Lesson 5 · 30 min

Formulas and composition

Mass percentages, empirical and molecular formulas — and recovering the formula of vitamin C from combustion data, the way Liebig did.

In this lesson

Calculate the mass percentage of each element from a formula
Distinguish the empirical (simplest) formula from the molecular formula, and find the latter from the molar mass
Derive an empirical formula from mass percentages, handling ratios such as 1.33 and 1.5 correctly
Use combustion-analysis data (masses of CO₂ and H₂O) to find the formula of an organic compound

The formula of vitamin C is $\ce{C6H8O6}$ . But chemists in the 1930s had only a pinch of white crystals — how did they know? They burned a few milligrams in oxygen and weighed the carbon dioxide and water produced: all the carbon ends up in $\ce{CO2}$ , all the hydrogen in $\ce{H2O}$ . From those two masses you can work back to the ratio of atoms and, with the molar mass, to the formula. This lesson walks the whole chain of reasoning.

Predict first

What do glucose, $\ce{C6H12O6}$ , and formaldehyde, $\ce{CH2O}$ , have in common?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Molar mass Mass percentage of each element Total atoms per molecule Chemical properties

Mass percentage

w(\mathrm{E}) = \frac{\nu_\mathrm{E}\,A_\mathrm{r}(\mathrm{E})}{M_\mathrm{r}} \times 100\%

The percentage by mass of element E in a compound.

$w(\mathrm{E})$: mass fraction (percentage) of element E
$\nu_\mathrm{E}$: atoms of E per formula unit
$A_\mathrm{r}(\mathrm{E})$: relative atomic mass of E
$M_\mathrm{r}$: relative formula mass of the compound

Empirical and molecular formulas

An empirical (simplest) formula gives only the smallest whole-number ratio of atoms; a molecular formula gives the actual number of each atom in one molecule. The molecular formula is always a whole multiple of the empirical one: $n = M / M(\text{empirical})$ . Benzene $\ce{C6H6}$ and ethyne $\ce{C2H2}$ share the empirical formula $\ce{CH}$ . Ionic compounds such as $\ce{NaCl}$ and $\ce{Fe2O3}$ have no discrete molecules, so the formula written is empirical.

From mass percentages to an empirical formula:

Take 100 g of sample, so percentages become grams.
Divide each by the atomic weight to get moles.
Divide all by the smallest.
If the ratios are not all near whole numbers (1.33, 1.5, 1.25 …), multiply them all by the smallest integer that makes them whole.

Mass percent → mole ratio → smallest whole numbers

Empirical-formula finder

Sample analysed

Total 100.00%

Molar mass (optional) / (g/mol)

Working (per 100 g of sample)
Element	Mass / g	÷ Ar → n / mol	÷ smallest	× 3	Whole
C 12.011	40.92	3.407	1.000	3.000	3
H 1.008	4.58	4.544	1.334	4.001	4
O 15.999	54.50	3.406	1.000	3.000	3

The ratios are not all whole numbers, so everything is multiplied by 3.

Empirical formula

\ce{C3H4O3}

88.06 g/mol

Molecular formula

\ce{C6H8O6}

M ÷ M(empirical) = 2.000 ≈ 2

For vitamin C the ratio is 1 : 1.33 : 1 and needs ×3; haematite’s 1 : 1.5 needs ×2. Then switch to “From combustion” and reach the same answer from the masses of CO₂ and H₂O.

Worked example

Combustion analysis: recovering the formula of vitamin C

10.00 mg of a compound containing only C, H and O burns completely to 14.99 mg of $\ce{CO2}$ and 4.09 mg of $\ce{H2O}$ . Its molar mass is separately found to be about 176 g/mol. Find the molecular formula.

Concept check

Benzene is $\ce{C6H6}$ . What is its empirical formula?

\ce{C6H6}

\ce{C3H3}

\ce{CH}

\ce{C2H2}

Try the calculation

What is the mass percentage of nitrogen in the fertiliser ammonium nitrate, $\ce{NH4NO3}$ ?

Key takeaways

Mass fraction $w(\mathrm{E}) = \nu_\mathrm{E} A_\mathrm{r}(\mathrm{E})/M_\mathrm{r}$ ; substances with the same empirical formula have the same composition.
The empirical formula is the simplest whole-number ratio; the molecular formula is a whole multiple of it: $n = M/M(\text{empirical})$ .
To find it: 100 g → moles → divide by the smallest → multiply by a whole number if needed (1.33 → ×3, 1.5 → ×2).
In combustion analysis all C goes to CO₂ and all H to H₂O; oxygen is found by difference.

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Data and references

IUPAC abridged standard atomic weights (2021), via the VisualChem element table.
J. Liebig, “Ueber einen neuen Apparat zur Analyse organischer Körper”, Annalen der Physik und Chemie 21, 1–47 (1831).
P. Atkins, L. Jones, L. Laverman, Chemical Principles, 7th ed. (W. H. Freeman, 2016), Fundamentals E–F.
人民教育出版社《普通高中教科书化学必修第一册》（2019），第二章物质的量。