deduce a molecular formula by counting the atoms shown in a displayed or skeletal structure.
- Explain why a molecular formula cannot distinguish between two isomers.
- Identify cases in which the molecular and empirical formulae of a compound are identical.
- Calculate the multiplier linking an empirical formula to a molecular formula from the relative molecular mass.
- Deduce a molecular formula by scaling every subscript of an empirical formula.
- Verify a deduced molecular formula by recalculating its relative molecular mass.
- Calculate an empirical formula from percentage composition where one element is given only as the remainder.
- Determine a molecular formula from an empirical formula and a supplied relative molecular mass.
- Decide when a ratio containing 1.5 must be doubled rather than rounded.
Empirical and molecular formula exam tips
Marking Points
- one mark for finding a missing percentage by subtracting the stated percentages from 100
- one mark for dividing each percentage or mass by the relative atomic mass of that element
- one mark for dividing all the relative amounts by the smallest value
- one mark for the whole-number ratio written as a formula, multiplying up where a half or a third remains
- one mark for the actual number of atoms of each element in one molecule
- one mark for recognising when the molecular formula and empirical formula are the same
- one mark for showing the molecular formula as a whole-number multiple of the empirical formula
- one mark for distinguishing a molecular formula from a structural formula
- one mark for the empirical formula mass obtained by summing the relative atomic masses in the empirical unit
- one mark for dividing the relative molecular mass by the empirical formula mass to find the multiplier
- one mark for multiplying every subscript in the empirical formula by that multiplier
- one mark for the multiplier given as a whole number
- one mark for the percentage of the element described as the remainder
- one mark for dividing each percentage by the relative atomic mass to obtain relative amounts
- one mark for the ratio divided by the smallest value and expressed as the empirical formula
- one mark for the multiplier found from relative molecular mass divided by empirical formula mass
- one mark for the molecular formula obtained by scaling every subscript
Examiner Tips
- 💡Set the work out as a table with one column per element; the process marks are awarded row by row.
- 💡Keep three significant figures in the relative amounts so a genuine 1.5 is not lost to rounding.
- 💡Check the finished formula by confirming no common factor is left in the subscripts.
- 💡Count atoms from a displayed formula systematically, element by element, rather than by eye.
- 💡State whether the question wants empirical, molecular or structural before you write anything.
- 💡Remember that isomers share a molecular formula, so a molecular formula alone never identifies a compound.
- 💡Write the multiplier down as a labelled value before you use it; it is worth a mark on its own.
- 💡Check the finished molecular formula by recalculating its relative molecular mass against the value given.
- 💡If the multiplier is not close to a whole number, recheck the empirical formula rather than forcing the answer.
- 💡Use a four-row table: percentage, divided by relative atomic mass, divided by the smallest, multiplied up.
- 💡Carry at least three significant figures until the final ratio so that 1.5 is not mistaken for 1 or 2.
- 💡Read the final line of the question again; empirical and molecular are often both required.
Common Mistakes
- dividing by the atomic number instead of the relative atomic mass
- rounding a ratio of 1.5 down to 1 rather than doubling the whole set
- omitting the element described as the rest of the sample
- giving the molecular formula when the empirical formula was asked for
- giving the empirical formula when the molecular formula was requested
- assuming the two formulae must always differ, so doubling an already simplest formula
- miscounting hydrogen atoms when reading a skeletal or displayed structure
- quoting a molecular formula for an ionic compound such as sodium chloride
- multiplying only the first element's subscript by the multiplier
- dividing the empirical formula mass by the relative molecular mass, inverting the ratio
- rounding a multiplier of 2.9 down to 2 rather than checking the earlier working
- forgetting that the multiplier can legitimately be 1
- not calculating the percentage of the element given only as the rest of the sample
- dividing by the largest relative amount instead of the smallest
- leaving the ratio as decimals such as 3 : 5 : 1.5 : 1 without doubling
- stopping at the empirical formula when the question also asked for the molecular formula