The molecular mass of the compound is 90 amu. What is the molecular formula? Insert subscripts as needed.

Question

A compound is 53.31% C, 11.18% H, and 35.51% O by mass.

Answer 1

Given

1. Carbon: 53.31 g
2. Hydrogen: 11.18
3. Oxygen: 35.51 g

We know that the respective atomic weights:

1. C - carbon - 12.0107
2. H - hydrogen - 1.00794
3. O - oxygen - 15.9994

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

Now convert the given Grams to moles to get the following:

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

Using the ratio to get empirical formula ( meaning, divide them by smaller # of 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 correspond to

C2H5O

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

Since the mass of C2H5O is exactly 45, that means is half of what you are given!

Molecular formula = (C2H5O)n

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

n = 90/45 = 2

Molecular formula = C4H10O2