Empirical and molecular formula

Every compound has two related but distinct formulas. The empirical formula records only the simplest whole-number ratio of atoms of each element — it says nothing about how many atoms are actually in one molecule. The molecular formula records the true, actual number of atoms of each element in one molecule. For a compound like glucose, the empirical formula $\text{CH}_2\text{O}$ and the molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ describe the same substance at two different levels of detail — the molecular formula is simply the empirical formula scaled up by a whole number. That scaling factor has a name and a formula:

where: $n$ is a positive integer, the molecular mass is the compound's experimentally measured molar mass (found separately, e.g. by vapour density or a mass spectrometer), and the empirical formula mass is the sum of atomic masses implied by the empirical formula's subscripts. Finding the empirical formula itself is a short, mechanical sequence: convert every element's mass (or assumed-100 g percentage) into moles using $n = m/M_{\text{atomic}}$, divide every mole value by the smallest one to get a ratio anchored at 1, and — if any ratio still isn't close to a whole number — scale the whole set up by a small integer until it is.

$0.246\ \text{g}$ of an organic compound of carbon, hydrogen and oxygen was burnt completely, producing $0.352\ \text{g}$ of $\text{CO}_2$ and $0.144\ \text{g}$ of $\text{H}_2\text{O}$. Its molar mass was separately found to be $180\ \text{g mol}^{-1}$. Find its empirical and molecular formula. (Atomic masses: $\text{C} = 12$, $\text{H} = 1$, $\text{O} = 16$; molar masses $\text{CO}_2 = 44$, $\text{H}_2\text{O} = 18$.)

Step 1. Find the mass of carbon from the $\text{CO}_2$ produced — all the carbon in the sample ends up there:

Step 2. Find the mass of hydrogen from the $\text{H}_2\text{O}$ produced — all the hydrogen ends up there:

Step 3. Find the mass of oxygen by difference, since only C, H and O are present:

Step 4. Convert each mass to moles: $n_{\text{C}} = 0.0960/12 = 0.00800$, $n_{\text{H}} = 0.0160/1 = 0.0160$, $n_{\text{O}} = 0.134/16 = 0.00838$.

Step 5. Divide by the smallest value ($0.00800$): C $: 1.00$, H $: 2.00$, O $: 1.05 \approx 1$. Small deviations like $1.05$ are expected experimental scatter and round to the nearest whole number. The ratio C : H : O is $1:2:1$, so the empirical formula is $\text{CH}_2\text{O}$.

Step 6. Scale up to the molecular formula. The empirical formula mass is $12 + 2(1) + 16 = 30\ \text{g mol}^{-1}$, so $n = 180/30 = 6$, and the molecular formula is $(\text{CH}_2\text{O})_6$.

The compound has empirical formula $\text{CH}_2\text{O}$ and molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ — the same result as for glucose, obtained here directly from combustion data rather than from a stated percentage composition.

• Dividing by the largest mole value instead of the smallest in the "divide by smallest" step — this always distorts the ratio.
• Rounding a ratio like $1.33$ straight to $1$ instead of recognising it as $4/3$ and scaling the whole set by $3$ to clear the fraction.
• Forgetting that in combustion analysis, all the carbon appears in the $\text{CO}_2$ produced and all the hydrogen in the $\text{H}_2\text{O}$ produced — the factors $12/44$ and $2/18$ come directly from this, and oxygen (if present) must be found by mass difference, not measured directly.
• Confusing empirical formula mass with molecular mass when finding $n$ — $n$ is molecular mass divided by empirical formula mass, and a correct calculation always lands close to a whole number.

Ask the tutor: if you have a percentage composition or combustion data set you want to convert into an empirical or molecular formula, ask the AI tutor to walk through the "divide by smallest" step with you.