Empirical and molecular formulae

Empirical formula is the simplest whole-number ratio of atoms of each element in a compound — it says nothing about the actual size of a molecule, only the proportions.

Molecular formula is the actual number of atoms of each element in one real molecule, and it is always a whole-number multiple of the empirical formula:

where: $n$ is a positive whole number, found from

Percentage composition runs the same idea forward, from formula to percentages, using relative atomic masses:

Two compounds can share an empirical formula while being completely different substances. Formaldehyde (CH₂O, $n=1$, molecular mass 30) and glucose (C₆H₁₂O₆, $n=6$, molecular mass 180) both simplify to the empirical formula CH₂O — same atom ratio, six times the size. Percentage composition alone can never tell them apart; only the molecular mass fixes $n$ and hence the real formula. If the molecular mass is not given directly, it may need to be found first from vapour density, using $\text{Molecular mass} = 2 \times \text{V.D.}$ It helps to keep the two directions of the calculation straight, since board questions test both. Going from formula to percentages is a single division using the formula's own atom counts and atomic masses — nothing to solve for. Going from percentages back to a formula is the harder, more error-prone direction, because you must first turn masses into moles, then moles into a ratio, and only then (if molecular mass is supplied) turn that ratio into the true formula. Every step in that reverse direction is testable separately, which is why examiners often stop a question after just the empirical formula, or occasionally give the empirical formula directly and ask only for the multiplier $n$.

Given only the percentage composition and the relative molecular mass of a compound, we can uniquely reconstruct its molecular formula; the goal here is to justify the four-step arithmetic chain — moles, ratio, empirical mass, multiplier — as a single reliable procedure rather than a set of disconnected tricks.

Step 1. Convert each percentage into relative moles by dividing by the atomic mass of that element. Since percentage composition is really "grams of the element per 100 g of compound," dividing by atomic mass converts grams into moles for each element, all on the same 100 g basis.

Step 2. Divide every mole value by the smallest one among them. This rescales the numbers so the smallest becomes exactly 1, revealing the simplest ratio in which the elements combine — because only the ratio between elements matters, not their absolute mole counts.

Step 3. If any ratio is not a whole number (such as 1.5 or 1.33), multiply every ratio by the smallest whole number that clears the fraction (by 2, or by 3), since atoms can only combine in whole-number amounts. Write these whole numbers as subscripts to get the empirical formula.

Step 4. Add the atomic masses in the empirical formula to get the empirical formula mass, then divide the given molecular mass by it to get $n = \text{molecular mass}/\text{empirical formula mass}$. Multiply every subscript in the empirical formula by $n$ to obtain the molecular formula — this final step is valid because the molecular formula is defined as a whole-number multiple of the empirical formula, and $n$ is precisely that multiple.

Problem: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and its relative molecular mass is 180. Find its empirical formula and its molecular formula. (RAM: C = 12, H = 1, O = 16)

Step 1. Divide each percentage by its atomic mass to get relative moles.

Step 2. Divide each value by the smallest, 3.33.

Step 3. These are already whole numbers, so the empirical formula is $\text{CH}_2\text{O}$.

Step 4. Empirical formula mass $= 12 + 2(1) + 16 = 30$. Then $n = \dfrac{180}{30} = 6$, so the molecular formula is $(\text{CH}_2\text{O}) \times 6$.

Answer: empirical formula CH₂O, molecular formula C₆H₁₂O₆ (glucose).

Competitive stretch

A related question skips straight to masses: 6.4 g of a compound contains 3.2 g of sulphur, the rest oxygen. Mass of oxygen $= 6.4 - 3.2 = 3.2$ g. Moles: $n(\text{S}) = 3.2/32 = 0.1$, $n(\text{O}) = 3.2/16 = 0.2$. Dividing by the smaller (0.1) gives S : O $= 1 : 2$, so the empirical formula is directly SO₂ — no scaling needed since the ratio was already whole.

• Stopping at the empirical formula when the question asks for the molecular formula — always check whether a molecular mass (or vapour density) is given, and compute $n$ if it is.
• Dividing by the largest mole value instead of the smallest when simplifying the ratio, which produces numbers less than 1 instead of a clean whole-number ratio.
• Rounding an awkward ratio like 1.5 down to 1 or up to 2, instead of multiplying the whole ratio by 2 to clear the fraction, e.g. turning $1 : 1.5$ into $2 : 3$.
• Forgetting that the percentages must account for all elements present (including oxygen "by difference" if it is not stated directly) before starting the mole-ratio calculation.

Ask the tutor: If your mole ratio won't simplify to whole numbers, ask the AI tutor to walk through the scaling step with your exact numbers.