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# Calculation of Empirical Formula from Quantitative Analysis and Percentage composition

"An empirical formula (or) simplest formula for a compound is the formula of a substance written with the smallest integer subscripts".

Calculation of Empirical Formula from Quantitative Analysis and Percentage composition

Empirical Formula

"An empirical formula (or) simplest formula for a compound is the formula of a substance written with the smallest integer subscripts".

For most ionic substances, the empirical formula is the formula of the compound. This is often not the case for molecular substances. For example, the formula of sodium peroxide, an ionic compound of Na+ and O22-, is Na2O2. Its empirical formula is NaO. Thus empirical formula tells you the ratio of numbers of atoms in the compound.

Steps for writing the Empirical formula

The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps.

1.     Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.

2.     Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.

3.     Multiply the figures, so obtained by a suitable integer of necessary in order to obtain whole number ratio.

4.     Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.

Solved Problem

1.A compound has the following composition Mg = 9.76%,S = 13.01%, 0 = 26.01, H2O = 51.22, what is its empirical formula?

[Mg = 24, S = 32, O = 16, H = 1]

Solution

Hence the empirical formula is Mg SO4.7H2O. Molecular Formula from Empirical Formula

The molecular formula of a compound is a multiple of its empirical formula.

Example

1.The molecular formula of acetylene, C2H2 is equivalent to (CH)2, and the molecular formula of benzene, C6H6 is equivalent to (CH)6. Therefore, the molecular weight is some multiple of the empirical formula weight, which is obtained by summing the atomic Weights from the empirical formula. For any molecular compound.

Molecular Weight = n x empirical formula weight.

Where `n' is the whole number of empirical formula units in the molecule. The molecular formula can be obtained by multiplying the subscripts of the empirical formula by `n' which can be calculated by the following equation

n  = Molecular Weight / Empirical formula Weight

Solved Problem

1. A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09% 0 = 36.36. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Solution

Calculation of empirical formula Empirical formula is C2 H4 O.

Calculation of Molecular formula

Empirical formula mass  =   12 x 2 + 1 x 4 + 16 x 1 = 44

Molecular mass  =  2 x Vapour density   = 2 x 44 = 88

n  = (Molecular mass / Empirical Formula mass ) = 88 / 44  =  2

Molecular formula =  Empirical formula x n

C2 H4 O x 2

C4 H8 O2

2.A compound on analysis gave the following percentage composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

Solution :- Calculation of empirical formula The empirical formula is Na2 SH20O14

Calculation of Molecular formula

Empirical formula mass  = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322

n   =  (Molecular mass / Empirical formula mass ) = 322 / 322 = 1

Hence molecular formula = Na2 SH20 O14

Since all hydrogens are present as H2O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.

Hence, molecular formula = Na2SO4.10H2O.

Problems for Practice

1. An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?

Ans:- C2H5NO

2. A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula. Ans:- C2H3O3, C4H6O6

An acid of molecular mass 104 contains 34.6% carbon, 3.85% hydrogen and the rest is oxygen. Calcualte the molecualr formula of the acid.

What is the simplest formula of the compound which has the following percentage composition: carbon 80%, Hydrogen 20%, If the molecular mass is 30, calcualte its molecular formula.

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11th 12th std standard Class Organic Inorganic Physical Chemistry Higher secondary school College Notes : Calculation of Empirical Formula from Quantitative Analysis and Percentage composition |