Question

A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% O. Determine the empirical formula and the molecular formula.

Answer 1

Solution :-

Mass of the each element in the compound is as follows

Mass of C = 65.45 g

Mass of H =5.45 g

Mass of O = 29.09 g

Now lets calculate the moles of the each element

Moles = mass / molar mass

Moles of C= 65.45 g / 12.01 g peer mol = 5.45 mol C

Moles of H= 5.45 g / 1.0079 g per mol = 5.41 mol H

Moles of O = 29.09 g / 16 g per mol = 1.82 mol O

Now lets find the mole ratio of the each element by dividing the moles of the each element by the smallest mole value

C= 5.45 mol / 1.82 mol = 3

H = 5.41 mol / 1.82 mo = 3

O = 1.82 mol / 1.82 mol = 1

So the empirical formula of the compound is C3H3O

Now need to find the ratio of the molar mass and empirical formula mass

Empirical formula mass of C3H3O = 55.0 g

N= molar mass / empirical formula mass

N= 100g /55 g =1.82 rounded to 2

So the molecular formula = N*empirical formula

                                              = 2*C3H3O

                                              = C6H6O2