Molecular Formula: Molecular formula and empirical formula are two significant aspects of chemistry. Both these phenomena contribute to helping students engage in different calculations and understand the subject better. It is one of the basic concepts of Chemistry and plays an important role in building a strong base for students.

This article aims to discuss different properties and examples of molecular formulas. Learn in detail about molecular formulas, calculating empirical formulas with solved examples. Scroll down to read more.

What is Molecular Formula?

Molecular formulas describe the exact number and type of atoms present in a single molecule of a compound. The constituting elements of a compound are represented by their chemical symbols followed by numeric subscripts describing the number of atoms of each element present in the molecule.

A molecular formula contains no words and is not a chemical name. Although a molecular formula may imply specific simple chemical structures, it is not the same as a complete chemical structural formula. They are more limiting than structural formulas. They are compact and easy to communicate; however, they lack the information about bonding and atomic arrangement provided in a structural formula.

Empirical Formulas and Molecular Formulas

The simplest types of chemical formulas are called empirical formulas, representing the simplest whole-number ratio of atoms in a compound. The molecular formula for a compound can be the same as or a multiple of the compound’s empirical formula.

 For example:

1. The molecular formula for glucose is \({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}\). The molecular formula indicates the exact number of atoms in the molecule.
2. The empirical formula expresses the smallest whole-number ratio of the atoms in the molecule. In the above case, the empirical formula of glucose is \({\rm{C}}{{\rm{H}}_2}{\rm{O}}\).

Empirical Formulas and Molecular Formula Table

Name of Compound	Molecular Formula of All Elements	Empirical Formula	n factor
Methane	\({\rm{C}}{{\rm{H}}_4}\)	\({\rm{C}}{{\rm{H}}_4}\)	\(1\)
Acetic acid	\({{\rm{C}}_2}{{\rm{H}}_4}{{\rm{O}}_2}\)	\({\rm{C}}{{\rm{H}}_2}{\rm{O}}\)	\(2\)
Glucose	\({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}\)	\({\rm{C}}{{\rm{H}}_2}{\rm{O}}\)	\(6\)
Benzene	\({{\rm{C}}_6}{{\rm{H}}_6}\)	\({\rm{ CH }}\)	\(6\)
Urea	\({\rm{C}}{{\rm{H}}_4}\;{{\rm{N}}_2}{\rm{O}}\)	\({\rm{C}}{{\rm{H}}_4}\;{{\rm{N}}_2}{\rm{O}}\)	\(1\)
Hydrogen peroxide	\({{\rm{H}}_2}{{\rm{O}}_2}\)	\({\rm{ HO }}\)	\(2\)

The table above shows the comparison between the empirical and molecular formulas of methane, acetic acid, glucose, benzene, urea, hydrogen peroxide and their respective n factors. The molecular formula of methane is \({\rm{C}}{{\rm{H}}_4}\). It is also its empirical formula.

It is because it contains only one carbon atom. Formic acid, an acid readily found in ant and bee stings, has the molecular formula \({\rm{C}}{{\rm{H}}_2}{\rm{O}}\).

Molecular Formula of all Elements

Acetic acid is the main component of vinegar with a molecule formula \({{\rm{C}}_2}{{\rm{H}}_4}{{\rm{O}}_2}\). The simplest of all carbohydrates,  glucose has the molecular formula \({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}\). However, all three have the same empirical formula of \({\rm{C}}{{\rm{H}}_2}{\rm{O}}\).

Structural Formula of Acetic Acid

Structural Formula of Formic Acid

Similarly, the compound dichlorine hexoxide has an empirical formula \({\rm{Cl}}{{\rm{O}}_3}\) and the molecule formula \({\rm{C}}{{\rm{l}}_2}{{\rm{O}}_6}\) and the compound hydrogen peroxide has the empirical formula HO and the molecular formula \({{\rm{H}}_2}{{\rm{O}}_2}\).

For organic compounds, carbon and hydrogen are listed as the first element in the molecular formula, and the remaining elements follow them in alphabetical order. For example, for Butane, the molecule formula is \({{\rm{C}}_4}{{\rm{H}}_{10}}\).

The molecule formula will always be some integer multiple \(\left( {\rm{n}} \right)\) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).

\({\rm{Molecular}}\,{\rm{Formula}} = {\rm{n}}({\rm{Empirical}}\,{\rm{Formula}})\)

\({\rm{n}} = \frac{{{\rm{ Molecular\, Formula }}}}{{{\rm{ Empirical\, Formula }}}}\)

The integer multiple, n, can also be obtained by dividing the molar mass, MM, of the compound by the empirical formula mass, EFM (the molar mass represented by the empirical formula).

