An organic compound x contains 40% carbon, 6.7% hydrogen the third compound oxygen. If the relative molecular mass of the compound is 60, calculate (i) the empirical formular of x (ii) the molecular formular of x

Answer 1

Take a 100 g sample which give you
40 g C
6.7 g H
100-40-6.7 = 53.3 g O

Convert these g to mols. mol = g/atomic mass.

40/12 = about 3.3 mols C
6.7/1 = 6.7 mols H
53.3/16 = about 3.3

Now find the ratio of these three elements to each other. The easy way to do that is to divide the smallest number by itself, then divide the other numbers by the same small number; i.e.,
C = 3.3/3.3 = 1.0
H = 6.7/3.3 = 1.988 whch rounds to 2.0
O = 3.3/3.3 = 1.0
NOTE. If you do those mols more accurately it comes out a little better. So the empirical formula is
C1H2O for an empirical mass of 1*12 + 2*1 + 1*16 = about 30

empirical mass x number = molar mass = 60
30*number = 60
number = 2
Therefore, the molecular formula is
(CH2O)2 or C2H4O2

Answer 2

Source very helpful