Atoms and Molecules : Percentage Composition

 

Empirical and Molecular Formula from Percentage Composition

Empirical Formula                                                                                                                                                                

The empirical formula of compound shows the simplest atomic ratio of the elements present in a molecule of the compound. For example, the empirical formula of hydrogen peroxide (H2O2) is HO, because the simplest atomic ratio of hydrogen and oxygen in a molecule of it is 1 : 1.

The empirical formula of a compound can be determined if the percentage composition of the elements in the compound is known.

The percentage of each element in the compound gives the mass (in grams) present in 100g of the compound. So, if we divide the percentage by the atomic mass of the element, we get the number of moles of atoms of the element present in 100 g of the compound. The ratio of the number of moles of the different elements thus calculated can be divided by a suitable common factor to obtain the simplest whole-number ratio of the atoms present in the compound. Thus the calculation is carried out in the following steps.

  • Tabulate the percentage composition of the different elements.
  • Divide the percentage of each element by the respective atomic mass to obtain the number of moles of atoms of the element.
  • Divide the numbers of moles by a suitable common factor to obtain them in the simplest whole-number ratio.
  • Deduce the empirical formula of the compound from the whole-number of the atoms.

 

Illustration 1:         Determine the empirical formula of compound if it contains 5.9% H and 94.1% O.

Solution:       Element         Percentage    Atomic mass  Moles of atoms   Atomic ratio

                        H                           5.9                         1                   5.9/1 = 5.9        5.9/5.9=1

                        O                          94.1                        16                 94.1/16=5.9      5.9/5.9 = 1

                        The empirical formula is HO.

 

Exercise 2: A hydrocarbon contain 20% H and 80% C. Deduce its empirical formula.

Ans.    CH3

Exercise 3: A compound contains 6.7% H, 40% C and 53.3% O. Determine its empirical formula.

Ans.    CH2O

Exercise 4: Deduce the empirical formula of a compound containing 1% H, 33% S and 66% O.

Ans.    H2SO4

Exercise 5: Ammonium chloride contains 31.8% NH3 and 66.3% Cl. Work out the empirical formula of the compound (H = 1, N = 14, O = 16, Cl = 35.5).

Ans.    NH4Cl

 

Molecular formula

The molecular formula can be derived from the empirical formula when the molecular mass of the compound is known.

As you know, the empirical formula gives the ratio of the atoms of the elements present in a molecule of a compound. The molecular formula gives the actual number of atoms of the different elements present in a molecule of the compound.

                        (Empirical formula)n = molecular formula,

                        where n is an integer.

 

∴ relative molecular mass = n x empirical formula mass.

So, if we know the empirical formula mass (which can be easily calculated) and the molecular mass (which can be experimentally determined), we can calculate n and thus determine the molecular formula of the compound.

If the substance is a gas or a compound which can be completely volatilised, the molecular mass can be easily calculated from the relative density (or the vapour density).

 

Illustration 6:         The empirical formula of a compound is HO and the relative molecular mass 34. Deducehe molecular formula of the compound.

Solution:       Let the molecular formula be (HO)n.

                        The empirical formula mass = 1 + 16 = 17.

                        We know that

                        n  x  empirical formula mass = relative molecular mass

                                    ⇒ n x 17 = 34

                                    ⇒    \frac  { 34 }{ 17 }   = 2.

   Therefore the molecular formula of the compound is (HO)2, i.e., H2O2.

 

Illustration 7:         Deduce the molecular formula of a compound whose empirical formula is CH3 and relative molecular mass 30.

Solution:       Let the molecular formula be (CH3)n.

                        The empirical formula mass = 12 + 3 = 15.

                        We know that

                        n  x  empirical formula mass = relative molecular mass

                                    ⇒ n x 15 = 30

                                ⇒ \frac {30 }{15} = 2

 Therefore the molecular formula of the compound is (CH)2 , i.e., C2H6.

 

Exercise 8: Acetylene and benzene have the same empirical formula CH, but vapour densities of 13 and 39 respectively. Deduce their molecular formulae.

Ans.    C2H2,  C6H6

Exercise 9: An organic compound of vapour density 29 contains 62.07% C, 10.34% H and 27.59% O. Deduce the molecular formula of the compound.

Ans.    C2H6O

Exercise 10: An oxoacid of sulphur, relative molecular mass 194, contains 33% S and 66%. Deduce the molecular formula of the acid.

Ans.    H2S2O8

Exercise 11: Anhydrous aluminium chloride contains 20.22% Al and 79.78% Cl. Though a solid, it volatilizes completely on heating. At 350oC, the vapour density is 133.5 and at 750oC, 66.75. Deduce the molecular formulae of the salt at the two temperatures.

Hint: Vapor density = Molecular weight / 2        

Ans.    Al2Cl6, AlCl3

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Atoms and Molecules : Reaction Based Questions