**The Mole Concept **

A chemical change involves atoms, molecules, ions and electrons. We will realise that depicting a chemical change on the basis of the number of particles (atoms, molecules, ions, electrons or any other elementary particles) is easier and at the same time of greater significance than doing so on the basis of their masses.

H_{2}(g) + Cl_{2}(g) 2HCl(g)

2 atoms 2 atoms 2 molecules

Ag^{+}(aq) + Cl^{–} Ag^{+}Cl^{–} (s) or AgCl

1 ion 1 ion 1 formula unit

Let us consider the first reaction in a little greater detail.

H_{2}(g) + Cl_{2}(g) 2HCl (g)

1 molecules 1 molecule 2 molecules

2 atoms 2 atoms 2 molecules

1 atom 1 atom 1 molecule

1 x N atoms 1 x N atoms 1 x N molecules

Since one atom of hydrogen reacts with one atom of chlorine to give one molecule of hydrogen chloride, one may say that, in terms of mass, 1 gram-atom of chlorine (35.5g) to give 1 gram-molecule of hydrogen chloride (36.5 g).

Thus, 1 g (i.e., 1 gram-atom) of hydrogen will contain the same number of atoms as 35.5 g (i.e., 1 gram-atom) of hydrogen will contain the same number of atoms as 35.5 g (1 gram-molecule) of hydrogen chloride.

Atomic and molecular masses are measured with respect to ^{12}C, which is considered to have a mass of exactly 12. So, the number of atoms in 1 gram-atomic mass of element and that of molecules in 1 gram-molecular mass of a substance must be the same as the atoms in exactly 12 g of ^{12}C. An assemblage of this number (N) of particles is known as a mole (mol when used as a unit), which may be defined as follows.

“A mole is the amount of a substance that contains the same number of particles – atoms, molecules, ions, electrons or any other elementary particles – as the number of atoms is exactly 12 g of carbon – 12.”

This number of particles (6.022 x 10^{23}), previously called the Avogadro number, is of great fundamental significance. Nowadays, the term ‘Avogadro constant’ is preferred, which is simply the Avogadro number suffixed by the unit of mol^{–}^{1}.

**Gram atomic mass and gram atom**

So, the gram-atomic mass or the gram-atom or a mole of an element is definated as the mass in grams of the same number of atoms as contained in exactly 12 g of ^{12}C. It is equal to the relative atomic mass expressed in grams.

We can easily arrive at the following important relationships.

**1.** 1 mole of atom of an element

= number of atoms in 1 gram-atom of the element

= 6.022 x 10^{23} atoms.

**2.** (a) The absolute mass of 1 atoms of an element

=

=

(b) The absolute mass of 1 atom of an element

= relative atomic mass x 1.66 x 10^{-24} g.

**3.** In a given mass of element,

(a) the number of gram-atoms (mole) = and,

(b) the total number of atoms =

**4.** No. of moles =

*Illustration 2: If you need 3.011 x 10 ^{23} atoms of Na for a chemical reactions, what is the mass of the metal required (relative atomic mass of Na = 23.0)?*

** Solution:** The gram-atomic mass of Na = 23.0 g.

For 6.022 x 10^{23} atoms of Na, the mass of the metal required = 23.0g

∴ for 3.011 x 10^{23} atoms of Na, the mass of the metal required

= = 11.5 gm

*Exercise 1: What masses of the following elements will contain 6.022 x 10 ^{23} atoms each?*

* (a) Hydrogen (b) Sodium *

* (c) Chlorine (d) Copper *

**Ans. ****(a) 1 gm **

** (b) 23 gm**

** (c) 35.5 gm**

** (d) 63.5 gms**

*Exercise 2: Calculate the absolute masses of 1 atom of H, N, O and Na*

**Ans. 1.67 x 10 ^{-27 }kg , 2.32 x 10^{ – 26} kg, 2.65 x 10^{ – 26} , 3.81 x 10^{ – 26}**

*Exercise 3: How many gram-atoms are there in 106.5g of chlorine (relative atomic mass of Cl =35.5)? *

**Ans. 3**

*Exercise 4: How many atoms of Na are present in 46.0 g of metal (A _{r} of Na = 23.0)?*

**Ans. 1.2 x 10 ^{24}**

**Gram Molecular Mass (GMM)**

“The gram-molecular mass or the mole of a substance is defined as the mass in grams of the same number of molecules or formula units of the substance as the number of atoms contained in exactly 12 g of ^{12}C”. It is equal to the relative molecular mass (M_{r}) expressed in grams.

As we have said earlier, the formula mass of a substance like an ionic solid or an associated liquid is considered as the molecular mass for purposes of calculation.

The following important relationships may be easily derived.

**1.** 1 mole of molecules of a substance

= number of molecules in 1 gram-mole of the substance

= 6.022 x 10^{23} molecules

**2.** The absolute mass of 1 molecule of a substance

= (in grams)

= x 10^{-3} (in kilograms)

3. In a given mass of a substance,

(a) the number of gram-moles =

(b) the total number of molecules = the number of gram-moles x 6.022 x 10^{23}

= x 6.022 x 10^{23}

(c) the number of atoms of a given element constituting the substance

= the number of gram-moles of the substance x the number of atoms of the element in 1 molecule of the substance x 6.022 x 10^{23}

(d) the total number of atoms= the number of gram-moles of the substance x the total number of atoms in a molecule of the substance x 6.022 x 10^{23}

*Illustration 3: How many moles of HCl molecules are there in 109.5 g of HCl? Calculate the number of molecules in the sample (A _{r} of H = 1.0, Cl = 35.5).*

** Solution:** The relative molecular mass of HCl = 1 + 35.5 = 36.5

The number of moles of HCl molecules in 109.5 g

=

The number of HCl molecules in 109.5 g

= number of moles x 6.022 x 10^{23}

= 3 x 6.022 x 10^{23}

= 18.066 x 10^{23}

= 1.8066 x 10^{24}

*Illustration 4: How many moles of water are there in 1 L of water? Assume a density of 1.0g mL ^{-1}.*

** Solution:** The mass of 1 L (i.e. 1000 mL) of water

= volume x density

= 1000 mL x

= 1000 g.

