Mole Concept MOLE AND AVOGADRO’S NUMBER In chemistry we deal with chemical reactions. Chemical reactions take place at molecular and atomic level. But in real practice we are using mass and volume of reactants. So, it is necessary that we know the number of atoms or molecules present in the reactants whether it is taken by mass or volume. Mole concept helps us in this matter. Mole is the SI unit of amount of a substance. It is abbreviated as mol. A mole of any substance is that much amount of it that contains as many entities (atom, molecule, ions, etc.) as there are C-12 isotopes in 12 g of carbon-12. 12 g of Carbon-12 is found to contain 6.0221367  1023 atoms of C-12. This number is called Avogadro's number (NA) in the honour of a great physicist, Amedeo Avogadro. In chemical calculation, this number is generally taken as 6.022  1023. As a corollary, we can say that one mole of any substance contains Avogadros number of it. i.e. 1 mole = 6.0221023 For instance, 1 mole of the element sodium contains 6.0221023 Na atoms and 1 mole of the element oxygen contains 6.0221023 molecules of oxygen. Similar is the case with ions and sub atomic particles. Mole is a chemist’s unit of counting particles such as atoms, molecules, ions, electrons, protons etc. which represents a value of 6.022  1023 just as a dozen for 12, score for 144 etc. are used to count different objects. Since the term mole is applicable to all sorts of particles, while using this term, it is very important to indicate the nature of the particles under consideration i.e. whether they are atoms, molecules or something else. For example, it is wrong to speak ‘one mole of hydrogen’. We must specify whether we are referring to hydrogen atoms or hydrogen molecules. Also, according to Avogadro’s hypothesis "Equal volumes of different gases under similar conditions of temperature and pressure contain equal number of molecules". It means that 6.022  1023 molecules of any gas at STP (i.e. standard temperature and pressure viz. 273 K temperature and 1 atm atmospheric pressure) must have the same volume. This volume has been experimentally found to be 22.4 L at STP and is called molar volume. And, as we know 6.0221023 represents one mole of that substance, we can say that the volume occupied by one mole of a gas at STP is called molar volume. As in case of gas 1 mole is equal to 22.4 L or 22400 mL of the gas at STP, for atoms 1 mole is equal to atomic mass, for molecule 1 mole is equal to molecular mass, for ionic compound, 1 mole is equal to formula weight of the compound, for ions 1 mole is equal to 6.0221023 ions and for subatomic particles 1 mole is equal to 6.0221023 sub atomic particles. This can be represented in the following diagram given below: Molecular weight of molecules

Avogadros number (6.0221023)

Avogadros number (6.0221023) of ions

Formula weight of ionic compounds

1 mole

22.4L or 22400 mL of a gas at STP

Avogadros number (6.0221023) of subatomic particles (e-, p+ and n0)

Problems involving Avogadro’s Number and Mole Concept Type I. On the calculation of the actual mass of a single atom or a single molecule. Example 1. Calculate the mass of (i) an atom of silver (ii) a molecule of carbon dioxide. Solution: (i)

Strategy: Mass of 1 mole of Ag atom mass of an atom of Ag

Atomic mass

Mass of avogadros number of Ag atom

1 mole of Ag atoms = 108 g = 6.022  1023 atoms of Ag Mass of one atom of silver = (ii)

Strategy: Mass of 1 mole of CO2 of one molecule of CO2

108 = 17.93  10–23 g 6.022  1023

Molecular mass of CO2

Mass of avogadros number of CO2

Mass

1 mole of CO2 = 44 g = 6.022  1023 molecules of CO2  Mass of 1 molecule of CO2 =

44 44  10–23 –23 g = 6.022 g = 7.307  10 6.022 1023

Type II. On the calculation of number of atoms or molecules in a given mass of the substance. Example 2. Calculate the number of molecules present in (i) 64.0 g of sulphur (S8) ? (ii) in one litre of water assuming that the density of water is 1 g/cm 3. Solution: (i)

Strategy: mass of molecule Mole Given Molecular formula of sulphur = S8



Molecular mass of sulphur (S8) = 32  8 = 256.0 amu



1 mole 6.0221023 molecules Number of S8 molecules in 64.0 g = 64.0 g  256g  1mole

Number of molecules

= 1.5061023 molecules of S8 Note: 1 molecule of sulphur (S8) contains 8 atoms of sulphur 

1.506  1023 molecules of sulphur will contain sulphur atoms = 8  1.506  1023 = 12.048  1023 atoms.

(ii)

Strategy: At first, get the mass of water from its volume and density and then similar to above.



Mass of 1 litre of water = volume  density = 1000  1 = 1000 g 1 mole 6.0221023 molecules of water 1000 g of water will contain =1000 g  18g  1 mole = 3.346  1025 molecules of water.

Type III. On the calculation of number of molecules present in a given volume of a gas under given conditions. Example 3. Calculate the number of molecules present in 350 cm 3 of NH3 gas at 273 K and 2 atmospheric pressure. Solution: Strategy: First of all, we have to determine the volume of the gas at STP. Then, Volume at STP mole number of molecules Given conditions At STP V1 = 350 cm3 V2 = ? T1 = 273 K

T2 = 273 K

P1 = 2 atmosphere

P2 = 1 atm

Applying gas equation:

320  2 1  V2 273 = 273

we get

350  2 273 V2 = 273  1 = 700 cm3

or 

P 1 V1 P 2 V2 T1 = T2

Number of NH3 molecules in700 cm3 of NH3 at STP 1 mole 6.0221023 molecules of NH3 = 700 cm  22400 cm3  = 1.882  1022 molecules of NH3. 1 mole 3

For Sub-atomic particles In case of sub-atomic particles like electrons, protons, neutrons etc. mole has its usual significance. Sub-atomic particles in an atom of an element are related to atomic number and atomic mass. Atomic number indicates number of protons or number of electrons in an atom while number of neutrons per atom is equal to mass number – atomic number. 32 Example 1. Calculate the number of electrons in 2 g of 16 S. Solution:

Strategy: Mass of S

mole of S

number of S atom

number of electrons

32 1 mole 6.0221023 atoms of S 16 electrons Number of electrons in 2 g of 16 S = 2g  32g   1 atom of S 1 mole = 6.022  1023 electrons. Example 2. Calculate the number of neutrons in 7 mg of 14C. Solution: Strategy: Mass of C mole of C atoms of C number of neutrons Neutrons per atom = mass number – atomic number = 14 – 6 = 8 1 mole 6.0221023atoms of C 8 neutrons 14  Number of neutrons in 7 mg of C = 7 mg  12000mg   1 atom of C 1mole C

= 2.4092  1021 neutrons. Some solved examples of past NEB questions: 1. Calculate the mass of the following: (i) 4 atom of C (ii) 3 molecules of hydrogen [NEB 2073 'D'] Solution: (i) Strategy: Number of atom number of gram atom mass of atoms We know, 1 gram atom of C = 12 g, and 1 gram atom of C = 6.0221023 atoms of C. 1 gram atom 12 g  Mass of 4 atoms of C = 4 atom   1 gram atom = 7.97  10-23 g or C. 23 6.02210 atoms (ii) Strategy: number of molecules mole mass We know, 1 mole of hydrogen = 2.016 g and 1 mole of hydrogen = 6.0221023 molecules of hydrogen 1 mole H2 2.016 g Mass of 3 molecules of hydrogen = 3 molecules   1 mole H = 1.004  10-23 g. 23 6.02210 molecules 2