Question

I have an unknown organic compound containing carbon, hydrogen and oxygen. Combustion analysis of 4.05 g sample of the unknown produced 8.07 g CO2 and 3.76 g H2O. The molar mass of the unknown was determined to be 180 +/- 4 g/mol. What is the molecular formula of the unknown?

Answer 1

From the given information first we have to find number of moles of C, H, o # weight of co- 8.079. moles of con- 8.018 = 0.183 mol 4uglmet. One more of co, is made up of a mol of c. 10. 183 mel co, contain 0.183 mele of c. mass of C= moles x molar mass. = 0.183 mel X12 g mer Mass of c = 2.1969 Moles o C = 0.183 mer # weight of Ho - 3.76 g. Morus of H₂O = 3.76 g = 0.209 mel. 18 g/mol = 0.209 mal 1 mer Moles H₂O of is made up of a moles of H. H = 2x 0.209 mel. moles of H= 0.418 mer. weight of H = 0.418 max 1g /mel. 0.418 g H

# Mass of oxygen = (mass of sample)-(Mass of C+H) = 4.058 - (2-1969 + 0.4189) = 4.059-2.614 g = 1.436 g Moles of o= 1.436 g = 0.09 mer of of 1691mal we have : 0.183 mol e 0.418 mel H 0.09 mel o. all metes by smaller number ie oog. c = aC. x2 = 40 "Divide 0.183 0.09 0.418 0.09 0.09 0.09 .H = 4.64 H. x2 - 9H 0.. o x2 = 20'. Cu Hg O2 Emperical formula = Molecular formula = nx Emperical formula n = Molecular formula mass – 180 + 4 = 2 - Emperical formula mass . 89 Molecular formula - Catig On (Mass = 178 g/met)