Question

If mass percent of a chemical

% C = 53.3 % % H = 11.19 % H % of O = 35.51 %


Calculate Empirical formula and Molecular formula for this compound.

Answer 1

Let X - molecular mass. Then:

12/X = 0.533 -> X = 22.5

1/X = 0.1119 -> X = 9

16/X = 0.3551 -> X = 45

The least common multiple is 45. So:

C: 45/22.5 = 2

H: 45/9 = 5

O: 45/45 = 1

So, the empirical formula is C₂H₅O

Taking into account valencies, the molecular formula is C₄H₁₀O₂

Answer: Empirical - C₂H₅O; Molecular - C₄H₁₀O₂