### Derivation of empirical and molecular formula from percentage composition

Derivation of empirical formula from percentage composition:
Empirical formula of a compound is a formula that shows the simplest whole number ratio between the atoms of the elements in the compound.

It can be calculated from the percentage composition of the elements present in a compound. The various steps involved in determining the empirical formula are:

1) Divide the percentage composition of the elements present in a compound by their relative atomic mass. The quotients will be the ratio of the number of atoms.

2) The quotients obtained are then divided by the smallest quotient. This gives the simplest ratio between the atoms.

3) After the simplest ratio is obtained, multiply by any convenient integer to give the simplest whole number ratio.

4) The empirical formula is derived by writing the symbol of the elements present in it, followed by the number of atoms of each, as subscript, to the right of the symbol.

Derivation of molecular formula from percentage composition:

Molecular Formula of a substance gives the names and number of atoms of the various elements present and the molecular mass of it.

To find the molecular formula from percentage composition we have to undertake the following steps:

1) First find the empirical formula, as illustrated earlier.

2) Then find the empirical formula mass: for this we have to add the atomic masses, of all the atoms in the empirical formula.

For e.g., assume the empirical formula of a compound is HO.

Empirical formula mass = Mass of 1 hydrogen atom + mass of 1 oxygen atom

= 1 + 16 = 17

3) The relative molecular mass or molar mass is obtained from experiment or from the vapour density relationship: molecular mass = 2 x vapour density.

In the example cited above let the molar mass be = 34.

4) We then have to find the ratio (n) of the relative molecular mass to the empirical formula mass. This ratio is calculated as

From the example above 'n'

5) The molecular formula is found from the above ratio by using the realtion:

Molecular formula = n x empirical formula

In the example, the relative molecular mass is found to be two times greater than the empirical formula mass,

the molecular formula = empirical formula x 2

= (HO) x 2 = H2O2
Problems

8. An organic compound weighing 0.2 g containing C, H and O gave on combustion 0.296 g CO2 and 0.12 g H2O. Its molecular mass is 180. Determine its empirical and molecular formulae.

### Solution

Mass of the compound taken = 0.2 g

Mass of CO2 formed = 0.296 g

Mass of H2O formed = 0.12 g

Therefore, the percentage of O = 100 - (40.0 + 6.67) = 53.33(by difference)

### Calculation of empirical formula

Therefore, empirical formula of the compound = CH2O

### Calculation of molecular formula

Empirical formula mass = (1 x 12 amu + 2 x 1 amu + 1 x 16 amu)

= 30 amu

Molecular mass = 180 amu

Hence,

Therefore, Molecular formulae of a compound = 6 x CH2O = C6H12O6

9. A compound (molecular mass 147) contains 49.0 % carbon and 2.72% hydrogen. 2.561 mg of the compound gave 5.00 mg of silver chloride in Carius estimation. Determine its molecular formula. Find its empirical formula also.

### Solution

The problem is done in steps:

Calculation of the percentage of chlorine:

We know that, Cl = AgCl

35.5g (108 +35.5)= 143.5? 5 mg

Then,

Percentage of chlorine in the sample =

Empirical formula = C3H2Cl

Empirical formula mass = (3 x 12 amu + 1 amu + 1 x 35.5 amu)

= 73.5 amu

Molecular mass (given) = 147 amu

Hence,

Therefore,

Molecular formula = 2 x Empirical formula

= 2 x C3H2Cl

= C6H4Cl2

10. About 0.45 g of an organic compound gave on combustion 0.792 g of CO2 and 0.324 g of water 0.24 g of the same substance was Kjeldahlised and the ammonia evolved was absorbed in 50.0cm3 of N/4 H2SO4. The excess of the acid required 77.0 cm3 of N/10 NaOH for complete neutralization. Calculate the empirical formula of the acid.

### Solution

Calculation of the percentages of different elements.

Percentage of nitrogen can be calculated thus,

Volume of H2SO4 taken = 50 cm3 N/4 of solution

Volume of NaOH required for excess acid = 77 cm3 N/10 of solution

To calculate volume of N/4 H2SO4 used for neutralization of NH3:

N1V1 (acid left unused) = N2V2 (alkali used)

Volume of N/4 H2SO4 used fro neutralizing NH3= 50 30.8 = 19.2 cm3

Now, 1000 cm3 of N/4 contain nitrogen = 14 g

Percentage of oxygen = 100 - (%C + %H + %N)

= 100 - (48 + 8 + 28) =16

### Calculation of empirical formula

The simplest ratio of C : H : N : O is 4 : 8 : 2 : 1

Empirical formula of the compound is C4H8N2O.