Molecular Mass and Mole Concept: Calculation, Examples

Molecular Mass and Mole concept: The molecular mass of a material is defined as the relative mass of its molecule when measured against the mass of a \(12\,{\rm{C}}\) atom divided by \(12\) units. In layman’s words, it refers to the number of times a molecule of the concerned substance is heavier than an atom.

The term mole is derived from a Latin word that means pile or heap. The mole of any substance represents \(6.022 \times {10^{23}}\) particles of that substance. Let us learn about molecular mass and mole concepts in detail. Scroll down to learn more!

Relative Molecular Mass

Like the atomic masses, the molecular masses of compounds are also very small, and these cannot be measured directly. The molecular masses of compounds are also expressed as relative molecular masses \(\left( {{{\rm{M}}_{\rm{r}}}} \right)\) The relative molecular mass of a compound is the average mass of its one molecule compared to \(1/12{\rm{th}}\) the mass of one carbon-\(12\) atom.

Relative molecular mass \(\left( {{{\rm{M}}_{\rm{r}}}} \right) = \frac{{{\rm{Average}}{\mkern 1mu} \,{\rm{mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} \,{\rm{one}}{\mkern 1mu} \,{\rm{molecule}}{\mkern 1mu} \,{\rm{of}}\,{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{compound}}}}{{\frac{1}{{12}} \times \left( {{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} \,{\rm{an}}{\mkern 1mu} \,{\rm{atom}}\,{\mkern 1mu} {\rm{of}}{\,^{12}}{\rm{C}}} \right)}}\)

The relative molecular mass \(\left( {{{\rm{M}}_{\rm{r}}}} \right)\) of a compound is a pure number, and it does not have a unit.

Molecular Mass (M)

The average mass of one molecule of a compound in an atomic mass unit is called molecular mass \(\left( {\rm{M}} \right)\) Hence,
Molecular mass \(\left( {\rm{M}} \right) = {{\rm{M}}_{\rm{r}}} \times 1{\rm{u}} = {{\rm{M}}_{\rm{r}}}{\rm{u}}\)
It has unit amu or \({\rm{u}}.\) Note that the magnitudes of molecular mass \(\left( {\rm{M}} \right)\) and relative molecular mass \(\left( {{{\rm{M}}_{\rm{r}}}} \right)\) are equal. They differ only in their units.

Calculation of Molecular Mass of a Compound

The molecular mass of a substance can be calculated as the sum of the atomic masses of all the atoms which constitute a molecule of that substance. For example,

The molecular mass of ammonia \(\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)\): The molecular formula of ammonia is \({\rm{N}}{{\rm{H}}_3}\). Hence, the molecular mass of ammonia \(= \left( {1 \times {\rm{Atomic}}{\mkern 1mu} \,{\rm{mass}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{nitrogen}}} \right) + \left( {3 \times {\rm{Atomic}}\,{\mkern 1mu} {\rm{mass}}{\mkern 1mu} \,{\rm{of}}\,{\mkern 1mu} {\rm{hydrogen}}} \right)\)

\( = (1 \times 14{\rm{u}}) + (3 \times 1{\rm{u}}) = 14 + 3 = 17{\rm{u}}\)

The molecular mass of carbon monoxide (CO): The molecular formula of carbon monoxide is \({\rm{CO}}\). Hence, molecular mass of carbon monoxide
\(= \left( {1 \times {\rm{Atomic}}\,{\mkern 1mu} {\rm{mass}}\,{\mkern 1mu} {\rm{of}}\,{\mkern 1mu} {\rm{carbon}}} \right) + \left( {3 \times {\rm{Atomic}}{\mkern 1mu} \,{\rm{mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} \,{\rm{oxygen}}} \right)\)
\( = (1 \times 12{\rm{u}}) + (1 \times 16{\rm{u}}) = 12 + 16 = 28{\rm{u}}\)

Percentage Composition of a Compound

A compound contains two or more elements combined in a certain fixed ratio. The percentage composition of a compound is the mass of each element of the compound, present in \(100{\rm{g}}\) of that compound, i.e. the mass percentage of each element present in the compound. The mass percentage of each element in a compound can be calculated using either of the following two equations:

1. When the masses of the compound and each element are given:

Mass percentage of an element can be obtained if we know the mass of that element in a known mass of the compound.

