A sample of an organic compound consisting of carbon, hydrogen and oxygen was subjected to combustion analysis. 0.5439g of this compound gave 1.039g of carbon dioxide and 0.63469g of water vapours. The empirical formula of this compound is:

Explanation

Convert g CO2 to g C.Convert g H2O to g H.g O = total g - g C - g Hg C = g CO2 x (C/CO2) = 1.039 x 12.01/44.01 = approx 0.2833g g H = g H2O x (2H/H2O) = 0.63469 x (2/18) = 0.07052g g O = .5439-0.07052-0.2833 = about 0.1900Now covert g C, g O, g H to mols.mols C = 0.2833/12 = about 0.0237molmols H = 0.0705/1 = about 0.0705molmols H = 0.0705/1 = about 0.0705molNow find the ratio of these elements to each other with the lowest number no less than 1.00 and round to whole numbers. The easy way to do that is to divide the smallest number by itself (making sure that is 1.00), then divide the other two numbers by the same small number. Here is what you have:C = 0.0237H = 0.0705O = 0.0119. Divide each by 0.0119.C = 0.0237/0.0119 = 1.99 = 2.0H = 0.0705/0.0119 = 5.92 = 6.0O = 0.0119/0.0119 = 1.0So the empirical formula is C2H6O.