Question

A 5.91 g unknown sample analyzed by elemental analysis revealed a composition of 37.51 % C. In addition, it was determined that the sample contains 1.4830 x 10^23 hydrogen atoms and 1.2966 x 10^23 oxygen atoms. What is the empirical formula?

Answer 1

Solution

The mass of carbon in 5.91 grams of sample is = 37.51 % X 5.91 = 2.22 grams

Moles of carbon = Mass / Atomic mass = 2.22 / 12 = 0.185 moles

Moles of oxygen = Number of atoms / Avagadro’s Number = 1.2966 x 10²³ / 6.023 X 10²³

Moles of oxygen = 0.215

Moles of hydrogen = Number of atoms / Avagadro’s Number = 1.4830 x 10²³ / 6.023 X 10²³

Moles of hydrogen = 0.246

Let us divide all the moles of given elements with the least number of moles

Mole ratio of carbon = 0.185 / 0.185 = 1

Mole ratio of oxygen = 0.215 / 0.185 = 1.16

Mole ratio of hydrogen = 0.246 / 0.185 = 1.33

Let us convert these fraction into whole number by multiplying with 6

Mole ratio of carbon 1 X 6 = 6

Mole ratio of oxygen = 1.16 X 6 = 6.96 = 7

Mole ratio of hydrogen = 1.33 X 6 = 7.98 = 8

So empirical formula will be: C₆H₈O₇