Empirical & Molecular Formula, and Hydrates

Percentage Composition

The proportion by mass of each element in a compound

The formula

% X = (# atoms of X)(atomic mass of X) × 100

molar mass of compound

A chemical formula such as CO₂ shows the ratio of atoms. Percentage composition tells you the proportion by mass. Chemists use it to assess purity (compare experimental to theoretical), verify compound identity from elemental analysis, and support stoichiometry calculations.

Compound explorer

Carbon dioxide

CO₂

Molar mass44.01 g/mol

Elements2

Type any formula

Supports brackets — Ca(NO3)2, Al2(SO4)3 — and hydrate dots — CuSO4.5H2O or CuSO4·5H2O.

Real-world applications

Forensics

Drug identification in a crime lab

A seized powder is burned in an elemental analyser. The measured percentages are matched against a reference database. Methamphetamine C₁₀H₁₅N has a unique signature of 80.48 % C, 10.13 % H, 9.39 % N — if the sample matches within experimental error, the identity is confirmed in court.

Nutrition

What's actually in a sugar cube

Table sugar is sucrose, C₁₂H₂₂O₁₁ (342.30 g/mol). By mass it is 42.11 % carbon, 6.48 % hydrogen and 51.42 % oxygen. Food scientists use percentage composition to calculate calorie content and to verify that a product's label matches what is really in the package.

Determining Empirical Formula

'Empirical' means 'by experiment' — the simplest whole-number ratio of atoms

Step 1

Percent to mass

Step 2

Mass to mole

Step 3

Divide by small

Step 4

Multiply 'til whole

Step-by-step solver

Presets

Data given as

The four steps

Step	What you do
1	Find the grams of each element. If given percentages, assume 100.0 g — then grams = the percentage.
2	Convert grams → moles using n = m / M. Write the mole amounts as subscripts.
3	Divide every subscript by the smallest value in the set.
4	If any subscript is not near a whole number, multiply all of them by the smallest factor that fixes it. (Not always needed.)

Real-world applications

Elemental analysis

Combustion analysis

A weighed sample is burned in excess O₂. All carbon becomes CO₂ and all hydrogen becomes H₂O. Two absorbent traps are weighed before and after — the mass gained gives the mass of CO₂ and H₂O, and from those the mass of C and H in the original sample. Oxygen is found by difference.

Instrumentation

Mass spectrometry

Molecules are ionised, accelerated, then bent by a magnetic field. Lighter ions bend more, so each mass-to-charge ratio lands at its own point on the detector. The tallest peak at high m/z is the whole molecule — giving the molar mass you need to turn an empirical formula into a molecular one.

Determining Molecular Formula

The true formula — always a whole-number multiple of the empirical formula

Empirical vs molecular

The empirical formula is the smallest whole-number ratio of atoms.
The molecular formula is the actual number of atoms in one molecule.
They are always multiples of one another.

Molecular	Empirical	Multiple
P₄O₁₀	P₂O₅	×2
C₁₀H₂₂	C₅H₁₁	×2
C₆H₁₈O₃	C₂H₆O	×3
C₅H₁₂O	C₅H₁₂O	×1
N₂O₄	NO₂	×2

Molecular formula calculator

Empirical Molar mass of molecule g/mol

The three steps

Step	What you do
1	Calculate the molar mass of the empirical formula (mm EF).
2	Divide: mm MF ÷ mm EF. This gives a simple whole number.
3	Multiply every subscript in the empirical formula by that factor.

If the division does not come out close to a whole number, something is wrong — recheck the empirical formula or the molar mass. A real answer is within about ±0.1 of an integer.

Real-world applications

Same ratio, different molecule

One empirical formula, three compounds

Formaldehyde CH₂O (embalming fluid), acetic acid C₂H₄O₂ (vinegar) and glucose C₆H₁₂O₆ (blood sugar) all share the empirical formula CH₂O. Percentage composition alone cannot tell them apart — you need the molar mass to find which multiple you actually have.

Pharmaceutical QC

Aspirin synthesis & verification

Salicylic acid + acetic anhydride → aspirin, C₉H₈O₄ (180.2 g/mol). Chemists confirm the product by combustion analysis plus mass spec. Because 9, 8 and 4 share no common factor, aspirin's empirical formula is already C₉H₈O₄ — the multiple is ×1.

Hydrates

Salts that carry water inside their crystal lattice

Hydration & dehydration

Heat 0 %

FormulaCuSO₄·5H₂O

Waters remaining5

ColourBlue

StateHydrated

In a hydrate the water sits inside the ionic crystal lattice, but it is only weakly bonded — heating drives it off. What remains is the anhydrous salt. The colour change of copper(II) sulfate (blue ⇌ white) is dramatic enough to be used as a chemical test for water.

Naming hydrates

1. Name the ionic compound first, ignoring the water — Ba(OH)₂·8H₂O → "barium hydroxide".
2. Add a Greek prefix (including mono) to the word "hydrate" for the number of waters → "octahydrate".
Note: separate the salt from the waters with a centred dot in the formula, but not in the name.

Click a prefix

Practice

Formula	Name
MgSO₄·7H₂O	magnesium sulfate heptahydrate
Fe(OH)₃·3H₂O	iron(III) hydroxide trihydrate
PbS₂·6H₂O	lead(IV) sulfide hexahydrate
Na₂SO₄·10H₂O	sodium sulfate decahydrate

% water by mass

% H₂O = (# H₂O molecules)(molar mass of H₂O) × 100

molar mass of hydrate

Find the formula of a hydrate

Anhydrate Mass of hydrate Mass of anhydrate

Real-world applications

Classic lab test

Heating blue copper(II) sulfate

Blue CuSO₄·5H₂O heated in a crucible turns white anhydrous CuSO₄ as steam escapes — a mass loss of 36.08 %. The reaction reverses instantly on contact with water, which makes white CuSO₄ a sensitive test for the presence of water in a liquid.

Packaging

Silica gel desiccant packets

Those "DO NOT EAT" packets protect electronics and medicines by pulling water vapour out of the air. Many contain CoCl₂ indicator beads that are blue when dry and turn pink as CoCl₂·6H₂O forms — a visual signal that the desiccant is spent. Baking it drives the water back off and regenerates it.

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Covers percentage composition, empirical formula, molecular formula & hydrates

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Summary

Quick reference

Concept	Definition	Key equation
Percentage composition	Proportion by mass of an element in a compound	% X = (# atoms × atomic mass) / molar mass × 100
Empirical formula	Smallest whole-number ratio of atoms	n = m / M, then divide by smallest
Molecular formula	Actual number of atoms in one molecule	factor = mm MF / mm EF
Hydrate	Salt with water in its crystal lattice	x = mol H₂O / mol anhydrate
Anhydrous	The salt after all water has been driven off	m(H₂O) = m(hydrate) − m(anhydrate)
% water in a hydrate	Mass fraction of the waters of hydration	% H₂O = (x × 18.02) / mm hydrate × 100

The rhyme: Percent to mass · Mass to mole · Divide by small · Multiply 'til whole
Watch out: ratios ending in .5 → ×2 · .33 or .67 → ×3 · .25 or .75 → ×4
Remember: the molecular formula is always a whole-number multiple of the empirical formula (sometimes ×1).

Greek prefixes for hydrates

1	2	3	4	5	6	7	8	9	10
mono	di	tri	tetra	penta	hexa	hepta	octa	nona	deca