Empirical formula and molecular formula

This page covers a single Form 4 content standard: empirical formula and molecular formula. Here the mole concept is put to work to discover the formula of a compound from experimental data, masses or percentage compositions, rather than being told it. This is a favourite Paper 2 question because it tests the whole chain of skills: relative mass, moles and ratio all in one problem, which makes it excellent preparation for the rest of SPM Chemistry.

Two kinds of formula

• The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound. For example, the empirical formula of glucose is CH₂O.
• The molecular formula is the actual number of atoms of each element in one molecule of the substance. The molecular formula of glucose is C₆H₁₂O₆.

The molecular formula is always a whole-number multiple of the empirical formula: C₆H₁₂O₆ is (CH₂O) × 6. Some compounds have identical empirical and molecular formulae, water is H₂O in both, while for ionic compounds we normally quote only the empirical formula (NaCl, MgO), because there are no discrete molecules.

Finding an empirical formula: the four steps

Whatever the data, the method is always the same, and it runs through the mole:

1. Write the mass (or percentage) of each element. If percentages are given, assume a 100 g sample so the percentages become masses in grams.
2. Divide each mass by that element’s Ar to get the number of moles of atoms.
3. Divide every mole value by the smallest of them, to get the ratio.
4. Round to the nearest whole numbers, multiplying up if you get a value like 1.5, and write the ratio as the formula.

The idea behind step 2 is the heart of the mole concept: the ratio of atoms is the same as the ratio of moles of atoms, and moles are what you get by dividing mass by Ar.

From empirical to molecular formula

To reach the molecular formula you need one extra piece of information: the relative molecular mass (Mr) of the compound. Then:

1. Work out the empirical formula mass (add the Ar values in the empirical formula).
2. Divide the given Mr by the empirical formula mass to get the whole number n.
3. Multiply every subscript in the empirical formula by n.

Worked example

Question. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Find its empirical formula and its molecular formula. [Ar: C = 12, H = 1, O = 16]

Step 1, Assume 100 g, so masses are 40.0 g C, 6.7 g H, 53.3 g O.

Step 2, Divide each by its Ar to get moles.

• C: 40.0 ÷ 12 = 3.33 mol
• H: 6.7 ÷ 1 = 6.7 mol
• O: 53.3 ÷ 16 = 3.33 mol

Step 3, Divide by the smallest (3.33).

• C: 3.33 ÷ 3.33 = 1
• H: 6.7 ÷ 3.33 = 2
• O: 3.33 ÷ 3.33 = 1

So the empirical formula is CH₂O.

Step 4, Find n and scale up. Empirical formula mass = 12 + (2 × 1) + 16 = 30. n = Mr ÷ empirical formula mass = 180 ÷ 30 = 6. Multiply CH₂O by 6.

Answer. Empirical formula CH₂O; molecular formula C₆H₁₂O₆.

Practice question

Question. In an experiment, 2.4 g of magnesium combines completely with 1.6 g of oxygen to form an oxide. Determine the empirical formula of the oxide. [Ar: Mg = 24, O = 16]

Answer. Moles of Mg = 2.4 ÷ 24 = 0.1 mol; moles of O = 1.6 ÷ 16 = 0.1 mol. Ratio Mg : O = 0.1 : 0.1 = 1 : 1. The empirical formula is MgO.

Exam tip

Set your working out as a table, element, mass, ÷ Ar, ÷ smallest, so each step is visible and earns its method mark. Two traps recur. First, if a ratio comes out as 1.5 (or another neat fraction), do not round it to 2; multiply every value by 2 to clear the fraction (for example 1 : 1.5 becomes 2 : 3). Second, never quote the molecular formula unless you are given the Mr, without it you can only reach the empirical formula. And keep the definitions crisp: “simplest whole-number ratio” for empirical, “actual number of atoms in a molecule” for molecular. Those exact phrases are what the marking scheme looks for.

Reading the data you are given

Empirical-formula questions arrive in several disguises, but every one reduces to a list of masses. Percentages become masses if you assume a 100 g sample. A question that gives the mass of a metal and the mass of its oxide hides the mass of oxygen as the difference between the two, for instance, if 4.0 g of a metal forms 5.6 g of oxide, the oxygen mass is 5.6 − 4.0 = 1.6 g. A reduction experiment that gives the mass of copper produced and the mass of copper oxide used works the same way. Training yourself to extract “mass of each element” from whatever scenario is described is the real skill this standard tests; once you have those masses, the four-step method is identical every time.

Why the mole makes this possible

It is worth pausing on why dividing by Ar works. Chemical formulae are ratios of atoms, but you cannot weigh single atoms, you weigh bulk samples in grams. Dividing each mass by that element’s relative atomic mass converts grams into a number proportional to the count of atoms, i.e. moles. Because every element is scaled by its own Ar, the mole values you get are directly comparable, and their ratio is the atom ratio you want. This is the same mole-as-a-bridge idea from standards 3.2 and 3.3, now used in reverse to uncover a formula, a neat demonstration of how central the mole is to the whole chapter.

Where this fits

This is content standard 3.5 of the The Mole Concept, Chemical Formula and Equation chapter, applying the mole and relative mass to real experimental data. Work through more determinations in the worked examples and review the pitfalls in common mistakes. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers show you how to lay out the ratio table cleanly and how to spot when a fraction must be scaled up rather than rounded.

Quick recap

• Empirical formula = simplest whole-number ratio of atoms; molecular formula = actual atoms per molecule.
• Method: mass (or % as mass) → ÷ Ar for moles → ÷ smallest → whole-number ratio.
• Molecular formula = empirical formula × n, where n = Mr ÷ empirical formula mass.
• Multiply up a 1.5 ratio rather than rounding; you need the Mr for the molecular formula.