Question

1. An organic compound containing carbon, hydrogen and oxygen was analyzed that the mass percentage of each was found to be 49.99 % for carbon, 5.48 % for hydrogen, and 44.53 % for oxygen. Find out the empirical formula of this organic compound. 2. For the question 1, if the molecular weight of the compound was found to be 112.13 g/mol, what is the molecular formula of this compound?

Answer 1

First, assume 100 g of sample

so

49.99 g of C

5.48 g of H

44.53 g of O

change to mol

MW of C = 12, MW of H = 1, MW of O = 16

mol of C = mass/MW = 49.99/12 = 4.16583

mol of H = mass/MW = 5.48/1 = 5.48

mol of O = mass/MW = 44.53/16 = 2.783125

ratio:

H:C = 5.48/4.16583 = 1.3 or 4:3

H:O 5.48/2.783125= 1.96 or 2

the nearest empiricalformula:

C3H4O2

if MW of molecular compound = 12*3 + 4*1 + 16*2 = 72 g/mol

ratio = 112.13/72 = 1.5x

so

C3H4O2 * 1.5 = C4.5H6O3

nearest vlaue will be C5H6O3