**Mole Concept and Molar Masses**

- Atoms and molecules are extremely small in size but their numbers are very large even in a small amount of any substance.
- In order to handle such large numbers, a unit of similar magnitude is required.
- We use unit dozen to denote 12 items; score for 20 items and so one. To count the entities at microscopic level mole concept was introduced.
- In SI system, mole was introduced as seventh base quantity for the amount of a substance.
- One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the
^{12}C isotope. - The mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to (1.992648 × 10
^{–23}) g. - We know that one mole of carbon weighs 12g,the number of atoms in it is equal to:-
- (12 g / mol
^{12}C)/( 1.992648 × 10^{–23}g/^{12}C atom) - = 6.0221367 ×10
^{23}atoms/mol. - This number of entities in 1 mol is so important that it is given a separate name and symbol.
- It is known as ‘Avogadro constant’, denoted by N
_{A}in honour of Amedeo Avogadro. - If a number is written without using the powers of ten 602213670000000000000000, so many entities (atoms, molecules or any other particle) constitute one mole of a particular substance.
- 1 mol of hydrogen atoms = 6.022 x 10
^{23} - 1 mol of water molecule = 6.022 x 10
^{23} - 1 mol of sodium chloride = 6.022 × 10
^{23}units of NaCl.

1 mole of various substances

Molar mass

- The mass of one mole of a substance in grams is called its molar mass.
- The molar mass in grams is numerically equal to atomic /molecular/formula mass in u.
- Molar mass of oxygen (O) =16.02g.
- Molar mass of water (H
_{2}O) = 18.02g. - Molar mass of carbon (C) = 12g.
- Molar mass of sodium chloride (NaCl) = 58.5g.

**Percentage Composition**

**Percentage Composition**

- The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.
- For example:-Consider H
_{2}O molecule. Molar mass of Hydrogen (H) = 2g, and Molar mass of Oxygen (O) = 16g.- Consider 18g of H
_{2}O it contains 2g of H and 16g of O. - Mass % of H = (2×1.008×100)/(18.02) = 11.18%
- Mass % of O = (16.00×100)/(18.02)= 88.79%

- Consider 18g of H
: – By using information from percentage composition we can calculate empirical formula.__Note__

- For example:-Consider H

**Empirical Formula for Molecular formula**

- An empirical formula represents the simplest whole number ratio of various atoms present in a compound.
- Molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
- If the mass percent of various elements present in a compound is known, then its empirical formula can be determined.
- Molecular formula can further be obtained if the molar mass is known.

__Problem:-__

A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

** Answer**:-

Step 1:- Conversion of mass per cent to grams.

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the

100 g sample of the above compound,

4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2:- Convert into number moles of each element

Divide the masses obtained above by respective atomic masses of various elements.

Moles of hydrogen = (4.07 g/1.008 g) = 4.04

Moles of carbon = (24.27 g/12.01g) = 2.021

Moles of chlorine = (71.65 g/35.453 g) = 2.021

Step 3:- Divide the mole value obtained above by the smallest number

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H: C: Cl.

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4:- Write empirical formula by mentioning the numbers after writing the symbols of respective elements.

CH_{2}Cl is, thus, the empirical formula of the above compound.

Step 5:- Writing molecular formula

(a) Determine empirical formula mass

Add the atomic masses of various atoms present in the empirical formula.

For CH_{2}Cl, empirical formula mass is

(12.01) + (2 × 1.008) + (35.453)

= 49.48 g

(b) Divide Molar mass by empirical formula mass

n = (Molar mass/Empirical formula)

n=2

(c) Multiply empirical formula by n obtained above to get the molecular formula

Empirical formula = CH_{2}Cl, n = 2. Hence molecular formula is C_{2}H_{4}Cl_{2}.

** Problem**:-

Calculate the mass percent of different elements present in sodium sulphate (Na_{2}SO_{4}).

__Answer:-__

The molecular formula of sodium sulphate is (Na_{2}SO_{4})

Molar mass of (Na_{2}SO_{4}) = [(2 × 23.0) + (32.066) + 4 (16.00)]

= 142.066 g

Mass percent of an element:-

(Mass of that element in the compound/ Molar mass of the compound) x 100

Therefore, Mass percent of sodium:

= (46.0g/142.066g)/100

=32.379

=32.4%

Mass percent of sulphur:

= (32.066g)/ (142.066g) x 100

=22.57

=22.6%

Mass percent of oxygen:

= (64.0g)/ (142.066g) x 100

=45.049

=45.05%

** Problem**:-

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

** Answer**:-

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given] Relative moles of iron in iron oxide:

Relative moles of oxygen in iron oxide:

(% of iron by mass)/ (Atomic mass of iron)

= (69.9)/ (55.85)

Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5

2: 3

Therefore, the empirical formula of the iron oxide is Fe_{2}O_{3}.