Mass percentage of an element \({\rm{X}} = \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}\,{\rm{element}}\,{\rm{in}}\,{\mkern 1mu} {\rm{the}}\,{\rm{given}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound}}}}{{{\rm{Total}}\,{\rm{mass}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} \,{\rm{the}}\,{\mkern 1mu} {\rm{compound}}}} \times 100\)

2. When the formula of the compound and the atomic masses of the elements are given:

When the formula of the compound and the atomic masses of the elements are given, the molecular mass of the compound can be calculated by adding the masses of all the elements present in the compound. Then the mass percentage of each element can be calculated using the following formula:

Mass percentage of an element \( = \frac{{{\rm{Total}}{\mkern 1mu} \,{\rm{mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{compound}}{\mkern 1mu} \,{\rm{element}}\,{\mkern 1mu} {\rm{in}}{\mkern 1mu} \,{\rm{one}}\,{\mkern 1mu} {\rm{molecule}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} \,{\rm{the}}{\mkern 1mu} \,{\rm{compound}}}}{{{\rm{Molecular}}{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{compound}}}} \times {\rm{100}}\)

Solved Examples on Molecular Mass of a Compound

Example 1: \(0.24\;{\rm{g}}\) sample of a compound of oxygen and boron was found by analysis to contain \(0.096\;{\rm{g}}\) of boron and \(0.144\;{\rm{g}}\) of oxygen. Calculate the percentage composition of the compound by mass?
Solution:
Mass percentage of boron \( = \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{born}}}}{{{\rm{Mass}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{compound}}}} \times 100 = \frac{{0.096\;{\rm{g}}}}{{0.24\;{\rm{g}}}} \times 100 = 40.0\)
Mass percentage of oxygen \(= \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{oxygen}}}}{{{\rm{mass}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{the}}{\mkern 1mu} \,{\rm{compound}}}} \times 100 = \frac{{0.144\;{\rm{g}}}}{{0.24\;{\rm{g}}}} \times 100 = 60.0\)
Hence, the boron and oxygen mass percentages in the given compound are \(40.0\% \) and \(60.0\% \), respectively.

Example 2: Calculate the percentage of water of crystallisation in washing soda whose formula is \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot 10{{\rm{H}}_2}{\rm{O}}\)
Solution:
Molecular mass of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot 10{{\rm{H}}_2}{\rm{O}}\)
\( = 2 \times 23{\rm{u}} + 12{\rm{u}} + 3 \times 16{\rm{u}} + 10(2 \times 1{\rm{u}} + 1 \times 16{\rm{u}})\)
\( = 46{\rm{u}} + 12{\rm{u}} + 48{\rm{u}} + 180{\rm{u}} = 286{\rm{u}}\)
\(286{\rm{u}}\) of washing soda contains \(180{\rm{u}}\) of water of crystallisation.
\(100{\rm{u}}\) of washing soda will contain \(\frac{{180{\rm{u}}}}{{286{\rm{u}}}} \times 100\;{\rm{g}}\) water of crystallisation \( = 62.94{\rm{u}}\)

The amount of water of crystallisation in washing soda \( = 62.94\% \) by mass.

