Question

Calculate the empirical and molecular formula of a compound containing carbon, 6.38%hydrogen and rest oxygen it's vapour density is 47.​

Answer 1

Here is your answer:

You missed the composition of carbon

I have did this question in 11th

Carbon % = 76.5%

Hydrogen% = 6.38%

Oxygen% = 100 - (76.5 + 6.38)

= 100 - 82.88

= 17.12%

Now divide the percentage composition by the atomic mass of the element

For carbon:  76.5/12 = 6.38

For hydrogen: 6.38/1 = 6.38

For oxygen: 17.12/16 = 1.07

Now divide the answers you got to the least composition to a rough and nearest integer:

For carbon: 6.38/1.07 = 6

For Hydrogen: 6.38/1.07 = 6

For Oxygen: 1.07/1.07 = 1

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

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Answer 2

Explanation:

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