Compound X contains 40% , Carbon 6.6% ,Hydrogen and 53.3% Oxygen its relative molecular mass is 180 What is the molecular formula ?

Given ,
Molecular mass of the said compound = 180 u
C = 40 %
H = 6.6 %
O = 53.3%
So in 100 g of this Compound -
C = 40 g
H = 6.6 g
O = 53.3 g
Dividing each element with their Molar Mass to get the no. of moles:-
C = 40 / 12.011 = 3.33
H = 6.6 / 1.008 = 6.6
O = 53.3 / 16.00 = 3.33
Dividing all of the Relative No. of Moles by the lowest term which is 3.33
C = 3.33/ 3.33 = 1 (SIMPLE WHOLE NO. RATIO)
H = 6.6  / 3.33 = 1.98 = 2 (SIMPLE WHOLE NO. RATIO)
O = 3.33 / 3.33 =  1 (SIMPLE WHOLE NO. RATIO)
Thus the Emperical Formula is CH₂O

Empirical Formula mass = ( 1 $\times$ C) + ( 2$\times$ H) + ( 1 $\times$ O)
= ( 1 $\times$12) + (2 $\times$1) + ( 1 $\times$ 1)
= 12 + 2 + 1
= 15 u
Molecular formula Mass = n $\times$ Empirical Formula mass
Thus n = 180 u / 15 u = 12
Thus Molecular Formula = 12 ( CH₂O )
= C₁₂H₂₄O₁₂