Calculate the empirical and molecular formula of a compound containing

1.	Calculate the empirical and molecular formula of a compound containing carbon, 6.38%hydrogen and rest oxygen it's vapour density is 47.
Answer» 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 Please mark me BRAINLIEST

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

Answer»

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

Please mark me BRAINLIEST