Vanillin, a flavoring agent, is made up of carbon, hydrogen, and Oxygen atoms. When a sample of Vanillin weighing 2.500g burns in Oxygen, 5.79 g of carbon dioxide and 1.18 g of water are obtained. What is the empirical formula of Vanillin?

To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that, we need to know the amount of carbon, hydrogen and oxygen atoms in the compound. We determine these from the given amounts of the products after burning.

Given:

2.5 g vanillin

5.79 g CO2

1.18 g H2O

Solution:

mol C = 5.79 g CO2 (1 mol CO2 / 44.01 g CO2) (1 mol C / 1 mol CO2) = 0.132 mol C

mol H = 1.18 g H2O (1 mol g H2O / 18.02 g H2O) (2 mol H / 1 mol H2O) = 0.131 mol H

We convert them into grams,

0.132 mol C (12.01 g / 1 mol) = 1.59 g

0.131 mol H (1.01 g / 1 mol) = 0.132 g

mass of C and H in vanillin = 1.722 g

mass of O = 2.5 - 1.722 = 0.778 g O

mol O = 0.778 g O (1 mol O / 16 g) = 0.049 mol O

Dividing the number of moles of each element with the smallest value of number of mole, we will have the empirical formula:

moles ratio

C 0.132 / 0.049 2.69 x 3 = 8

H 0.131 / 0.049 2.67 x 3 = 8

O 0.049 / 0.049 1 x 3 = 3

The empirical formula would be C8H8O3.