\({\rm{n}} = \frac{{{\rm{MM}}({\rm{Molar}}\,{\rm{Mass}})}}{{{\rm{EFM}}({\rm{Empirical}}\,{\rm{Formula}}\,{\rm{Mass}})}}\)

If the vapour density of the substance is known, its molecular mass can be calculated by using the equation.

\({\rm{Molecular}}\,{\rm{mass}} = 2 \times {\rm{Vapour}}\,{\rm{density}}({\rm{VD}})\)

Empirical formulas can be determined from the percentage composition of a compound. However, to determine its molecular formula, it is necessary to know the molar mass of the compound.

Steps to Calculate Molecular formula of all Elements

The following steps can determine the molecule formula of a compound-

• 1st Step: Calculate the empirical formula from percentage composition.
• 2nd Step: Calculate the Empirical Formula mass (EFM) by adding up the molar atomic masses of all atoms constituting the formula.
• 3rd Step: Determination of the Molar Mass of the given compound.
• 4th Step: Determination of the value ‘n’. It is done by dividing the molar mass of the compound by the empirical Formula Mass of the compound calculated in step 2.
• 5th Step: Multiply all the subscripts in the empirical formula by the whole number found in step 4. The result is the molecule formula.

Solved Examples of Molecular Formula

1. A compound contains carbon, hydrogen and nitrogen in the ratio \(9:1:3.5\). Calculate the empirical formula. If its molecular mass is \(108\), what is the molecular formula?
Solution:

Element	Element ratio	Atomic mass	Relative number of atoms	Simplest ratio
Carbon	\(9\)	\(12\)	\(\frac{9}{{12}} = 0.75\)	\(\frac{{0.75}}{{0.25}} = 3\)
Hydrogen	\(1\)	\(1\)	\(\frac{1}{1} = 1\)	\(\frac{1}{{0.25}} = 4\)
Nitrogen	\(3.5\)	\(14\)	\(\frac{{3.5}}{{14}} = 0.25\)	\(\frac{{0.25}}{{0.25}} = 1\)

Empirical formula \( = {{\rm{C}}_3}{{\rm{H}}_4}\;{\rm{N}}\)
Empirical formula mass (EFM) \( = (3 \times 12) + (4 \times 1) + 14 = 54\)

\({\rm{n}} = \frac{{{\rm{MM}}({\rm{Molar}}\,{\rm{Mass}})}}{{{\rm{EFM}}({\rm{Empirical}}\,{\rm{Formula}}\,{\rm{Mass}})}} = \frac{{108}}{{54}} = 2\)

\({\rm{Molecular}}\,{\rm{Formula}} = {\rm{n}}({\rm{Empirical}}\,{\rm{Formula}})\)

Thus, molecule formula of the compound \( = 2{\rm{x}}\) empirical formula \( = 2{\rm{x}}\,{{\rm{C}}_3}{{\rm{H}}_4}\;{\rm{N}} = {{\rm{C}}_6}{{\rm{H}}_8}\;{{\rm{N}}_2}\)

2. A carbon compound on analysis gave the following percentage composition. Carbon \(14.5%\), hydrogen \(1.8%\), chlorine \(64.46%\), oxygen \(19.24%\). Calculate the empirical formula of the compound.

Solution: Step 1: Percentage composition of the elements present in the compound.

\(\begin{array}{*{20}{c}}
{\rm{C}}\\
{{\rm{14}}{\rm{.5}}}
\end{array}\,\,\,:\,\,\,\,\begin{array}{*{20}{c}}
{\rm{H}}\\
{{\rm{1}}{\rm{.8}}}
\end{array}\,\,\,\,{\rm{:}}\,\,\,\begin{array}{*{20}{c}}
{\rm{C}}\\
{{\rm{64}}{\rm{.46}}}
\end{array}\,\,\,{\rm{:}}\,\,\,\begin{array}{*{20}{c}}
{\rm{O}}\\
{{\rm{19}}{\rm{.24}}}
\end{array}\)

Step 2: Dividing by the respective atomic weights

\(\begin{array}{*{20}{c}}
{\frac{{14.5}}{{12}}}\\
{1.21}
\end{array}\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\frac{{1.8}}{1}}\\
{{\rm{1}}{\rm{.8}}}
\end{array}\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\frac{{64.46}}{{35.5}}}\\
{1.81}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\frac{{19.24}}{{16}}}\\
{{\rm{1}}{\rm{.2}}}
\end{array}\)

Step 3: Dividing the values in step 2 among them by the smallest number.

