Question

A molecule with molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. What are the empirical and molecular formulas of the molecule?

Answer 1

The empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆.

Assume we have 100 g of the compound.

Then we have 40.00 g of C, 6.72 g of H, and 53.28 g of O.

Moles of C = 40.00 g C × `(1"mol C")/(12.01"g C")` = 3.331 mol C

Moles of H = 6.72 g H × `(1" mol H")/(1.008"g H")` = 6.67 mol H

Moles of O = 53.28 g O × `(1" mol O")/(16.00"g O")` = 3.330 mol H

Moles of C:Moles of H:Moles of O = 3.331:6.67:3.330=1.000:2.00:1 ≈ 1:2:1

The empirical formula is CH₂O.

The empirical formula is the simplest formula of a compound. The actual formula is an integral multiple of the empirical formula.

If the empirical formula is CH₂O, the actual formula is `"(CH₂O)"_n` or `C_nH_(2n)O_n`, where `n` = 1, 2, 3, … Our job is to determine the value of `n`.

The empirical formula mass of CH₂O is 30.03 u. The molecular mass of 180.18 u must be some multiple of this number.

`n = (180.18" u")/(30.03" u")` = 6.000 ≈ 6

∴ The molecular formula = `C_nH_(2n)O_n` = C₆H₁₂O₆.

Hope this helps.