Example 3: Find the percentage composition of potassium permanganate.
Solution:
Formula unit mass of \({\rm{KMn}}{{\rm{O}}_4}\)
\( = 39{\rm{u}} + 55{\rm{u}} + 16{\rm{u}} \times 4 = 39{\rm{u}} + 55{\rm{u}} + 64{\rm{u}} = 158{\rm{u}}\)
Percentage of potassium \( = \frac{{39}}{{158}} \times 100 = 24.68\% \)
Percentage of manganese \( = \frac{{55}}{{158}} \times 100 = 34.81\% \)
Percentage of oxygen \( = \frac{{64}}{{158}} \times 100 = 40.51\% \)
The percentage composition of potassium permanganate is as follows:
Potassium \( = 24.68\% \) Manganese \( = 34.81\% \) Oxygen \( = 40.51\% \)

Mole Concept

A mole (Latin: \( = \) pile or heap) of any substance (atoms, molecules or ions) represents \(6.022 \times {10^{23}}\) particles of that substance. This number is called the Avogadro’s constant or Avogadro’s number \(\left( {{{\rm{N}}_0}} \right)\). The name Avogadro’s number was given in honour of the Italian scientist Amedeo Avogadro. The value\(6.022 \times {10^{23}}\) is experimentally obtained.

The term mole applies to various items such as atoms, molecules, ions, electrons, protons, photons or chemical bonds. The SI unit of the amount of substance is the mole (symbol- mol). Thus, to express one mole, we write it as \({\rm{1 mol}}\). We shall be using the word mole in the text, but when it is used as a unit, we shall denote it as mol.

A Mole of Electrons

The Avogadro’s number of electrons is called one mole of electrons. The charge on one electron \( = – 1.6022 \times {10^{ – 19}}{\rm{C}}\). The charge on Avogadro’s number of electrons (i.e. one mole of electrons) \(= 6.022 \times {10^{23}} \times 1.6022 \times {10^{ – 19}}{\rm{C}}\; \sim 96500\,{\rm{C}}\). This quantity of charge is called one faraday \(\left( {\rm{F}} \right)\). The charge possessed by the Avogadro’s number of electrons is called one faraday.

One mole of hydrogen atoms
\( = 6.022 \times {10^{23}}\) atoms of hydrogen
One mole of hydrogen molecules
\( = 6.022 \times {10^{23}}\) molecules of hydrogen
One mole of carbon monoxide molecules
\( = 6.022 \times {10^{23}}\) molecules of carbon monoxide
One mole of electrons \( = 6.022 \times {10^{23}}\) electrons
One mole of protons \( = 6.022 \times {10^{23}}\) protons
One mole of photons \( = 6.022 \times {10^{23}}\) photons
One mole of sodium ions \(\left( {{\rm{N}}{{\rm{a}}^ + }} \right) = 6.022 \times {10^{23}}{\rm{N}}{{\rm{a}}^ + }\) ions.

How Large is One Mole?

Avogadro’s number is very large. Let us try to find out how big this number is with the help of the following example.

Let us calculate the number of years which one would take to spend Avogadro’s number of rupees at the rate of \(1\) million \(\left( { = {{10}^6}} \right)\) rupees per second.

The number of years required to spend the Avogadro’s number of rupees at the rate of \(1\) million \(\left( { = {{10}^6}} \right)\) rupees per second is calculated as shown below:

\( = \frac{{6.022 \times {{10}^{23}}}}{{{{10}^6} \times 60 \times 60 \times 24 \times 365\,/{\rm{Year}}}}\)

\( = 1.91 \times {10^{10}}\) years (since \(1\) year \( =365 \) days \( = 60 \times 60 \times 24 \times 365\) seconds)
Thus, it will take \(1.91 \times {10^{10}}\) years to spend Avogadro’s number of rupees at the rate of \(1\) million rupees per second! Isn’t it an amazing number!

Molar Mass

The mass of one mole of any substance is called its molar mass. Alternatively, the average mass of one mole of any substance is called its molar mass.
It is given by,
Molar mass \(\left( {\rm{M}} \right) = \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{substance}}\,{\mkern 1mu} {\rm{in}}\,{\mkern 1mu} {\rm{grams}}}}{{{\rm{Amount}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{the}}\,{\mkern 1mu} {\rm{substance}}\,{\mkern 1mu} {\rm{in}}{\mkern 1mu} \,{\rm{moles}}}} = \frac{{\rm{m}}}{{\rm{n}}}\)

The unit of molar mass is \({\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\).
The molar mass of a substance \( = \) mass of \(6.022 \times {10^{23}}\) chemical units of that substance
The molar mass of hydrogen atom \( = \) mass of \(6.022 \times {10^{23}}\) atoms of hydrogen
The molar mass of hydrogen molecule \( = \) mass of \(6.022 \times {10^{23}}\) molecules of hydrogen

How Many Molecules are there in a Certain Mass of a Substance?

