Question

A compound is 53. 31% c, 11. 18% h, and 35. 51% o by mass. What is its empirical formula? insert subscripts as needed.

Answer 1

C2H5O

Step-by-step explanation:

In a 100 g sample we would have

53.31 g of C

11.18g of H

35.51g of O

First, we find the relative number of atoms of each element by dividing the number of grams the element has in the compound by its atomic mass.

Atomic mass of carbon is 12.011

Relative number of carbon atoms = 53.31 / 12.011 = 4.4

Atomic mass of hydrogen = 1.007

Relative number of hydrogen atoms : 11.18/1.007 = 11.1

Atomic mass of oxygen : 15.999

Relative number of oxygen atoms : 35.51 / 15.999 = 2.2

Now we find a ratio of the relative number of atoms by dividing the # of relative atoms of each element by the element's relative number of atoms that had the lowest number. ( oxygen which had 2.2 ) The outcome of each will be the subscript or number of atoms of each element.

Carbon : 4.4 / 2.2 = 2

Hydrogen : 11.1 / 2.2 = 5

Oxygen : 2.2 / 2.2 = 1

The answer is C2H5O