Empirical, Molecular Formula And Limiting Reagents

A.Percentage composition.

The mass percentage of each constituent elements present in any compound is called its Percentage composition

$$Mass\; percentage\; of element=\frac{Total\; mass\; of\; particular\; element}{Molecular\; mass\; of\; compound}\times\;100\% $$

For example % composition of potassium nitrate (KNO₃) can be calculated as,

1 mole of KNO₃ contains 1 mole of K, 1-mole of N and 3-mole of O .

The molecular mass of KNO₃ = 39 + 14 + 98 = 101 amu
Now,
% of K =\( \frac {39}{101}\) x 100% = 38.6%
% of N = \(\frac {14}{101}\) x 100% = 13.9%
And % of O = \(\frac{98}{101}\) x 100% = 47.5%
Therefore, the % composition of KNO₃ is K = 38.6%, N = 13.9% and O = 47.5%

B. Empirical Formula :

The empirical formula of a compound is defined as the simplest formula that shows the simplest whole number ratio of the different atoms present in one molecule of the compound.
Benzene has a molecular formula C₆H₆. The simplest whole no. the ratio of C and H is 1:1 so empirical formula of C₆H₆ is CH.

C. Molecular formula :

It is defined as the symbolic representation of a molecule that shows the actual number of atoms present in that molecule of the compound.
The molecular formula may be the same or an integral multiple of its empirical formula. Therefore the relation between empirical and molecular formula is given as.

Molecular Formula = ( empirical formula ) x n

Where n = 1, 2, 3, 4 ......... (Interger)

And the value of 'n' can be obtained by the following relation,

n = \(\frac{Molecular formula mass(mole. wt.)}{Empirical formula mass(emp. wt.)}\)

Derivation of empirical and molecular formula from percentage composition

I. For Empirical formula:

This process involves the following rules:

1. Determine that % of each element present in the compound by the chemical analysis.
2. If the sum of the percentage of the given elements is not equal to 100% then the remaining % will be the percentage of oxygen.
   i.e. % of O = ( 100% - Sum of % of all other elements )
3. Divide the % of each element by their respective atomic weight to get relative moles of atoms (atomic ratio).
4. Divide the relative moles of atoms by their lowest value to get the simplest ratio of atoms.
   If it is in the whole number then this represents an empirical formula. But if it is not in the whole number then all values should be multiplied

II. For molecular formula:

This process involves the following rules :

1. Determine the empirical formula of a compound as outlined above.
2. Determine the empirical formula mass.
3. Find the value of 'n' through the relation.
   n = \(\frac{molecular\ mass}{emp.\ formula\ mass}\)
   The molecular mass of the compound is given.
   If V.D. is given then mol. wt. = 2 x V.D. ( V.D = vapour density)
4. Finally, molecular formula is determined by
   Molecular formula = ( empirical formula ) x n

Solved Example 1

3.0-gram sample of copper chloride on analysis gave 1.42 gm copper and 1.5 gm chlorine. What is the Empirical formula of copper chloride?

Solution :

i. calculation of % of elements
wt. of copper chloride = 3.0 gm
wt. of copper = 1.42 gm
wt. of chlorine = 1.58 gm
Now, % of Cu = \(\frac{wt.\ of \ copper}{wt.\ of\ copper\ chloride}\) x 100% = \(\frac{1.42}{3.0}\) x 100% = 47.33%
And % of Cl = \(\frac{wt. of chlorine}{wt. of copper cholride}\) x 100% = \(\frac{1.58}{3.0}\) x 100% = 52.66%

ii. Claculation of empirical formula :

Therefore, empirical formula of the compound is CuCl₂.

Solved question 2
An inorganic compound contains 28.16 % potassium and 25.63% chlorine. Determine its empirical and molecular formula, if its molecular mass is 138.5 amu.

Solution:(i) Calculation of % of elements
% of K = 28.16 %
% of Cl = 25.63 %
And % of O = 100 - (28.16 + 25.63) = 43.21 %
(ii) Calculation of empirical formula :

So, Empirical formula = KCLO₄

(iii) Calculation of molecular formula
Now, empirical formula (KCLO₄) = 39 + 35.5 + 64 = 138.5 amu
Given, molecular mass = 138.5 amu
We have, n = \(\frac{molecular\ mass}{empirical\ formula\ mass} = \frac{138.5}{138.5}\) = 1
Hence, molecular formula = (empirical formula ) x n
= (KCLO₄) x 1
= KCLO₄
Therefore, molecular formula of compound = KCLO₄

D. Limiting reagent ( Limiting reactant) :

The reactant in accordance with the stoichiometry indicated by the balanced equation. Generally, one of the reaction mixtures is present in the lesser amount than other reactants as required by the balanced chemical equation. So the amount of product formed during the reaction depends on the mass of reactant which is consumed completely.

Limiting reagents is defined as the chemical reagent present in the reaction mixture which is in deficit and completely consumed at first during the chemical reaction.

Limiting reagent controls or limits the formation of the product.

Essential of limiting reagent :

Limiting reagent is essential in stoichiometric calculation because of the following reasons:

1. It gives the information about the given chemical reaction and helps in a number of chemical calculations.
2. It helps to calculate a number of products formed.
3. It helps to calculate the amount of excess reagent left as non-reacted portion.
4. It helps to calculate the amount of one reagent required for a chemical reaction with respect to other reactant.

Chemical Calculations based on chemical equations:

The general method of calculations for all the problems of reactions consists of the following steps:

1. Write down the balanced chemical equation.
2. Write a relative number of moles or relative masses of the reactants and the products below their formula.
3. In case od gases substance, write down 22.4 litres at NTP below the formula in place of I-mole.
4. Find limiting reagents by using the unitary method if required.
5. Apply Unitary method to make the required calculations.

To tackle the problems on chemical equations, we should be able to relate the given quantities in terms of,

1. Mole and mole relation:
   For the reaction, 2H₂(g) + O₂(g) → 2H₂O (V)
   2-moles 1-moles 2-moles
   2-moles of hydrogen reacts with 1-moles of oxygen to give two moles of water vapour.
2. Mass and Mass relation :
   For the reaction, 2H₂(g) + O₂(g) → 2H₂O (V)
   2-moles 1-moles 2-moles
   4gm 32gm 36gm
   4gm of hydrogen reacts with 32 gm of O₂ to give 36 gm of H₂O (V).
3. Volume-volume relation:
   For the reaction, 2H₂(g) + O₂(g) → 2H₂O (V)
   2-moles 1-moles 2-moles
   2 x 22.4 litres 22.4 litres 2 x 22.4 litres
   44.8 gm litres of H₂ gas reacts with 22.4 litres of O₂ gas to give 44.8 litres of H₂O vapour at NTP.
   
   

References :-

Ghosh, A.K. Chemical Calculations. 15th Edition. India: Scientific Book Company, 1991.

M.L Sharma & P.N. Chaudhary. Advanced Level Chemistry. 2nd Edition . Vol. I. Kathmandu, Nepal: Ekta Books, 2011.

Palak, K.R. Fundamentals of Chemistry. Kathmandu, Nepal: Ratna Pustak Bhandar, 2000.

Pathak, Prof. Dr Tulsi Prasad. Rectified Chemistry. Kalikasthan, Kathmandu : Airawati Prakashan (P.) Ltd., 2014.