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Determination of Formula of Compounds from Experimental and Given Data

The molecular formula of a compound may be defined as the formula which gives the actual number of atoms of various elements present in the molecule of the compound. For example, the molecular formula of the compound glucose can be represented as C6H12O 6 A molecule of glucose contains six atoms of carbon, twelve atoms of hydrogen and six atoms of oxygen.

In order to find out molecular formula of a compound, the first step is to determine its empirical formula from the percentage composition.

The empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the compound. The empirical formula of the compound glucose (C6H12O 6), is – CH2O which shows that C, H and 0 are present in the simplest ratio of 1 : 2 : 1. Empirical formula mass of substance is equal to the sum of atomic masses of all the atoms in the empirical formula of the substance.


Relation between the Two Formulae

Molecular formula is whole number multiple of empirical formula. Thus,

Molecular formula = Empirical formula x n where n = 1, 2, 3 …

n  = Molecular formula / Empirical formula

n =Molecular mass / Empirical formula mass

 

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:

(i)   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.

(ii    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.

(iii)    Multiply the figures, so obtained by a suitable integer if necessary in order to obtain whole number ratio

(iv)   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 corner of each symbol. This will represent the empirical formula of the compound.

 

STEPS FOR WRITING THE MOLECULAR FORMULA

(i)                 Calculate the empirical formula as described above.

 

(ii)               Find out the empirical formula mass by adding the atomic masses of all the atoms present in the empirical formula of the compound.

 

(i)                 Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass and find out the value of n.

 

(iii)             Multiply the empirical formula of the compound with n so as to find out the molecular formula of the compound

The determination of the empirical formula and molecular formula of a compound involving the above steps is well illustrated by the following examples.

 

SOLVED EXAMPLES

Example  12.21. Write the empirical formula of the compounds having molecular formulae:

(i) C6H6           (ii) C6H12       (iii) Hp2

(iv) Hp             (v) Na2C03      (vi) Bp6

(vii) N204         (viii) Hf104     (ix) Fep3

(x) C2H2


Solution. Empirical formula is a simplest whole number ratio of atoms in the molecule, therefore the empirical formula of given compounds m:

(i)CH               (ii) CH2           (iii) HO

(iv)H2O           (v) NH2CO3   (vi) BH3

(vii) NO2         (viii) H3PO 4   (ix) Fe 3

(x) CH.

 

Example 12.22 A compound of nitrogen and oxygen is analyzed and a sample weighing 1.587g is found to contain 0.483gN and l.l04g0. What is the empirical formula of the compound?

 Solution .

  % age of N in compound = 0.483 / 1.587 x 100 = 30.43

% age of O in compound  = 1.104 / 1.587 x 100 = 69.57

Example 12.23.  A compound has the following composition: J1.g = 9. 76%, S = 13.01%, 0 = 26.01%, H2O = 51.22%. What is its empirical formula? [Mg = 24, S = 32, 0 = 16, H = 1]

SOLUTION.

Example 12.24.What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, Hydrogen 20%? If the molecular mass is 30, calculate its molecular formula.

Solution. Calculation of empirical formula:

Empirical formula is CH3

Calculation of molecular formula:

Empirical formula mass

= 12 x 1 + 1 x 3 = 15

n  = Molecular mass / Empirical formula mass

Molecular formula

= Empirical formula x 2

= CH2 X 2 = C2H6.