Please, please, please help with emperical formula. helllllp!!!!

Determine the emperical formulae of a compound with the following composition by mass: 18% C, 2.5% H, 63.5% I and 16% O. If this compound has a molar mass of 400g/mol, what is the molecular formulae? Please do show the steps and what is meant by 400g/mol.

Consider 100 g of the compound.
Number of moles of C = Mass/molar mass = 18/12 = 1.5
Number of moles of H = 2.5/1 = 2.5
Number of moles of I = 63.5/127 = 0.5
Number of moles of O = 16/16 = 1

If you multiply the number of moles of each element by 2, you will get C as 3, H as 5, I as 1 and O as 1. Hence empirical formula is C3H5IO2.
Molecular mass = 36 + 5 + 127 + 32 = 200
But, the molar mass is 400. Thus, there are two C3H5IO2 units in one molecule of the compound.
Thus, molecular formula = C6H10I2O4.

BEST ANSWER:

Assume a 100-gram sample:

mole of C = 18.0g x (1 mol C/12.01g) = 1.50 moles

mole of H = 2.50g x (1 mol H/1.01g) = 2.48 moles

mole of I = 63.5g x (1 mol I/126.90g) = 0.500 mole

mole of O = 16.0g x (1 mol O/16.00g) = 1.00 mole

C: 1.50/.500 = 3

H: 2.48/.500 = 4.96 ≈ 5

I: .500/.500 = 1

O: 1.00/.500 = 2

Empirical Formula is : C3H5IO2

n = (400 g/mol)/199.98 g/mol = 2

Molecular Formula = C3H5IO2 x (2) = C6H10I2O4

Thank you...