The number of molecules present in a certain mass of a substance is calculated using the following formula:
No. of molecules of a substance \(= \frac{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{the}}{\mkern 1mu} \,{\rm{substance}}}}{{{\rm{Molar}}{\mkern 1mu} \,{\rm{mass}}{\mkern 1mu} \,{\rm{of}}\,{\mkern 1mu} {\rm{the}}{\mkern 1mu} \,{\rm{substance}}}} \times {\rm{Avogadro’s}}\,{\rm{number}}\)

\( = \frac{{\rm{m}}}{{\rm{M}}} \times 6.022 \times {10^{23}}{\rm{molecules}}\)
where \({\rm{m}}\) is the mass of the substance and \({\rm{M}}\) is the molar mass of the substance.

The molar mass of the first \(20\) elements are as follows:

Solved Examples on Molar Mass

Example: How many moles are \(9.033 \times {10^{24}}\) atoms of helium \(({\rm{He}})\)?
Solution. We know that:
\(6.022 \times {10^{23}}\) atoms of helium \( =1 \) mole atoms of helium \( = 1\) mole
So, \(9.033 \times {10^{24}}\) atoms of helium \( = \frac{1}{{6.022 \times {{10}^{23}}}} \times 9.033 \times {10^{24}}\) 
\( = 15\) moles.
Thus, \(9.033 \times {10^{24}}\) atoms of helium are \(15\) moles of atoms.

Summary

The study of atoms and their characteristics is the foundation of chemistry. Atomic mass is the mass of a single atom, whereas molecular mass is the mass of a group of atoms. If you want to learn more about molecular mass and the mole concept, you will need to learn everything there is about know about them. The concepts of relative molecular mass, molecular mass calculation, percentage composition, and molar mass were discussed in this article.

FAQs on Molecular Mass and Mole Concept

Q.1. How do you find the molar mass of an element?
Ans: Mass of one mole of any substance is called its molar mass. Alternatively, the average mass of one mole of any substance is called its molar mass.
It is given by,
Molar mass \(({\rm{M}}) = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance}}\,{\rm{in}}\,{\rm{grams}}}}{{{\rm{Amount}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance}}\,{\rm{in}}\,{\rm{moles}}}} = \frac{{\rm{m}}}{n}\)

Q.2. What is the molar mass of carbon?
Ans: The molar mass of the \({\rm{C}}\) atom is \(12\;{\rm{g}}/{\rm{mol}}\).

Q.3. How many grams is a mol?
Ans: A mole is an amount of a substance with mass equal to gram molecular mass (or gram atomic mass) of the substance. For example, \(1\) mole of nitrogen atoms \( = \) gram atomic mass of nitrogen \( = 1\;{\rm{g}}\) atom of \({\rm{N}} = 14.0\;{\rm{g}}\).

Q.4. Is molar mass expressed in grams?
Ans: Yes, the molar mass is the mass of Avogadro’s number of molecules expressed in grams.

Q.5. What is meant by molecular mass?
Ans: The average mass of one molecule of a compound in an atomic mass unit is called molecular mass \(({\rm{M}})\). Hence,
Molecular mass \(\left( {\rm{M}} \right) = {\rm{Mr}} \times 1{\rm{u}} = {{\rm{M}}_{\rm{r}}}{\rm{u}}\)
It has unit \({\rm{amu}}\) or \({\rm{u}}\). Note that the magnitudes of molecular mass \(({\rm{M}})\) and relative molecular mass \({{\rm{M}}_{\rm{r}}}\) are equal. They differ only in their units.

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