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Learning Objectives

Percent Composition

Empirical Formula

Molecular Formula

Let’s start

Formula

➢ Formula is the symbolic representation of a chemical substances.

➢ It gives information regarding the chemical composition i.e. in what ratio , different elements are present in the compound.

Formula

Example :Formula of Water i.e. H2OIndicates that :

1. Hydrogen and oxygen present in 2 : 1 mole ratio in water

2. One molecule of water is made up of two atoms of hydrogen & one atom of Oxygen.

Steps to be followed while writing Formula

(i) Write the symbols of the constituent elements of the compound side by side.


(ii) Write the valency of each atom on the top of symbol.


(iii) Reverse the Order of the valency written on top, below the symbol of each element at its right.


Write it in the simplest ratio

Will you try this ?

Write the formula of the a) Sodium Sulphate b) Aluminium Nitride

PERCENTAGE COMPOSITION


The percent composition is the percent by mass of each element in a compound. It is calculated as

How to calculate percent composition

1. Find the molar mass of the compound by adding up the masses of each atom

2. Calculate the mass due the component by the total molar mass of the compound & multiply by 100



Percentage composition of water

Mass % of of hydrogen=(2/18)*100=11.18

Mass % of oxygen=(16/18)*100 =88.79

Percentage composition of water

What is the mass percent composition of FeSO4 ?

C2H5OH

C = 24/46 *100% = 52 %

H = 6/46 * 100% = 13 %

O =16/46 * 100 % = 35 %

What is the mass percent composition of C2H5OH ?

Molecular mass of Washing Soda = 286Molecular mass of water = 18g/mol

Mass of water present in Washing soda =10*18=180 g

Find the percentage of water of crystallization in washing Soda crystals

Na2CO3.10H2O

Mass percent of water = 180/286 *100= 62.9 %

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Empirical Formula of a compound

The empirical formula doesn't indicate the actual number of the elements present in the compound.

The empirical formula of a compound which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.

Empirical formula of Glucose :

CH2O

Molecular formula of Glucose :C6H12O6


An empirical formula of acetic acid :

CH2O

Molecular formula of Acetic Acid :

CH3COOH or C2H4O2


The empirical formula mass is the sum of atomic masses of various elements present in the empirical

formula.

Empirical formula mass

An empirical formula of acetic acid :

CH2O

An empirical formula mass of acetic acid :

Empirical Formula for Molecular Formula

Note :

The empirical formula of two compounds may be same but molecular formula can’t be same for two compounds

Steps to calculate empirical formula

1. The percentage weight of Oxygen, carbon or hydrogen is sometimes not given.Hence it is first determined by subtracting the sum of other elemental percentage from 100.


2. The percentage weight of each element is divided by its atomic weight. This gives the ratio of the number of atoms in a molecule of the compound.


3. If the molecule contains water of crystallization,its percentage weight is divided by the molecular weight of water. i.e. 18.

This gives the ratio of the number of water molecules in one molecule of the compound


4. To get a whole number ratio of atoms of different elements & water molecules the numbers obtained in steps 2 & 3 are divided by the smallest number.


5. The empirical formula is now derived by writing the symbols of various elements side by side with the number of atoms of each one as the subscript to the lower right of its symbol.

Find the empirical formula of the compound with the following percentage

compositions:

a) Zn = 47.8 , Cl = 52.2

Element Mass percent

Atomic mass

Number of atoms

Simplest Ratio

Zn 47.8 65 47.8/65=0.73 0.73/0.73=1

Cl 52.2 35.5 52.2/35.5=1.46 1.46/0.73=2

Thus, the ratio of Zn:Cl = 1:2

An Organic compound has oxygen 26.24 % & hydrogen 4.92 % .Find it's

empirical formula.

Every Organic compound contains Carbon.Find the percentage of Carbon

An Organic compound has oxygen 26.24 % & 4.92 % .Find it's empirical

formula.

