Answer: Molecular formula will be 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 40 g
Mass of H = 6.71 g
Mass of O = 100 - (40+6.71 ) = 53.29 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H =
For O = 
The ratio of C: H: O= 1 : 2: 1
Hence the empirical formula is 
The empirical weight of
= 1(12)+2(1)+1(16)= 30g.
The molecular weight = 90.08 g/mole
Now we have to calculate the molecular formula:

The molecular formula will be=
Molecular formula will be 