Chemical Formulae

Empirical and Molecular Formulas

Empirical Formula: The simplest ratio of atoms in a compound. It shows the relative number of atoms of each element. For example, the empirical formula of hydrogen peroxide is $HO$ .
Molecular Formula: This shows the actual number of atoms of each element in a molecule. For hydrogen peroxide, the molecular formula is $H_2O_2$ , which indicates two hydrogen atoms and two oxygen atoms.

Calculating Empirical Formulas

From Percentage Composition by Mass:

Convert percentage composition to grams (assume 100 g of substance).
Convert grams to moles by dividing by the atomic mass of each element.
Find the simplest ratio by dividing the number of moles of each element by the smallest number of moles calculated.
Adjust the ratio to whole numbers if needed.

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Example: A compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

$40$ g $C$ , $6.7$ g $H$ , $53.3$ g $O$

Moles of $C$

= \frac{40}{12} = 3.33

Moles of $H$

= \frac{6.7}{1} = 6.7

Moles of $O$

= \frac{53.3}{16} = 3.33

Simplest ratio:

C:H:O = \frac{3.33}{3.33}:\frac{6.7}{3.33}:\frac{3.33}{3.33} = 1:2:1.

Empirical formula = $CH_2O$

From Masses of Reactants and Products:

Find the mass of each element in the compound.
Convert these masses to moles.
Use the mole ratio to find the empirical formula as outlined above.

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Example: A sample has 6 g of magnesium and 4 g of oxygen.

Moles of $Mg$

= \frac{6}{24} = 0.25

Moles of $O$

= \frac{4}{16} = 0.25

The ratio of $Mg:O$ = $1:1$

Therefore the Empirical formula = $MgO$

Molecular Formulas from Empirical Formulas

To determine the molecular formula:

Calculate the molar mass of the empirical formula.
Divide the given molecular mass by the empirical formula mass.
Multiply the subscripts in the empirical formula by this value to get the molecular formula.

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Example: If the empirical formula is $CH_2O$ and the molecular mass is 180 g/mol:

Molar mass of $CH_2O$

= 12 + (2 × 1) + 16 = 30 g/mol.

\frac{180}{30} = 6

Molecular formula:

(CH_2O)6 = C_6H{12}O_6 (glucose).

Percentage Composition by Mass

This represents the percentage by mass of each element in a compound. It can be calculated from the molecular formula.

Formula:

\text{Percentage of element} = \frac{\text{mass of element in 1 mole}}{\text{molar mass of compound}} \times 100

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Example: For water $H_2O$

Molar mass $= 2(1) + 16 = 18 g/mol$

Percentage of $H$ :

= \frac{2}{18} \times 100

= 11.11\%

Percentage of $O$ :

= \frac{16}{18} \times 100

= 88.89\%

Structural Formulas

A structural formula shows how atoms are arranged and bonded in a molecule. It provides more information than a molecular formula by indicating the connectivity between atoms.

Examples:

Methane ( $CH₄$ ): A single carbon atom bonded to four hydrogen atoms in a tetrahedral arrangement.
Glucose ( $C₆H₁₂O₆$ ): Contains multiple hydroxyl ( $-OH$ ) groups and a six-membered ring with alternating single and double bonds.

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Exam Tip:

In calculations, ensure to check units and round your answers appropriately.
Show all steps clearly, especially for empirical and molecular formula derivations.

Chemical Formulae Simplified Revision Notes for Leaving Cert Chemistry

Chemical Formulae

Empirical and Molecular Formulas

Calculating Empirical Formulas

From Percentage Composition by Mass:

From Masses of Reactants and Products:

Molecular Formulas from Empirical Formulas

Percentage Composition by Mass

Formula:

Structural Formulas

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