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What is Empirical formula?

CBSE CHEMISTRY IB DIPLOMA CHEMISTRY IGCSE CHEMISTRY Stoichiometric Relationships

What is Empirical 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:   

C6H6 is the molecular formula of benzene. 

CH is the simplest ratio of atoms of C6H6

Molecular formula = n X Empirical formula 

where n=Molecular formula mass/Empirical formula mass

  molecular formula mass of C6His 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= C6H6

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:  

Step1Iron(Fe)=44.09Chlorine(Cl)=100-44.09=55.91
Step244.09/5655.91/35.5
Step30.79/0.79=11.57/0.79= 1.98=2
Step412

So the emp formula is FeCl so now to find out molecular formula, 

use Molecular formula = n X Empiricalformula where n=1 

hence Empiricalformula = Molecular formula

Practice questions:  

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.

Read Mole Concept

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