Empirical and Molecular Formulas

Two gases have exactly the same elemental analysis: 92.3% carbon, 7.7% hydrogen. One is benzene, a liquid at room temperature that ring-shaped molecules make famously stable. The other is acetylene, a gas that burns hot enough to cut steel. Chemical analysis cannot tell them apart, because analysis measures ratios and these two substances have the same ratio — one carbon to one hydrogen.

That is the whole reason this topic exists as a two-part problem. Analysis gives you the ratio. Something else has to give you the size.

"Empirical" here does not mean approximate

The word trips people up because in ordinary English "empirical" suggests rough, observational, provisional. In this context it means something precise and slightly old-fashioned: derived from measurement rather than from theory.

• The empirical formula is the simplest whole-number ratio of atoms. Benzene's is CH.
• The molecular formula is how many atoms are actually in one molecule. Benzene's is C₆H₆, acetylene's is C₂H₂.
• The structural formula shows how those atoms are connected, and it is the only one of the three that identifies a substance uniquely.

The gap between the second and third matters more than students expect. C₂H₆O is the molecular formula of ethanol and also of dimethyl ether — the same atoms, differently joined, one a drinkable liquid and one a gas that boils at −24 °C. Isomers share a molecular formula and share nothing else. No amount of counting will separate them.

The four steps, worked on a sugar

An analysis reports a compound as 40.00% carbon, 6.71% hydrogen and 53.29% oxygen. Its molar mass, measured separately, is 180.2 g/mol.

Step 1 — assume 100 g. Percentages become grams with no arithmetic at all. Forty per cent carbon becomes 40.00 g of carbon. This is the step that makes everything afterwards easy, and it works because the ratio you are chasing does not depend on how much you have.

Step 2 — grams to moles, element by element, dividing by each atomic weight:

• carbon: 40.00 ÷ 12.011 = 3.3303 mol
• hydrogen: 6.71 ÷ 1.008 = 6.6567 mol
• oxygen: 53.29 ÷ 15.999 = 3.3308 mol

Step 3 — divide every result by the smallest of them. Here the smallest is carbon's 3.3303:

• carbon: 3.3303 ÷ 3.3303 = 1.000
• hydrogen: 6.6567 ÷ 3.3303 = 1.999
• oxygen: 3.3308 ÷ 3.3303 = 1.000

The empirical formula is CH₂O, with an empirical formula mass of 30.026 g/mol.

Step 4 — scale up to the measured molar mass. Divide the real molar mass by the empirical one:

180.2 ÷ 30.026 = 6.00

Multiply every subscript by six and the molecular formula is C₆H₁₂O₆ — glucose, or one of its many isomers, which is exactly as far as this method can take you.

Note where each piece of information did its job. The percentages fixed the ratio and could never have fixed the size; the molar mass fixed the size and could never have fixed the ratio. A question that gives you only percentages is asking only for the empirical formula, and writing a molecular formula in that case means you have assumed something you were not told.

When the ratio comes out at 3.5

Step 3 does not always produce integers, and the correct response is never to round aggressively.

Take an analysis reading 26.57% potassium, 35.36% chromium, 38.07% oxygen:

• potassium: 26.57 ÷ 39.098 = 0.6796
• chromium: 35.36 ÷ 51.996 = 0.6801
• oxygen: 38.07 ÷ 15.999 = 2.3795

Dividing through by 0.6796 gives 1.000, 1.001 and 3.501. Rounding that oxygen to 4 would give KCrO₄, which is not a real compound and is not what the numbers say. Multiplying all three by two gives 2, 2, 7 — K₂Cr₂O₇, potassium dichromate, which is.

The thirds are the other common case. An iron oxide analysing at 72.36% iron and 27.64% oxygen gives 1.2957 mol and 1.7276 mol; dividing by the smaller yields 1.000 and 1.3333. Two thirds of the way to an integer is not a rounding error, it is a factor of three: multiply both by three to get Fe₃O₄.

The multipliers worth recognising on sight:

• .50 → multiply by 2
• .33 or .67 → multiply by 3
• .25 or .75 → multiply by 4
• .20, .40, .60, .80 → multiply by 5

The judgement nobody writes down: what to do with 2.7

Textbooks present step 3 as though ratios always land on something recognisable. Real numbers do not, and the unspoken skill is deciding whether a value is a slightly noisy integer, a genuine fraction, or evidence that something is wrong.

A workable standard is that anything within about 0.05 of a whole number is that whole number, and the deviation is measurement error. Anything within about 0.02 of a half, a third or a quarter is a genuine fraction and needs the multiplier. A value like 2.7 is neither, and it means one of three things: the analysis is poor, an element was missed, or the sample was impure. Multiplying 2.7 by ten to get 27 is technically permitted by the method and is essentially always wrong.

There is also a limit built into the method that examiners rarely mention. Empirical formulas with large subscripts cannot survive it. A protein might be C₁₆₃₈H₂₅₈₀N₄₄₀O₄₉₈S₁₈; distinguishing that from a ratio one atom different would require analysis accurate to a hundredth of a per cent, which the technique does not deliver. This is why molecular formulas for large molecules come from mass spectrometry and sequencing rather than from combustion figures.

A combustion report, read as arithmetic

For compounds of carbon, hydrogen and oxygen, the analysis usually arrives as two masses rather than as percentages: how much carbon dioxide and how much water the sample produced. The logic is that every carbon atom ends up in one CO₂ and every two hydrogen atoms end up in one H₂O.

Suppose a report gives, for a 0.5000 g sample, 1.1366 g of CO₂ and 0.4653 g of H₂O.

Carbon. Moles of CO₂ = 1.1366 ÷ 44.009 = 0.025827 mol, and that is also the moles of carbon. Its mass is 0.025827 × 12.011 = 0.3102 g.

Hydrogen. Moles of H₂O = 0.4653 ÷ 18.015 = 0.025828 mol, so moles of hydrogen is twice that, 0.051656 mol — this doubling is the single most-missed step on the page. Its mass is 0.051656 × 1.008 = 0.05207 g.

Oxygen by difference. Oxygen cannot be measured this way, because the products contain oxygen from two sources at once. It is found by subtraction:

0.5000 − 0.3102 − 0.05207 = 0.1377 g, which is 0.1377 ÷ 15.999 = 0.008608 mol.

Dividing all three mole figures by the smallest, 0.008608, gives carbon 3.000, hydrogen 6.001, oxygen 1.000 — an empirical formula of C₃H₆O. If a separately measured molar mass comes back at 58.1 g/mol, then 58.1 ÷ 58.08 = 1.00, and here the molecular formula and the empirical formula are the same thing. That is a legitimate answer, not a sign that you have missed a step.

The compounds that have no molecular formula at all

Step 4 assumes molecules exist. For a great many substances they do not, and the empirical formula is the end of the road rather than a stepping stone.

Sodium chloride is written NaCl because the crystal contains equal numbers of sodium and chloride ions in a continuous lattice — there is no NaCl molecule to have a molecular formula. The correct term for what NaCl describes is a formula unit. Every ionic compound works this way.

Giant covalent solids are the same. Quartz is SiO₂ and diamond is C, and in both cases the whole crystal is a single connected structure. Asking for the molecular formula of quartz is asking how many atoms are in one grain of sand.

Metals extend the pattern one step further: their formulas are bare symbols, and alloys such as brass have no fixed formula at all because their composition is adjustable by design. Whenever a question hands you a molar mass and expects step 4, it is telling you implicitly that the substance is molecular.