A sample of liquid consisting of only C,H and O and having a mass of 0.5438g was burned in pure oxygen and 1.039g of CO2 and 0.6369g H2O were obtained. what is the empirical formula of the compound?

Answer 1

12g/molC / 44g/molCO2 = x / 1.039g CO2
12/44 (1.039) = 0.2834 g C
2g/mol H2 / 18g/molH2O = x / 0.6369gH2O
2/18 (0.6369) = 0.071 g H
0.5438 - 0.2834 - 0.071 = 0.1939 g O
0.2824gC/12g/molC = 0.0236 mol C
0.071gH / 1g/molH = 0.071 mol H
0.1939gO/16g/molO = 0.0121 mol O
Divide each of these numbers by lowest moles.
0.0236/0.0121 = 2
0.071/0.0121 = 6
0.0121/0.0121 = 1
This number represents the ratio of each element (the subscript of each)
Therefore empirical formula = C2H6O

Answer 2

C2H6O