Question

A certain compound has the following percent composition: 57.1% C, 4.8% H, and 38.l% O.

Answer 1

Answer:

C₆H₆O₃

Explanation:

Calculation sequence:

% => grams => moles => reduce => empirical Ratio

Molecular multiple = Molecular Mass / Empirical Mass

  C: => 57.1% => 57.1 g => 57.1/12 = 4.7583

  H: =>  4.8% =>  4.8 g =>   4.8/1  = 4.8000

  O: => 38.1% => 38.1 g => 38.1/16 = 2.3813

TTL => 100%       100 g

Reduced Mole values =>

C : H : O => 4.7583/2.3813 : 4.8000/2.3813 : 2.3813/2.3813 => 2 : 2 :  1

∴ empirical formula => C₂H₂O

empirical formula weight => 2C + 2H + 1O = [2(12) + 2(1) + 1(16)] amu = 42 amu

molecular formula weight (given in problem) = 126 g/mole

The molecular formula is a whole number multiple of the empirical formula.

molecular multiple = 126 amu / 42 amu = 3

∴ molecular formula => (C₂H₂O)₃ => C₆H₆O₃