The gram-molecular mass of water (H_{2}O) = 2 x 1 + 16 = 18 g.

∴ the number of moles of water 1 L = =55.55.

*Exercise 5: For 196g of pure H _{2}SO_{4}, calculate*

* (a) the number of moles of H _{2}SO_{4}*

* (b) the total number of H _{2}SO_{4} molecules,*

* (c) the total number of atoms,*

* (d) the number of atoms of each kind, and *

* (e) the absolute mass of an H _{2}SO_{4} molecule. (A_{r} of H = 1, O = 16, S =32.)*

**Ans5. **

**(a) 2**

** (b) 1.2 x 10 ^{24}**

** (c) 8.43 x 10 ^{24}**

** (d) H = 2.4 x 10 ^{24} , S = 1.2 x 10^{24} , O = 4.81 x 10^{24}**

** (e) 1.62 x 10 ^{-23} kg**

*Exercise 6: The mass of an atom of an element is 3.986 x 10 ^{-26} kg. Find the relative atomic mass of the *

*ement and identify it.***Ans 6. **

**24, Magnesium**

* *

*Exercise 7: How many moles of carbon atoms are there in 540g of glucose(C _{6}H_{12}O_{6})?*

* (A _{r} of C = 12, H = 1, O = 16.)*

**Ans. 18 Moles**

** **

**A mole of ions**

Ions are formed by the loss or gain of electron(s) from or by an atom. The mass of an electron is negligible in comparison to that of an atom. So, for all practical purposes, atoms have the same mass as do the ions derived from them.

A mole of ions is defined as the assemblage of the same number of ions as the number of atoms in exactly 12g of ^{12}C.

In other words, mole ions is an assemblage of 6.022 x 10^{23} ions. Its mass, called the gram-ion, is the same as the gram-atom of the element from the which it is derived.

For example, a mole of Na^{+} ions is a collection of 6.022 x 10^{23} Na^{+} ion, which collectively weight 23 g. Similarly, a mole of Cl^{–} ions is an assemblage of 6.022 x 10^{23} Cl^{–} ions, which collectively weight 35.5 g. 1 mole of NaCl will furnish 1 mole of Na^{+}and Cl^{–} ions.

NaCl → Na^{+} + Cl^{–}

*Exercise 8: How many moles of Ca ^{2+} ions and Cl*

^{–}

*ions will be furnished by 444 g of CaCl*_{2}

*(A*_{r}of Ca = 40, Cl = 35.5)?**Ans. ****4 Moles of Ca ^{+2} and 8 Moles of Cl ^{–}**

**Mole of electrons:**

Electrons have negligible mass and hence they are reckoned only by number and not by mass.

6.022 x 10^{23} electrons constitute a mole.

As you know, all electrons carry an electrical charge.

The charge carried by 1 electron = 1.602 x 10^{–}^{19} C.

∴ the charge carried by 1 mole of electrons = 1.602 x 10^{–}^{19} c x 6.022 x 10^{23 }≈ 96,500 C.

The charge carried by 1 mol of electrons, i.e., 96,500 C, is called a faraday.

**Molar Volume**

The volume occupied by one mole i.e., one gram-molecular mass of a gas, at a given temperature and pressure is called the molar volume of the gas.

As the volume of a gas is dependent on temperature and pressure, they must be mentioned along with the molar volume. If the molar volume is taken at STP, it is called the standard molar volume or the molar volume at STP.

It has been found that one gram-mole of every gas occupies 22.4 L at STP. In other words, a collection of 6.022 x 10^{23} molecules of any gas occupies 22.4 L at STP.

So, the standard molar volume of a gas = 22.4 L.

Thus, as the molecular mass of H_{2} is 2, 2 g of H_{2} occupies 22.4 L at STP. Similarly 2 x 16 = 32 g of O_{2} will also occupy 22.4 L at STP. Some more examples are given below.

Gas |
Molecular Mass |
Gram–molecular mass |
Standard Molar volume |

H_{2} |
2 x 1 = 2 | 2 g | 22.4 L |

N_{2} |
2 x 14 = 28 | 28 g | 22.4 L |

O_{2} |
2 x 16 = 32 | 32 g | 22.4 L |

Cl_{2} |
2 x 35.5 = 71 | 71 g | 22.4 L |

CO_{2} |
12 + 2 x 16 = 44 | 44 g | 22.4 L |

CH_{4} |
12 + 4 x 1 = 16 | 16 g | 22.4 L |

SO_{2} |
32 + 2 x 16 = 64 | 64 g | 22.4 L |

* *

*Illustration 5: What is the volume occupied by 17.75 g of Cl _{2} at STP?*

** Solution:** The relative molar mass of Cl

_{2}= 2 x 35.5 = 71.

The number of moles of Cl_{2} in 17.75 g of the gas

=

The volume of 1 mol of Cl_{2} at STP = 22.4 L.

∴ the volume of ¼ mol of Cl_{2} at STP = ¼ x 22.4 = 5.6 L.

*Exercise 9: What will be the mass of 1.4 L of CO _{2} at STP?*

**Ans. 2.75 gm**