The elemental mass percent composition of succinic acid is 40.68% CC, 5.12% HH, and 54.19% OO. Determine the empirical formula of succinic acid.

C2H3O2

Explanation:

Empirical formula is defined as the simplest whole-number ratio of atoms present in a molecule. To solve this question we need to convert the percentage of each atom to moles using molar mass. With the moles of each atom we can find the ratio:

Moles C -Molar mass: 12.01g/mol-

40.68g * (1mol/12.01g) = 3.387 moles C

Moles H-Molar mass:1g/mol-:

5.12g * (1mol/1g) = 5.12 moles H

Moles O -Molar mass: 16g/mol-

54.19g * (1mol/16g) = 3.387 moles O

Ratio of atoms (Dividing in moles of C that are the lower number of moles):

C = 3.387 moles C / 3.387moles C = 1

H = 5.12moles H / 3.387moles C = 1.5

O = 3.387moles O / 3.387 moles C = 1

This ratio twice (To have only whole-numbers):

C = 2

H = 3

O = 2

Empirical formula of succinic acid: