made by ritik tatia class 11

1. Empirical formula
and molecular
formula

2. What is empirical and molecular formula ?
 Basically ,
 An empirical formula represents the simplest whole number ratio of various atoms
present in a compound whereas the molecular formula shows the exact number of
different types of atoms present in a molecule of a compound.
 We can calculate molecular formula by empirical formula by using the following steps :-
1) draw a table of 5 columns , including first column as element , then percentage
composition , atomic mass , no of moles and then finally simplest ratio.
After that we will get the empirical formula mass
Calculate it and then divide by the molar mass given. We will get the molecular formula.

3. Questions
 Given the percentage composition as H- 4.07 , C-24.27% AND Cl-71.65% calculate
the empirical and the molecular formula given the molar mass of the compound
as 98.98g.
 Answer,
element % composition Atomic mass No of moles Simplest ratio
H 4.07 1 4.07 4.07/2.02=2
C 24.27 12 2.02 2.02/2.02=1
Cl 71.65 35.5 2.01 2.02/2.01=1

4.  Therefore empirical formula becomes CH2Cl
 Now calculating the molar mass of CH2Cl we get 49.5
n= molar mass/ empirical formula mass = 98.96/49.5=2
Therefore the formula becomes 2(CH2Cl)=C2H4Cl2
Q.2) Given Na=14.3% , S=9.97% ,H=6.22% ,O=69.5% and all the H2 is present as water of crystallisation find
the empirical and molecular formula.
element % composition Atomic mass No of moles Simplest ratio
Na 14.3 23 0.622 2
S 9.97 32 0.311 1
H 6.22 1 6.22 20
O 69.5 16 4.34 14

5.  So after solving the compound will be Na2SH20O14
 Now since all the water is present as wate rof crystallisation the formula
becomes Na2SO4.10H20.
Q.3) Find the molecular formula of the compound that contains 56.36g of oxygen
and 54.6 grams of P. if the molar mass of the compound is said to be 189.5g/mol.
Hint:- 1) Find the empirical formula.
2)divide by molar mass to find the common factor.
3) multiply to get the molecular formula