\(\frac{{1.21}}{{1.20}}\,\,\,\,\,\,\,\frac{{1.8}}{{1.2}}\,\,\,\,\,\,\,\frac{{1.81}}{{1.2}}\,\,\,\,\,\,\frac{{1.2}}{{1.2}}\)

Step 4: Multiplication by a suitable integer to get the whole number ratio.

\(\begin{array}{*{20}{c}}
{1 \times 2}\\
2
\end{array}\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{1.5 \times 2}\\
3
\end{array}\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{1.5 \times 2}\\
3
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{1 \times 2}\\
2
\end{array}\)

\(\therefore \) The simplest ratio of the atoms of different elements in the compound
\({\rm{C: H: C l: O = 2: 3: 3: 2}}\)
\(\therefore \) Empirical formula of the compound \({{\rm{C}}_2}{{\rm{H}}_3}{\rm{C}}{{\rm{l}}_3}{{\rm{O}}_2}.\)

Molecular and Structural Formulas

Molecular formulas do not specify any information regarding the arrangement of atoms. Hence, one molecule formula can describe a number of different chemical structures. A structural formula indicates the number of atoms and their arrangement in space, and the bonding between atoms. Compounds with the same molecules formula but different chemical structures are known as isomers.

For example- The molecular formula of Butane is \({{\rm{C}}_4}{{\rm{H}}_{10}}\) and is the smallest hydrocarbon alkane that has isomers.

The four carbon atoms of Butane can arrange themselves in two different manners resulting in a straight-chain structure and a branched-chain structure. The straight-chain structure is called Butane (or n-butane), whereas its only isomer, the branched structure, is called isobutane. The IUPAC name of isobutane is \(2\)-methylpropane.

Both molecules have four carbon atoms and ten hydrogen atoms \(\left( {{{\rm{C}}_4}{{\rm{H}}_{10}}} \right)\) but the atoms are arranged differently in the both molecules.

Summary

The molecular formula is essential in enhancing the readability and understanding of chemical equations. Hence, it is important to learn its concept and how it is represented. We have learned the molecule formula of certain compounds and their relation to the empirical formula through this article. We also learnt the similarities and differences it has with empirical formulas through solved examples.

FAQs on Molecule Formula

Frequently asked questions related to the formula of the molecule are listed as follows:

Q.1: What is a molecular formula? Give an example.
Ans: The molecular formula represents the exact number and type of atoms present in a single molecule of a compound. The constituting elements of a compound are represented by their chemical symbols followed by numeric subscripts describing the number of atoms of each element present in the molecule. The molecular formula for glucose is \({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}\). The molecular formula indicates the exact number of atoms in the molecule.

Q.2: Difference between the molecular formula and empirical formula?
A: The molecular formula represents the exact number and type of atoms present in a single molecule of a compound, whereas empirical formulas represent the simplest whole-number ratio of atoms in a compound. The molecular formula of glucose is \({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}\) and that of formic acid is \({\rm{C}}{{\rm{H}}_2}{\rm{O}}\) but the empirical formula of both glucose and formic acid is \({\rm{C}}{{\rm{H}}_2}{\rm{O}}\). The molecular formula can be the same as the empirical formula. For example, methane has the same molecular and empirical formula as \({\rm{C}}{{\rm{H}}_4}\).

Q.3: What is the difference between chemical and molecular formulas?
A: The chemical formula represents the formula of ionic compounds, whereas the molecules formula represents the formula of covalent compounds. For example- Table salt has a chemical formula because it is made up of \({\rm{N}}{{\rm{a}}^ + }\) and  \({\rm{C}}{{\rm{l}}^ – }\) ions whereas glucose have the molecule formula \(\left( {{{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}} \right)\) because it is a covalent molecule.

Q.4: Is empirical formula related to molecular formula?
A: The molecules formula is always an integral multiple of the empirical formula. It can be expressed in terms of empirical formula as follows-
\({\rm{Molecular}}\,{\rm{Formula}} = {\rm{n}}({\rm{Empirical}}\,{\rm{Formula}})\)

Q.5: What is the value of oxygen molar mass?
A: The value of oxygen molar mass is 15.999 u.

Q.6: What is the molecular formula of water?
A: The formula of water is H2O.

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