Question

Fumaric acid is an organic compound composed of 41.39%C , 3.47% H, and the rest oxygen. If 0.145 mole of fumaric acid has a mass of 16.8 g, what are the empirical and molecular fomulas of fumaric acid? Empirical formula: Molecular formula:

Answer 1

Solution :-

Lets assume we have 100 g sample

then mass of C= 41.39 g

mass of H = 3.47 g

mass of O = 100 g - (41.39 g + 3.47 g) = 55.14 g

now lets calculate moles of each element

moles of C= 41.39 g / 12.01 g per mol = 3.446 mol C

moles of H = 3.47 g / 1.0079 g per mol = 3.442 mol H

moles of O =55.14 g /16 g per mol = 3.446 mol O

now lets find the mole ratio by dividing the moles by the smallest mole value

C= 3.446 / 3.442 = 1

H= 3.442/3.442 = 1

O = 3.446 / 3.442 = 1

so the empirical formula = CHO

now lets calculate the molar mass

molar mass = mass / moles

                     = 16.8 g/ 0.145 mol

                     = 116 g per mol

now lets find the ratio of the molar mass to empirical formula mass

CHO = 12.01 + 1.0079 + 16 = 29

ratio = 116 g / 29 g = 4

so the molecular formula = 4*emprical formula

                                                = 4* CHO

                                               = C4H4O4

So the molecualr formla is C4H4O4