Question

Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.

Answer 1

Given conditions,

Carbon % = 76.5%
Hydrogen% = 6.38%
Oxygen% = 100 - (76.5 + 6.38)
  = 100 - 82.88
  = 17.12%

Now, Refers to the attachment for the Empirical Formula Chart.

Empirical Formula is C₆H₆O.

Empirical Formula Mass = 12 × 6 + 1 × 6 + 16
 = 94 g

Molecular Mass = 2 × Vapor Density
∴ Molecular Mass = 2 × 47
∴ Molecular Mass = 94 g.

∴ n = Molecular Mass/Empirical Formula Mass
   = 94/94
   = 1

∴ Molecular Formula = 1 × C₆H₆O
   = C₆H₆O


Hope it helps.

Answer 2

hope it helps you
mark as brainliest if it helped you