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Emperical Formula and Molecular Formula

Aug 20, 2022
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Key Concepts

• Empirical formula

• Molecular formula

• Percentage yield

• Molecular mass

Introduction and Explanation:

Studying the qualities of molecules found in plants or other materials that have been used as medicines for generations is one method that which scientists find new pharmaceuticals. Chemists aim to identify the molecular structure of a physiologically active molecule taken from a natural product so that it can be enhanced or made in larger quantities. This section focuses on the first step in determining the molecular structure of a molecule, which is determining the “empirical formula” and “molecular formula.” 

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Before going into the topic of molecular formula or empirical formula one has to understand what is the percentage yield of percentage composition of an element. 

Percentage yield or percentage composition: 

Percentage yield is the theoretical yield that explains the amount of product formed from the complete conversion of the given amount of the reactant to the products. It is always calculated for the stoichiometric equation which is a balanced chemical equation. Some reactions are inherently inefficient due to the presence of side reactions that produce additional products and some others are by nature. Because some products are impossible to gather without losing some, a less-than-perfect recovery will lower the actual yield. The percentage yield of a reaction is a measure of how close to the theoretical yield it achieves: 

Percentage yield =

actual yieldTheortical yield ×100actual yieldTheortical yield ×100

Actual yield and theoretical yield both are expressed as masses or molar amounts. 

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Example: 

Zinc metal which is excess is reacted with 1.274grams of copper sulphate giving 0.392grams of copper metals. What is the percentage yield? 

Answer: From this reaction, Copper sulphate is confirmed as the limiting reactant. 

Percentage yield =

actual yieldTheortical yield×100actual yieldTheortical yield×100

Theoretical yield: 

1.274g CuSO4×1 mol  CuSO4159.62g  CuSO4×1 mol Cu1 mole  CuSO4×63.55g Cu1 mol Cu=0.5072g Cu1.274g CuSO4×1 mol  CuSO4159.62g  CuSO4×1 mol Cu1 mole  CuSO4×63.55g Cu1 mol Cu=0.5072g Cu

Percentage yield =

Actual yieldTheortical yield ×100Actual yieldTheortical yield ×100

                                 =

(0.392 g Cu0.5072g Cu)×1000.392 g Cu0.5072g Cu×100

                                 = 77.3% 

The percentage composition of a compound is also known as the relative mass of each of the elements present in 100 parts in it. The percentage composition of a compound is something that is to verify the purity of a compound. Percentage composition is only calculated when the chemical formula of a compound is known. A compound’s molecular mass is calculated by summing the masses of all the atoms in the constituent elements present in the molecule.  

Percentage of element =

 Mass of elementM ×100 Mass of elementM ×100

=XM×100=XM×100

Or 

Percentage composition of element=

(no. of atoms of that element)(atomic weight)Formula mass of a compound ×100%no. of atoms of that element(atomic weight)Formula mass of a compound ×100%

For irreversible reaction, the percentage yield is less than 100 which means the reactant is converted to a product and waste. 

Example:  

E.g.1: Calculate the percentage composition of calcium nitrate.  

The molecular formula of calcium is

 Calcium atomic mass= 40 

Nitrogen atomic mass=14 

Oxygen atomic mass= 16 

Molecular mass = 40 + 2(14 + 3 × 16) 

    =164 

Percentage composition of element= Percentage of element =

 Mass of elementM ×100 Mass of elementM ×100

Percentage composition of Ca=

40164×100=24%40164×100=24%

Percentage composition of N=

28164×100=1728164×100=17

Percentage composition of O=

96164×100=58.3%96164×100=58.3%

E.g.2: 16.5g of metal combined with oxygen to form 35.60g of metal oxide. Calculate the percentage of metal and oxygen in the compound. 

Answer: Given that 

Metal mass =16.5g 

Mass of metal oxide=35.60g 

Now, mass of oxygen=(35.60 – 16.5)gm= 19.10gm 

Percentage of metal =

16.5035.60×100=46.3

gm 

 Percentage of oxygen=

19.1035.60×100=53.7gm

Empirical formula and molecular formula:                                

The empirical formula of a compound explains the ratio of different atoms present in a molecule. It represents the exact number of atoms present in a molecule. as we even know that the empirical formula of a substance will even explain the relative number of atoms of each element in a molecule.  The empirical formula of glucose (C6H12O6) is CH2O which shows that it has one carbon,  
2 hydrogens and one Oxygen. In this way, the empirical formula of glucose contains one mole of carbon and two moles of hydrogen and one mole of oxygen. As in this way mole concept has provided a way for calculating the empirical formula of a substance. 

The followings steps are followed to determine the empirical formula: 

  1. Percentage composition is determined bu quantitative analysis. 
  1. Percentage of each element is divided by its atomic mass as it gives the ratio of elements that are present in the compound. 
  1. To get the simplest ratio of an atoms of elements present in a compound, atomic ratio of each element is divided by the minimum value of atomic ratio. 
  1. If the simplest ratio is in fraction, it has to be multiplied by the smallest integer to get the simplest whole number. 
  1. To obtain the empirical formula, symbols of the several elements present are written side by side, with their appropriate whole number ratios written as a subscript to the symbol’s lower right-hand corner. 

Molecular formula of a compound is determined from the empirical formula, if the molecular formula is known. 

Molecular formula is always the multiple of empirical formula, the value of simple multiple of empirical formula is known by dividing molecular mass with empirical formula 

Empirical formula: The formula which defines the number of atoms in the simplest ratio. 

Molecular formula: The formula defines the exact number of atoms that are present in one molecule of compound. 

Relationship between Empirical formula and molecular formula: 

M.F = E.F × n 

[Equation] 

Vapor density: 

Vapor density is the ratio of half of the molecular mass of a compound. 

Vapor pressure = [Equation] 

Example:  

E.g.2 A compound contains 34.8% oxygen, 52.2% carbon and 13.0% hydrogen. What is the empirical formula mass of the compound? 

Answer:  

Fig 1

The empirical formula is C2H6

The empirical formula mass = (2 × 12) + (6 × 1) + 16 = 46 

Summary

• Percentage yield is the theoretical yield that explains the amount of product formed from the complete conversion of the given amount of the reactant to the products.

• Actual yield and theoretical yield both are expressed as masses or molar amounts. • The percentage composition of a compound is also known as the relative mass of each of the elements present in 100 parts in it.

• The empirical formula of a compound explains the ratio of different atoms present in a molecule.

• Empirical formula:The formula which defines the number of atoms in a simplest ratio.

• Molecular formula: The formula defines the exact number of atoms that are present in one molecule of compound.

• Relationship between Empirical formula and molecular formula: M.F=E.F x n

N = Molecular mass / Emperical formula mass

• Vapor density: Vapor density is the ratio of half of the molecular mass of a compound.

Vapor pressure — Molecular mass / 2

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