Question

1 An organic compound on analysis gave the following percentage composition:
C=57. 8%: H=3.6% and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound​

Answer 1

Answer:

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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