Chemistry · Lesson

Composition, empirical formulas and molecular formulas

Convert measured mass proportions into atom ratios. Use a molar mass to choose a molecular formula, then identify the structural information still missing.

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By the end, you can

Distinguish mass percent from atom or amount ratios.
Derive an empirical formula from supplied element masses.
Use a molecular molar mass to choose an integer multiple.

Mass fractions count mass, not atoms

A mass percentage tells you how much of a sample’s mass belongs to a constituent: constituent mass divided by total mass, multiplied by 100%. In water, the two hydrogen atoms do not contribute two thirds of the mass. Hydrogen atoms are much lighter than oxygen atoms. With the supplied molar-mass contributions 2.016 and 16.00 g/mol, hydrogen contributes about 11.19% of the total 18.016 g/mol.

Oxygen contributes the remaining about 88.81%. The atom ratio is still H:O = 2:1. Mass percentages and atom ratios answer different questions. When composition data are given as mass percentages, you must convert each element’s mass to amount before identifying an atom ratio.

Convert each element separately

Consider an invented composition containing 12.01 g carbon, 2.016 g hydrogen and 16.00 g oxygen. Using C = 12.01, H = 1.008 and O = 16.00 g/mol, the element amounts are 1 mol C atoms, 2 mol H atoms and 1 mol O atoms. Dividing all amounts by the smallest gives the ratio 1:2:1.

The corresponding empirical formula is CH₂O. You can start from an assumed 100 g total if percentages are provided, because that choice turns the numerical percentages into constituent grams without changing the ratios. This is a calculation convenience, not a claim that the original measured sample necessarily weighed 100 g.

Mass proportions become an atom ratio

Element	Mass / g	Molar mass / (g/mol)	Atom amount / mol	Normalized ratio
C	12.01	12.01	1	1
H	2.016	1.008	2	2
O	16.00	16.00	1	1

Invented element masses and supplied rounded molar masses. Dividing each mass by the matching element’s molar mass produces 1:2:1, not the original mass ratio.

Look for a justified integer ratio

Measured ratios may not land on exact integers. Values close to 1:2:1 within the stated measurement quality can support CH₂O. A ratio near 1:1.5 must not be rounded casually to 1:2; multiply both entries by two to get 2:3. The goal is a common whole-number ratio that fits the data, not arbitrary independent rounding.

Uncertainty, sample purity and the assumed element list matter. If the data do not support a simple ratio, investigate before forcing a formula. Some materials are mixtures or have variable composition. The familiar empirical-formula procedure has a stated scope and should not be treated as proof that every analysed sample is one pure molecular compound.

Molar mass supplies the molecular multiplier

The empirical formula CH₂O has a calculated molar-mass sum of 12.01 + 2(1.008) + 16.00 = 30.026 g/mol using these inputs. If independent evidence gives a molecular molar mass of 180.156 g/mol for the intended pure compound, the ratio is six. Multiply every empirical subscript by six: C₆H₁₂O₆.

The resulting molecular formula specifies the atom counts in a molecule. It still need not specify connectivity, stereochemistry or which substance name is correct. Several structures can share a molecular formula. Composition evidence and molecular-mass evidence solve part of the identification problem; structure determination adds more.

Worked examples

From composition to molecular formula

The invented C/H/O composition above gives CH₂O. An independently supplied molecular molar mass is 180.156 g/mol. What molecular formula follows?

Calculate the empirical-formula molar-mass sum: 30.026 g/mol.
Divide the molecular value by the empirical value: 180.156 / 30.026 = 6.
Multiply all atom counts in CH₂O by 6.

C₆H₁₂O₆. This molecular formula does not by itself identify a unique structure or substance.

Try it yourself

Does H₂O mean hydrogen contributes two thirds of the mass?
Show the answer and reasoning
No.
The subscript counts atoms. Atomic masses determine their mass contributions.
An amount ratio is 1:1.5. What whole-number ratio has the same proportions?
Show the answer and reasoning
2:3.
Multiply both entries by two; independently rounding 1.5 would change the ratio.
An empirical formula is CH₂ and its molar-mass sum is 14.026 g/mol. A molecular molar mass is 42.078 g/mol. What formula follows?
Show the answer and reasoning
C₃H₆.
The molecular-to-empirical mass ratio is three, so multiply both subscripts by three.
Do composition and molecular mass always establish one unique molecular structure?
Show the answer and reasoning
No.
Different connectivities or spatial arrangements can share the same molecular formula.

What to carry forward

Mass percentage and atom ratio are different quantities.
Convert each element mass to atom amount before comparing ratios.
Use a common multiplier for fractional ratios.
A molecular formula needs the molecular multiplier and still leaves possible structural ambiguity.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.