Sales Toll Free No: 1-855-666-7446

# Empirical Formula

Top
 Sub Topics 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

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

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

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

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

 S.No Compounds Empircal formula 1 Empirical formula of magnesium oxide MgO 2 Empirical formula of copper chloride CuCl2 3 Ethyl butyrate empirical formula C6H12O2 4 Empirical formula for glucose C6H12O6 6 Empirical formula for tartaric acid C2H3O3 7 Empirical formula of hydrocarbon CH2 8 Ascorbic acid empirical formula C3H4O3 9 Empirical formula of silver oxide Ag2O 10 Empirical formula of benzene CH 11 Empirical formula of zinc chloride No empirical formula Molecular formula is ZnCl2 12 Empirical formula of ibuprofen C13H18O2 13 Empirical formula of menthol C10H20O 14 Empirical formula for copper sulfide CuS 15 Silicon tetrachloride empirical formula SiCl4 16 Empirical formula of caffeine C8H10N4O2 17 Vanillin empirical formula C8H8O3 18 Empirical formula for naphthalene C10H8 19 Empirical formula of iron oxide FeO 20 Empirical formula of aspirin C12H24N2O4

## Empirical Formula Problems

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?
Solution:

 Element Magnesium Chlorine Symbol Mg Cl Masses 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.
Solution:

 Element Mg S O Mass 23.3 30.7 46.0 Relative atomic mass 24 32 16.0 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 1 1 3

The empirical formula is MgSO3.