Empirical and Molecular formula calculation

Day 2
22.4.20

Good Morning Boys,
Today we will be learning about empirical and molecular formula calculation.

Learning Outcomes :
students will be able to

• Define empirical and molecular formula.
• learn formula
• apply formula

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 H2O molecule. 

Molar mass of Hydrogen (H) = 2g, and Molar mass of Oxygen (O) = 16g.

Consider 18g of H2O it contains 2g of H and 16g of O.

Mass % of H = (2×1×100)/(18) = 11.11%

Mass % of O = (16×100)/(18)= 88.88%

Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Answer:

The molecular formula of sodium sulphate is.

Molar mass of  = [(2 × 23.0) + (32.066) + 4 (16.00)]

= 142.066 g

Mass percent of an element 

∴ Mass percent of sodium:

Mass percent of sulphur:

Mass percent of oxygen:

Empirical Formula And 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.

steps

Calculation of Empirical formula :

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

Step 2:- Convert into number moles of each element

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

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.

Calculation of Molecular Formula :

(a) Determine empirical formula mass

b) Divide Molar mass by empirical formula mass

n = (Molar mass/Empirical formula)

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

Numerical Questions :

 1. 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:

Simplest molar ratio of iron to oxygen  

= 1: 1.5

 2: 3

∴ The empirical formula of the iron oxide is Fe₂O_3.

Question 2 :

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol^–1.

Answer:

Mass percent of iron (Fe) = 69.9% (Given)

Mass percent of oxygen (O) = 30.1% (Given)

Number of moles of iron present in the oxide   = 1.25

Number of moles of oxygen present in the oxide  = 1.88

Ratio of iron to oxygen in the oxide,

= 1 : 1.5

= 2 : 3

The empirical formula of the oxide is Fe₂O_3.

Empirical formula mass of Fe₂O₃ = [2(55.85) + 3(16.00)] g ,  Molar mass of Fe₂O₃ = 159.6g

Molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe₂O₃ and n is 1.

Hence, the molecular formula of the oxide is Fe₂O_3.

_{Assignment :}

_{solve the numerical related to % composition and Empirical and Molecular Formula From Ncert.}

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