Answer : The empirical formula of a compound is,
and the molecular of the compound is, 
Solution : Given,
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 = 44.4 g
Mass of H = 6.21 g
Mass of S = 39.5 g
Mass of O = 9.86 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of S = 32 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = 
Moles of H = 
Moles of S = 
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 S = 
For O = 
The ratio of C : H : S : O = 6 : 10 : 2 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = 
The empirical formula weight = 6(12) + 10(1) + 2(32) + 1(16) = 162 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :

Molecular formula = 
Therefore, the molecular of the compound is, 