Question

Calculate the empirical formula of the compound having the following percentage composition:
(a) C 6.7 %, H=74 %, N=25.9 %

Answer 1

Answer:  4(C2HO) = C8H4O4

Explanation: Composition of C = 57.8%

Composition of H = 3.6%

Total composition of organic compound = 100%

Composition of oxygen = 100 - (57.8 + 3.6)

                                      = 38.6%

Element    composition   Atomic mass    Ratio of atoms     Simplest ratio

  C                 57.8%                12              57.8/12 = 4.82        4.82/2.41 = 2      

  H                  3.6%                  1                3.6/1 = 3.6              3.6/2.41 = 1

  O                 38.6%                16              38.6/16 = 2.41        2.41/2.41 = 1

Therefore the empirical formula of the compound = C2HO

Molecular weight = 2 * vapour density

                            = 2*83

                            = 166

Emperical formula weight = 24+1+16 = 41

n = Molecular weight/Emperical weight

n = 166/41

n = 4 (nearly)

Molecular formula = 4(C2HO) = C8H4O4

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