A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula? The molecular mass of the compound is 132 amu. What is its molecular formula?

Answer 1

The empirical formula is the one with the lowest whole number of moles of each element.

When given %s, we can just assume that we have 100 g of the unknown, and then the percents given are the same as grams. We can then find moles of each element as follows:

mols C = 54.53 g C x 1 mol C / 12.01 g = 4.540 mols C

mols H = 9.15 g H x 1 mol H / 1.01 g = 9.059 mols H

mols O = 36.32 g O x 1 mol O / 16.0 g = 2.27 mols O

Dividing through by 2.27 to try to get whole numbers, we end up with

mols C = 4.54 / 2.27 = 2 mols C

mols H = 9.06 / 2.27 = 4

mols O = 2.27 / 2.27 = 1

EMPIRICAL FORMULA = C₂H₄O

To find the molecular formula, we need to determine how many of the empiricals it takes to make 1 formula.

molar mass of the empirical is 24 + 4 + 16 = 44 g/mol

132 / 44 = 3

MOLECULAR FORMUAL = C₆H₁₂O₃