Element Mass percent

Atomic mass

No of atoms

Simplest ratio

C 68.84 12 68.84/12=5.74

5.74/1.64=3.5

H 4.92 1 4.92/1=4.92 4.92/1.64=3

O 26.24 16 26.24/16=1.64 1.64/1.64=1

Multiply 3.5 ,3 & 1 by 2 in order to change them into whole number. Thus the ratio of C:H:O atoms = 7:6:2

The empirical formula is C7H6O2


Step 1. Conversion of mass percent to grams

in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.

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


Step 2. Convert into number moles of each element

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


Step 2. Convert into number moles of each element

Moles of hydrogen = 4.07g /1.008g = 4.04

Moles of carbon =24.27g /12.01g = 2.021

Moles of chlorine = 71.65g / 35.453 g =2 021


Step 3. Divide each of the mole values obtained above by the smallest number amongst them

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



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

CH2Cl is, thus, the empirical formula of the above compound.


Determination of Molecular formula

The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the Compound

Determination of Molecular formula

It gives the information that a molecule of blue vitriol is made of -

1. One atom of copper2. One atom of Sulphur3. Four atoms of Oxygen4. Five molecules of water of crystallization

Molecular formula of Blue vitriol is CuSO4.5H2O

Steps to find the molecular formula of a compound

➢ Divide it's molecular weight by Empirical weight which gives the number to get the molecular formula.

➢ Multiply the empirical formula buy this number to get the molecular formula

1. Calculate the empirical weight of the compound from its Empirical formula.

Calculate the empirical formula of a compound whose molecular formula is C8H6O4 & empirical formula weight

is 83

N = 166/83=2

Molecular weight of Compounds = 166

Empirical formula = M.F./2= (C8H6O4)/2

=C4H3O2

A compound is found to possess 40% Carbon, 6.7 % Hydrogen & 53.3 % Oxygen.Its molecular mass is 60. Find the molecular formula of the compound ?

The empirical formula of the compound is CH2OE.F.Mass = 30

element Mass percent

Atomic mass

Number of atoms

Simplest ratio

C 40 12 40/12=3.3 3.3/3.3=1

H 6.7 1 6.7/1=6.7 6.7/3.3=2

O 53.3 16 53.3/16=3.3 3.3/3.3=1

The empirical formula of the compound is CH2On = M.F. mass/ E.F.mass

= 60/30=2

Molecular formula = (E.F.)*2(CH2O)*2=C2H4O2


Atomic mass

Number of atoms

Simplest ratio

C 40 12 40/12=3.3 3.3/3.3=1

H 6.7 1 6.7/1=6.7 6.7/3.3=2

O 53.3 16 53.3/16=3.3 3.3/3.3=1

Calculate the empirical & molecular formula of the compound having the following percentage compositions.

C= 26.59 %; H= 2.22 %; O = 71.19 %If the vapour density is 45


Atomic mass

Number of atoms

Simplest ratio

C 26.59 12 26.59/12=2.216 2.216/2.216=1

H 2.22 1 2.22/1=2.22 2.22/2.216=1

O 71.19 16 71.19/16=4.45 4.45/2.216=2

The empirical formula is CHO2

Molecular wt = 2 * V. D.= 2* 45= 90

n = M.F. Wt/E. F mass= 90/45=2


Atomic mass

Number of atoms

Simplest ratio

C 26.59 12 26.59/12=2.216 2.216/2.216=1

H 2.22 1 2.22/1=2.22 2.22/2.216=1

O 71.19 16 71.19/16=4.45 4.45/2.216=2


Atomic mass

Number of atoms

Simplest ratio

C 26.59 12 26.59/12=2.216 2.216/2.216=1

H 2.22 1 2.22/1=2.22 2.22/2.216=1

O 71.19 16 71.19/16=4.45 4.45/2.216=2

M. F. = (E.F.)*n= (CHO2)*2=C2H4O4

Howe-work

Give the examples of two Compounds

which have Same Empirical Formula.

In today’s session we learnt:

Percentage composition Empirical & molecular formula

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