Analysis tells you what a compound is made of, by mass. Divide by the relative atomic masses and you have the ratio of atoms; the relative molecular mass then turns that ratio into the real formula. Extended content.
The empirical formula is the simplest whole-number ratio of the different atoms or ions in a compound. The molecular formula is the number and type of each atom in one molecule. To find the empirical formula, divide each mass (or percentage) by Ar, then divide by the smallest answer. To find the molecular formula, divide Mr by the empirical formula mass and multiply the empirical formula by the answer.
Fig. 1: from percentages to glucose, step by step
Section 01
Four steps to an empirical formula.
Extended content. Enter the masses of each element in the simulator and follow its working: divide by Ar, divide by the smallest, then make whole numbers.
1. Mass or %: write down the mass of each element (a percentage is the mass in 100 g).
2. Divide by Ar: this gives the moles of atoms of each element.
3. Divide by the smallest: this makes the smallest number 1.
4. Whole numbers: if a value ends in .5, double everything.
Magnesium oxide: 2.4 g Mg and 1.6 g O. Mg: 2.4 ÷ 24 = 0.1; O: 1.6 ÷ 16 = 0.1. Ratio 1 : 1, so MgO.
A hydrocarbon: 1.2 g C and 0.4 g H. C: 1.2 ÷ 12 = 0.1; H: 0.4 ÷ 1 = 0.4. Ratio 1 : 4, so CH4.
Stage 1 · Learn
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Four quick checks. Each correct answer earns XP and records this skill on your profile.
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The empirical formula of a compound is:
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What is the empirical formula of glucose, C6H12O6?
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A compound has atoms of iron and oxygen in the ratio 1 : 1.5. Its empirical formula is:
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A compound has the empirical formula CH2 and Mr = 56. Its molecular formula is:
Section 02
From percentages to molecular formulae.
Percentages work exactly like masses. Different compounds can share one empirical formula, and the Mr tells them apart.
85.7% C, 14.3% H: C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3. Dividing by 7.14 gives 1 : 2, so the empirical formula is CH2 (mass 14). If Mr = 56, then 56 ÷ 14 = 4 and the molecular formula is C4H8.
Ionic compounds such as NaCl only have an empirical formula: it gives the ratio of ions in the lattice.
Section 03
Ratios that are not whole.
Sometimes step 3 gives a number like 1.5. That is a real ratio, not a rounding error.
An oxide of iron with 70% Fe: Fe: 70 ÷ 56 = 1.25; O: 30 ÷ 16 = 1.875.
Divide by the smallest: 1.25 ÷ 1.25 = 1 and 1.875 ÷ 1.25 = 1.5, a ratio of 1 : 1.5.
Double: 2 : 3, so the formula is Fe2O3.
Examiner trap
Never round 1 : 1.5 to 1 : 2 (or to 1 : 1). A ratio ending in .5 means you must double every number. Rounding the iron oxide ratio gives FeO2, which is wrong; doubling gives Fe2O3. And never divide the masses themselves: 2.4 g Mg to 1.6 g O is a 1 : 1 ratio of atoms, not 3 : 2.
Stage 2 · Exam
Exam-style questions
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2.4 g of magnesium combines with 1.6 g of oxygen. What is the empirical formula of the oxide? (Ar: O 16, Mg 24)
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A hydrocarbon contains 1.2 g of carbon and 0.4 g of hydrogen. What is its empirical formula? (Ar: H 1, C 12)
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A compound contains 40.0% C, 6.7% H and 53.3% O by mass, and its Mr is 180. What is its molecular formula? (Ar: H 1, C 12, O 16)
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An oxide of iron contains 70% iron by mass. What is its empirical formula? (Ar: O 16, Fe 56)
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