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Empirical Formula


The empirical formula of a compound is the simplest formula that gives the correct relative numbers of atoms of each kind in a compound. The empirical formula of a compound can be determined from either the masses or mass percentage of the elements in a sample.

The empirical formula is usually all we need to describe the composition of an ionic compound. Empirical is a term that means "experimentally determined." Though the formula of a chemical compound can be written in the simplest way called empirical formula using chemical symbols and numerical subscripts.

Empirical Formula Definition

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Define Empirical Formula

The empirical formula of a compound is defined as that "formula of a substance which gives the simplest whole number ratio between the atoms of the various elements present in one molecule of the substance."

In other words empirical formula is the simplest formula calculated form the percent composition data.

Empirical and Molecular Formula

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From the formula of the compound the mass of each element can be determined. The reverse is also true; from the mass of each element present in a sample of the compound, the formula of the compound can be determined.

In general,
  • Number of mole of substance = $\frac{Mass\ of\ substance}{Mass\ of\ one\ mole\ of\ the substance}$
  • Mass of one mole of an element = Relative atomic mass in grams
  • Mass of one mole of a compound = Relative formula mass in grams
Empirical formula to molecular formula or molecular formula from empirical formula is given below.

Molecular formula = (Empirical formula) n

Where n = 1, 2, 3

$n = \frac{Molecular\ formula}{Empirical\ formula}$

Finding Empirical Formula

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In determine empirical formula three things are very important.
  1. The elements that are combined.
  2. Their atomic masses and
  3. The ratio by mass or percentage in which they are combined.

The three steps involved in calculating empirical formula are listed below.

  1. Find the amount in moles of each element present by dividing the mass of each element by its molar mass.
  2. Find the ratio of the number of atoms of each element by dividing the amounts in moles by the smallest value found in previous step.
  3. Convert these numbers into whole numbers, because atoms combine together in whole number ratios.

Empirical Formula Examples

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The simplest formula representing the composition of a compound is called empirical formula. Some of the examples are given below.

Empircal formula
1 Empirical formula of magnesium oxide MgO
Empirical formula of copper chloride CuCl2
Ethyl butyrate empirical formula C6H12O2
Empirical formula for glucose C6H12O6
Empirical formula for tartaric acid C2H3O3
Empirical formula of hydrocarbon CH2
Ascorbic acid empirical formula C3H4O3
Empirical formula of silver oxide Ag2O
Empirical formula of benzene CH
Empirical formula of zinc chloride No empirical formula
Molecular formula is ZnCl2
Empirical formula of ibuprofen C13H18O2
Empirical formula of menthol C10H20O
Empirical formula for copper sulfide CuS
Silicon tetrachloride empirical formula SiCl4
Empirical formula of caffeine C8H10N4O2
Vanillin empirical formulaC8H8O3
Empirical formula for naphthalene C10H8
Empirical formula of iron oxide FeO
Empirical formula of aspirin C12H24N2O4

Empirical Formula Problems

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Solved problems based on empirical formula are given below.

Solved Examples

Question 1: Given that 0.96g of magnesium combines with 2.84g of chlorine, what is the empirical formula for magnesium chloride?
Element Magnesium Chlorine
0.96g  2.84g
Relative atomic mass
24 35.5
Number of moles
$\frac{0.96}{24}$ = 0.04 $\frac{2.84}{35.5}$ = 0.08
Divide through by small number
1 mole to  2 mole 
Number of atoms
1 atom to 2 atoms

The empirical formula isMgCl2.

Question 2: Determine the empirical formula of a compound of magnesium which contains 23.3% of magnesium 30.7% of sulfur and 46% of oxygen by mass.
Mg S
Mass 23.3
Relative atomic mass
24 32
Mole number
$\frac{23.3}{24}$ = 0.971
$\frac{30.}{32}$ = 0.959
$\frac{46.0}{16.0}$ = 2.88
Divide by smallest number
$\frac{0.971}{0.959}$ = 1 $\frac{0.959}{0.959}$ = 1
$\frac{2.88}{0.959}$ = 3
Ratio of atoms

The empirical formula is MgSO3.