Empirical and Molecular formulae: Difference and Conversions

An empirical formula gives the simplest or reduced ratios (whole numbers) of atoms present in a compound. Whereas, a molecular formula tells the total number of atoms (whole numbers) of each element in a compound.

Let’s take the example of ethane to understand the empirical and molecular formulae. There are two carbon and four hydrogen atoms are present in ethane. So, the molecular formula of ethene is C2H4, while its empirical formula is CH2.

empirical vs molecular formulas: explanation

The main differences between Empirical and Molecular Formulae are given below:

Empirical Formula Molecular Formula
It is the simplest or reduced ratio of atoms in a compound It is the total number of atoms in a compound
An empirical formula cannot give the exact molecular mass. It gives the empirical formula mass A molecular formula can give the exact molecular mass
It can predict the type of atoms in ionic compounds and polymers It can predict the oxidation states of atoms, their reactions, and products
Empirical formulae are found by combustion analysis Molecular formulae are found by a mathematical procedure using empirical formulae
For example, Benzene (CH), Ethene (CH2), Glucose (CH2O), etc
For example, Benzene (C6H6), Ethene (C2H4), Glucose (C6H12O6), etc

To explain the molecular and empirical formulas of a compound, it is necessary to know the percentage of each element present in the compound. For this purpose,

  • First, identify the elements present in a compound. This step is termed qualitative analysis.
  • The second step is the quantitative analysis, where the mass of each element is determined.

The percentage composition is the number of grams of an element present in 100 grams of a compound.

Percentage composition of an element = Mass of an element in compound / Mass of the compound x 100%

Empirical Formula

It is a formula that gives the whole number ratio between the atoms of different elements present in a compound. For example,

  • Benzene has six carbons and six hydrogens atoms (C6H6), its empirical formula becomes CH.
  • Glucose has six carbons, twelve hydrogens, and six oxygens (C6H12O6). its empirical formula is, therefore, CH2O, etc.

How to write Empirical Formula

The empirical formula can be calculated by the following steps;

  1. First of all, the percentage elemental composition needs to be determined.
  2. Now, the number of grams of atoms of each element are calculated. For this, the percentage of each element is divided by its atomic mass.
  3. The gram atoms of each element are then divided by the smallest gram atoms value. This gives the atomic ratios of each element.
  4. An empirical formula can easily be obtained when the atomic ratios are simple whole numbers. Otherwise, those values are multiplied to get a whole numeric atomic ratio.

For example, vitamin C contains 40.92% carbon, 4.58% hydrogen, and 54.5% oxygen by mass. Its empirical formula can be calculated by the following steps;

To get the number of gram atoms, the percentage compositions are divided by the atomic masses of individual elements.

  • Number of gram atoms of carbon = 40.92 g / 12.0 gmol-1 = 3.41 gram atoms
  • Number of gram atoms hydrogen = 4.58 g / 1.008 gmol-1 = 4.54 gram atoms
  • Number of gram atoms oxygen = 54.5 g / 16 gmol-1 = 3.406 gram atoms

To get the atomic ratios, each value of gram atoms is divided by the lowest one i.e. 3.406,

C : H : O

3.41 / 3.106 : 4.54 / 3.406 : 3.406 / 3.406

1 : 1.33 : 1

For converting into whole numbers, multiply with the sum of all i.e. 3,

3 x (1 : 1.33 : 1)

3 : 4 : 3

So, the empirical formula for vitamin C is, therefore, C3H4O3.

Empirical Formula from Combustion Analysis

Compounds that contain only carbon, hydrogen, and oxygen can be analyzed by combustion. Carbon dioxide and water are produced during combustion. The masses of these products are then used up to make the empirical formulae.

Process of combustion analysis

Organic compounds are placed in the combustion tube as a sample. This tube is then fitted in the furnace. Oxygen is supplied as a source for burning, and the compound is burnt there. In this process, hydrogen is converted into water, and carbon is converted into carbon dioxide. Water and carbon dioxide are absorbed in the Mg (ClO4) absorber and 50% KOH absorber. Produced amount of water and carbon dioxide is obtained by the difference in the masses of absorbers. The oxygen is determined by the method of difference.

Combustion analysis: empirical formulas

To obtain the percentage of carbon, hydrogen, and oxygen following formulas are used;

  • Percentage of carbon = Mas of carbon dioxide / Mass of organic compound x 12.00 / 44.00 x 100
  • Percentage of hydrogen = Mas of water / Mass of organic compound x 2.016 / 18 x 100

The percentage of oxygen can be obtained by;

Percentage composition of oxygen = 100 – (% of carbon + % of hydrogen)

Obtaining an Empirical formula

A sample of the liquid containing carbon, hydrogen, and oxygen was subjected to combustion analysis. 0.5439 of the organic compound gave 1.039 g of carbon dioxide and 0.6369 g of water. Determine the empirical formula of the compound.

