GivenCarbon: 53.31 gHydrogen: 11.18 Oxygen: 35.51 g

We know that the particular atomic weights:

Using the given information and adding them up, we know it consists of 100g total:

Now transform the provided Grams to moles to get the following:

For Carbon: 53.31 g, ----> 53.31 / 12.0107 = 4.4385 molesFor Hydrogen: 11.18 g, ----> 11.18 / 1.00794 = 11.0919 molesFor Oxygen: 35.51 g. ----> 35.51 / 15.9994 = 2.2194 moles

Using the proportion to get empirical formula ( meaning, divide them by smaller sized # that moles)

Carbon: 4.4385 / 2.2194 = 1.9999

Hydrogen: 11.0919 / 2.2194 = 4.9977

Oxygen: 2.2194 / 2.2194 = 1

Using the obtained results you get that the empirical Formula exchange mail to

C2H5O

But because you are given 90 AMU over there is an extra step!!!!!

Since the mass of C2H5O is specifically 45, that method is half of what you are given!

Molecular formula = (C2H5O)n

Empirical fixed = 2X12+5+16 = 45 Amu

n = 90/45 = 2

Molecular formula = C4H10O2


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