Subject > Chemistry > Chemistry III > Organic Compounds > Empirical and Molecular Formulae

Empirical and Molecular Formulae

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Empirical Formula

The simple formula that gives the simplest whole number ratio between the atoms of the various elements present in the compound is called its empirical formula. For example, in a molecule of benzene (C₆H₆), carbon and hydrogen are in the ratio 1:1. So, the empirical formula of benzene is CH.

Molecular Formula

The simplest formula that gives the actual number of atoms of the various elements present in a molecule, of any compound is called its molecular formula. For example, one molecule of water contains two atoms of hydrogen and one atom of oxygen. Therefore, the molecular formula of water is H₂O.

Calculation of Empirical and Molecular Formulae

Empirical and molecular formulae of a compound can be determined from the composition of the compound, which is expressed in terms of the percentage of each element present in it.

To determine the formula of an unknown compound the following steps are taken:

Empirical and molecular formulae of a compound can be determined from the composition of the compound, which is expressed in terms of the percentage of each element present in it.

To determine the formula of an unknown compound the following steps are taken:

Step 1

Determine the percentage of each element present in the compound from the mass of each element present in a certain known mass of the compound.

Step 2

If the sum of the percentages of all these elements is not 100, then the difference gives the percentage of oxygen. The percentage of oxygen in the given compound is calculated by using the following relationship.

Percentage of oxygen = 100 - (sum of the percentages of all other elements)

Step 3

The percentage of each element is divided by the atomic mass of the respective element. The ratio so obtained is called 'atomic ratio'.

Step 4

The atomic ratios is divided by the lowest value, and converted into the nearest whole numbers. These whole numbers give the simplest ratio between the number of atoms of the various elements present in the compound.

Step 5

The empirical formula of the compound is written.

Step 6

The empirical formula mass is calculated by adding the atomic masses of all the atoms present in the empirical formula.

Step 7

The molecular mass (or molar mass) is obtained either from the experimental data or from the vapor density of the compound by using the relationship,

Molecular mass = 2 x Vapor density

Step 8

The value of n is obtained through the relation,

where, n is a pure number such as 1, 2 etc.

Step 9

The molecular formula is then obtained by

Molecular formula = n x Empirical formula

Problems

8. An organic compound weighing 0.2 g containing C, H and O gave on combustion 0.296 g CO₂ and 0.12 g H₂O. Its molecular mass is 180. Determine its empirical and molecular formulae.

Solution

Mass of the compound taken = 0.2 g

Mass of CO₂ formed = 0.296 g

Mass of H₂O formed = 0.12 g

Therefore, the percentage of O = 100 - (40.0 + 6.67) = 53.33(by difference)

Calculation of empirical formula

Therefore, empirical formula of the compound = CH₂O

Calculation of molecular formula

Empirical formula mass = (1 x 12 amu + 2 x 1 amu + 1 x 16 amu)

= 30 amu

Molecular mass = 180 amu

Hence,

Therefore, Molecular formulae of a compound = 6 x CH₂O = C₆H₁₂O₆

9. A compound (molecular mass 147) contains 49.0 % carbon and 2.72% hydrogen. 2.561 mg of the compound gave 5.00 mg of silver chloride in Carius estimation. Determine its molecular formula. Find its empirical formula also.

Solution

The problem is done in steps:

Calculation of the percentage of chlorine:

We know that, Cl = AgCl

35.5g (108 +35.5)= 143.5? 5 mg

Then,

Percentage of chlorine in the sample =

Empirical formula = C₃H₂Cl

Empirical formula mass = (3 x 12 amu + 1 amu + 1 x 35.5 amu)

= 73.5 amu

Molecular mass (given) = 147 amu

Hence,

Therefore,

Molecular formula = 2 x Empirical formula

= 2 x C₃H₂Cl

= C₆H₄Cl₂

10. About 0.45 g of an organic compound gave on combustion 0.792 g of CO₂ and 0.324 g of water 0.24 g of the same substance was Kjeldahlised and the ammonia evolved was absorbed in 50.0cm³ of N/4 H₂SO₄. The excess of the acid required 77.0 cm³ of N/10 NaOH for complete neutralization. Calculate the empirical formula of the acid.