Empirical, Molecular Formula and Limiting Reagents

Every chemical substance is known by a specific name. But many a times these names are cumbersome, confusing, and do not provide information about its chemical composition. To overcome this, each chemical compound is represented by a chemical formula that gives its composition (constituent elements present) and the number of elements of each type present. There are two types of chemical formula. They are:

Molecular Formula

The formula that gives the symbolic representation of the actual number of atoms of various elements present in one molecule of the compound is called the molecular formula. Discrete molecules can be described by this formula. As it represents one molecule of the substance giving the names and number of atoms of the various elements present, it denotes the molecular mass of the substance.

For example, the molecular formula of water is H₂O, which means that one molecule of water contains two atoms of hydrogen and one atom of oxygen. This also represents the molar mass, which is the sum of the gram atomic mass of all the atoms.

Gram atomic mass of 2 hydrogen atoms = 2 x 1.008 g

Gram atomic mass of oxygen atom = 16 g

Total molecular mass = 18 g

Empirical Formula

Empirical formula is defined as the simplest formula of the substance, which gives the relative number of atoms of each element present in the molecule of that substance. Substances which have no discrete molecules such as ionic and network covalent compounds are described by this empirical formula. This is also called as stoichiometric formula. It gives the simplest whole number ratio between the number of atoms of all the elements present in the compound.

For example, in the compound benzene, C₆H₆ there are six carbon atoms and six hydrogen atoms. The lowest whole number ratio between them is 1:1 (6:6 can be simplified to 1:1). Therefore, the empirical formula of benzene having molecular formula of C₆H₆ is CH.

Empirical formula mass or formula mass is equal to the sum of atomic masses of all the atoms present in the empirical formula. The empirical formula of benzene is CH. So,

The formula mass of CH is (12+1) = 13 amu or 13 g/mol.

Relationship between empirical and molecular formulae

The two formulas are related as Molecular formula = n x empirical formula

where 'n' may have whole number values 1,2,3… . The value of 'n' can be obtained by the following relationship

For example, the molecular mass of benzene is 78 and its empirical formula is CH and therefore, its empirical formula mass is 13.

Therefore, the molecular formula of benzene is 6 x (CH) = C₆H₆.

Determination of the empirical formula of a compound

The empirical formula of a compound is determined from the percentage composition of different elements and atomic masses of the elements. The various steps involved in determining the empirical formula are:

• The percentage of each element is divided by its atomic mass. This gives the relative number of atoms of various elements in the molecule of the compound.

• The result obtained in the above step is divided by the smallest value to get the simplest ratio of various elements.

• The values obtained are made to the nearest whole number ratio (multiplied if necessary by a suitable integer to make the values whole numbers).

• The symbols of various elements are written side by side and the numerical value at the right hand lower corner of each symbol is inserted.

Example

How can we differentiate a molecular formula from an empirical formula?

If the subscripts in the formula have a common divisor, it is usually a molecular formula. Generally the empirical formula is multiplied by this common divisor to get the molecular formula.

Example:

Empirical formula of acetic acid is CH₂O.

Molecular formula is CH₃ COOH = C₂H₄O₂

C₁H₂ O₁ x 2 = C₂H₄O₂ [Molecular formula]

Numericals based on empirical formula

Example:1

An oxide of iron contains 72.41% of iron. Calculate the empirical formula for the oxide of iron [Fe = 56; O=16].

Solution

Therefore simple ratio = Fe₃O₄.

Empirical formula = Fe₃O₄.

Steps for calculation

• Calculate the percentage by weight of each element.

• Find out relative number of atoms by dividing percentage of weight by atomic weight.

• Choose the simplest ratio and the smallest, divide all the ratios by it.

• If whole numbers are not obtained, then multiply it by a smallest integer to make it whole.

Example: 2

The percentage composition of a compound is 71.8% antimony (Sb) and 28.2% sulphur. What is the empirical formula of this compound?

Solution:

In 100g of the compound, masses of elements are as follows:

In 100 grams of compound, 71.8 g are antimony and 28.2% sulphur.

Empirical formula is Sb₂S₃.

Determination of molecular formula from empirical formula

Molecular formula is the chemical formula, which represents the actual numbers of atoms of each element present in a compound.

Example: 1

Calculate the molecular formula of a compound with vapor density of 30 having 40% carbon; 6.67% of hydrogen and the rest is oxygen.

Empirical formula = C₁H₂O₁

Empirical formula weight = 12 x 1 + 2 x 1 + 1 x 16

= 12 + 2 + 16

= 30 g

Molecular weight = 2 x vapor density

= 2 x 30

= 60

Molecular weight = n x empirical weight

60 = n x 30

Molecular formula = n x empirical formula

= 2 x CH₂O = C₂H₄O₂

Steps

• Calculate empirical formula.

• Use vapor density if given.

• If molecular weight given, calculate 'n' using this formula.

• Molecular formula = n x empirical formula

Example: 2

A compound has molecular formula C₅H_10. What is its empirical formula?

Solution:

Ratio of C atoms to H atoms is 5 : 10 = 1:2.

Empirical formula is C₁H_2.

_{Limiting reagents:}

_{In a desired chemical reaction some reactants may be present in lesser or greater proportions than the stoichiometry as indicated by the balanced chemical equation. The reactant, which is completely used up first as per the stoichiometry, limits the amount of product that can be formed and does not allow the reaction to go further. This is the limiting reagent. The excess reactants are left behind as unconsumed reagents, being limited by the limiting reagent. In such cases the desired reaction does not go to 100% completion.}