Question

a sample of liquid consisting of carbon , hydrogen,oxygen was subjected to combustion analysis 0.5439 g of the compound gave 1. 039 g of co2 0.63469g of h2o dertermine the emperical formula

Answer 1

Convert g CO2 to g C.
Convert g H2O to g H.
g O = total g - g C - g H.
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g C = g CO2 x (C/CO2) = 1.039 x 12.01/44.01 = approx 0.2833. Look up those numbers and confirm that.
g H = g H2O x (2H/H2O) = 0.63469 x (2/18) = 0.07052. Look up those numbers and confirm that.
g O = .5439-0.07052-0.2833 = about 0.1900
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Now covert g C, g O, g H to mols.
mols C = 0.2833/12 = about 0.0237
mols H = 0.0705/1 = about0.0705
mols O = 0.1900/16 = about0.0119

Now 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 = 0.0237
H = 0.0705
O = 0.0119. Divide each by 0.0119.

C = 0.0237/0.0119 = 1.99 = 2.0
H = 0.0705/0.0119 = 5.92 = 6.0
O = 0.0119/0.0119 = 1.0
So the empirical formula is C2H6O.

Answer 2

Thanks .