A compound consists of 65.45% c, 5.492% h and 29.06%o. it has a molar mass of approximately 110 amu. determine the empirical and molecular formula of the compound

The empirical and molecular formula of the compound is calculated as follows

the empirical formula calculation step

find the mole of each element = % composition of an element / molar mass of an element

carbon (C) = 65.45/12=5.454 moles

Hydrogen (H) = 5.492/1 = 5.492 moles

oxygen (O) = 29.06 / 16 = 1.816 moles

find the mole ratio by diving each mole by the smallest number of mole (1.816 moles) and round off to the nearest integer

that is Carbon (C) = 5.454/1.816=3

Hydrogen (H) = 5.492/1.816 = 3

oxygen (O) = 1.816/1.816=1

the empirical formula is therefore = C3H3O

molecular formula calculation

110 amu is the total molar mass of C3H3O

therefore (C3H3O) n = 110

find for n

={ (12 x3) + (1 x3) + (16 x1) }n = 110

55n = 110

divide both side by 55

n=2

the molecular formula = (c3H3O) 2 = C6H6O2