1.2 **The mole concept** is about the quantitative aspects of a substance.

**What is 1Mole?**

1 mole is an amount of a substance which contains as many particles as there are in 12 g of C-12 isotopes that is 6.022X10^{23}

**6.022X10 ^{23 }is known as Avogadro’s number.It is important part of mole concept.**

**Mole= Mass of a substance/Molar Mas**s

**Molar Mass is the mass of 1 mole of a substance in grams.**

For calculating Molar mass, you need to add the relative atomic masses of all the atoms in a

compound or ion. For example Mr of Na_{2}O = (23×2)+16 = 62 ,

because Ar( relative atomic mass) of Na=23,

Ar( relative atomic mass) of O= 16

Molar mass of H_{2}SO_{4 }= 98 g/mol

because Ar( relative atomic mass) of H=1,

Ar( relative atomic mass) of O= 16

Ar ( relative atomic mass) of S= 32

Since 1 mole atoms = gram atomic mass = 6.022X10^{23 }atoms

Thus 1 mole oxygen atoms = gram atomic mass of oxygen =16 g = 6.022X10^{23 }atoms

Hence 1 mole molecules = gram molecular mass = 6.022X10^{23 }molecules

Example –

1 mole oxygen molecules = gram molecular mass of oxygen =32 g =

6.022X10^{23} oxygen molecules

1 mole ions = formula unit mass = 6.022X10^{23 }ions

Example – 1 mole( SO_{4})^{2-} ions = formula unit mass of ( SO_{4})^{2-}=96g = 6.022X10^{23}

( SO_{4})^{2}ions

The mass (nucleon) number, *A*, is the sum of the protons and the neutrons. (= 12 for Carbon)

The atomic (proton) number, *Z*, of an element is the number of protons in the nucleus of the atom. In the neutral atom it is equal to the number of electrons. ( 6 for Carbon)

The Relative Atomic Mass, A(r), is the mass of one atom of an element compared to a scale in which one atom of carbon-12 has a mass of 12.0000.

The Relative Molecular Mass, M(r),is the ratio of the mass of one molecule of a substance to the mass of one atom of carbon-12. It is calculated by adding the relative atomic masses of all the atoms of all elements in the molecule.

**Relative molecular mass**, M(r), or *relative formula mass**:*

*relative formula mass*

It * (*for those compounds that do not exist as molecules) is calculated by adding together the **Relative Atomic Mass**A(r) ‘s of the elements in the molecular formula.

### What is **Empirical formula**?

** **Chemical Formula with simplest ratio of atoms of a compound.

**Molecular formula: **

Chemical Formula with definite number of atoms of a compound.

**Example: **

C_{6}H_{6} is the molecular formula of benzene.

CH is the simplest ratio of atoms of C_{6}H_{6}

#### Molecular formula = n X Empirical formula

where n=Molecular formula mass/Empirical formula mass

molecular formula mass of C_{6}H_{6 }is 78 g/mol

Empirical formula mass of CH is 13 g/mol

n= 78/13= 6

Molecular formula = n X Empirical formula = 6 X CH= C_{6}H_{6}

**Determining Empirical formula**

**Step 1: **Write the element symbols along with % given. Consider % as mass in gram.

**Step 2: **Divide the given mass of each element by relative atomic mass of that element.

( mole=mass/molar mass) so you get moles of each element here.

**Step 3: **Divide the moles of each element by the smallest moles obtained in step 2.

**Step 4: **Use the obtained values as number of atoms per element in empirical formula.

**Note : if you did not get the whole number values, do not simply round off the values but multiply by suitable integers to make the value whole number. Example if you get 2.5, do not round simply to 3 but multiply 2.5 by 2 to get 5 which is a whole number.**

**Worked Problem on empirical formula: **

A compound has 44.09 % iron and rest chlorine. Determine its emp formula. If the molar mass of the compound is 127 g/mol, determine its Molecular formula as well.

Solution:

Step1 | Iron(Fe)=44.09 | Chlorine(Cl)=100-44.09=55.91 |

Step2 | 44.09/56 | 55.91/35.5 |

Step3 | 0.79/0.79=1 | 1.57/0.79= 1.98=2 |

Step4 | 1 | 2 |

So the emp formula is FeCl_{2 } so now to find out molecular formula,

use Molecular formula = n X Empiricalformula where n=1

hence Empirical formula = Molecular formula

**Practice questions on 1.2 The mole concept: **

1. A binary oxide has 74.19 % sodium and rest oxygen. Determine its emp formula. If the molar mass of the compound is 62 g/mol, determine its Molecular formula as well.

2. A compound has 43.39 % sodium, 11.32% carbon and rest oxygen. Determine its emp formula. If the molar mass of the compound is 106 g/mol, determine its Molecular formula as well.

3. A compound has 34.46 % iron and rest chlorine. Determine its emp formula. If the molar mass of the compound is 162.5 g/mol, determine its Molecular formula as well.

4. A binary oxide has 43.66 % phosphorus and rest oxygen. Determine its emp formula. If the molar mass of the compound is 284 g/mol, determine its Molecular formula as well.

5. A compound has 40 % carbon, 6.6% hydrogen and rest oxygen. Determine its emp formula. If the molar mass of the compound is 180 g/mol, determine its Molecular formula as well.

I hope that by now you have become familiar about the subtopic 1.2 The Mole Concept.