Question

An unknown compound contains only C, H, and O. Combustion of 6.80 g of this compound produced 16.6 g of CO2 and 6.80 g of H2O.

What is the empirical formula of the unknown compound?

CHO

Answer 1

CHO

m = 6.8

mol oCO2 = mass/MW = 16.6/44 = 0.37727 mol of CO2

mol of H2O = mass/MW = 6.8/18 = 0.3777 mol of H2O

mol of C = 0.37727

mol of H = 2*0.3777 = 0.7554

mass of C = 0.37727 *12 = 4.52 g of C

mass of H = 0.7554*2 = 1.508 g of H

mass of O = 6.8-(4.52 +1.508 ) = 0.772

mol of O = 0.772/16 = 0.04825

ratios:

molC = 0.37727 molH = 0.7554 molO = 0.04825

C:O = 0.37727/0.04825 = 7.819 mol of C per O ... 8

H:O = 0.7554/0.04825 = 15.65 mol of H per mol of O .. 16

C:H = 0.7554/0.37727 = 2mol of H per C .. 2

C8H16O2

divide by 2

C4H8O