Mass of organic compound used = 0.5439 g

Mass of carbon dioxide released = 1.039 g

Mass of water produced = 0.6369 g

To get percentage use above formula of combustion analysis

  • Percentage of the carbon = 1.039 g / 0.543 g x 12.00 / 44.00 x 100 = 52.108 g
  • Percentage of the hydrogen = 0.6369 g / 0.543 g x 2.016 / 18 x 100 = 13.11 g
  • Percentage of the oxygen = 100 – (52.108 + 13.11 ) = 34.77 g

Now, calculating number the gram atoms;

  • Number of gram atoms of carbon = 52.108 / 12 = 4.34
  • Number of gram atoms of hydrogen = 13.11 / 1.008 = 13.01
  • Number of gram atoms of oxygen = 34.77 / 16.00 = 2.17

To calculate the atomic ratio, all values of gram atoms are divided by the smallest one (2.17 in this case);

  • Atomic ratio of carbon = 4.34 / 2.17 = 2
  • Atomic ratio of hydrogen = 13.01 / 2.17 = 6
  • Atomic ratio of oxygen = 2.17 / 2.17 = 1

The empirical formula for that organic compound is therefore C2H6O.

Molecular Formula

A molecular formula gives the total number of atoms present in a molecule of a compound. For example;

  • There are six carbons, twelve hydrogens, and six oxygens present in a molecule of glucose, so, the molecular formula is C6H12O6.
  • In a molecule of benzene, six carbons and six hydrogens are involved. The molecular formula of benzene is therefore C6H6.

The empirical formulae for benzene and glucose are CH and CH2O. The molecular formulas for these compounds are a multiple of their empirical formulas. Hence, it is concluded that;

Molecular formula = n (Empirical Formula)

where,

    • n = number of integers

There are many compounds whose molecular and empirical formulas are the same. For example;

  • Water (H2O)
  • Carbon dioxide (CO2)
  • Ammonia (NH3)
  • Sucrose (C12H22O11)
  • Triphophorus pentnitride (P3N5), etc.

In these cases, the multiple (n) is unity.

The value of n is the ratio of molecular mass and empirical formula mass of the substance.

n = Molecular mass / Empirical formula mass

Obtaining Molecular formula

Let’s say that, the combustion analysis of a compound gives 65.44% carbon, 5.50% hydrogen, and 29.06% oxygen. Now, what is the empirical formula of the compound? The molecular mass of this compound is 110.5 g/mol. Calculate the molecular formula of the compound.

First, obtain the gram atoms or moles by dividing the percentages by their atomic masses.

  • Number of gram atoms of carbon = 65.44 g / 12 gmol-1 = 5.45
  • Number of gram atoms of hydrogen = 5.50 g / 1.008 gmol-1 = 5.46
  • Number of gram atoms of oxygen = 29.06 g / 16.00 gmol-1 = 1.82

The molar ratio of this compound is, therefore;

Carbon : Hydrogen : Oxygen

5.45 : 5.46 : 1.82

Dividing by the smallest number, we get;

Carbon : Hydrogen : Oxygen

5.45 / 1.82 : 5.46 / 1.82 : 1.82 / 1.82

3 : 3 : 1

This compound has a ratio of 3 : 3 : 1. So, the empirical formula is C3H3O.

To determine the molecular formula, we have to calculate the empirical formula mass.

  • Empirical formula mass = 12 x 3 +1.008 x 3 + 16 x 1 = 55.05 g/mol
  • Molar mass of the compound = 110.15 gmol-1

The molecular formula is calculated as;

n = molecular mass / empirical formula mass

= 110.15 / 55.05

n = 2

Molecular formula = n (empirical formula)

= 2 ( C3H3O)

Molecular formula = C6H6O2    —    [Hydroquinone]

Key Takeaway(s)

Empirical and Molecular formulae Difference and Conversions Difference table

Concepts Berg

How to convert a molecular formula to an empirical formula?

We can convert a molecular formula into an empirical formula by getting the simple ratio of the atoms of elements present in that compound. For example, the molecular formula of benzene is C6H6 and its empirical formula is CH.

What is an empirical formula?

It is the simplest or reduced ratio of the atoms in a molecule of a compound. For example, there are six carbon atoms and six hydrogen atoms present in the benzene molecule. So, its empirical formula is CH, as (6/6 = 1).

What is a molecular formula?

A formula that gives the exact number of atoms present in a molecule of a compound. For example, six carbons and six hydrogen atoms are present in benzene and its molecular formula is C6H6.

How are the molar mass and empirical formula mass for a compound related?

Both are related to each other by a relationship that is

n = Molecular mass / Empirical formula mass

What’s the purpose of empirical formulas in chemistry?

It gives a simple ratio of the atoms of the element present in a molecule of a given sample compound.

Reference books

  • Chemistry in Quantitative Language: Fundamentals of General Chemistry, bChristopher O. Oriakhi (MIT, Nike, hp, RSC, etc)

Reference links

Was this article helpful?