# 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?

Jun 20, 2014

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 × $\left(1 \text{mol C")/(12.01"g C}\right)$ = 3.331 mol C

Moles of H = 6.72 g H × $\left(1 \text{ mol H")/(1.008"g H}\right)$ = 6.67 mol H

Moles of O = 53.28 g O × $\left(1 \text{ mol O")/(16.00"g O}\right)$ = 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 ${\text{(CH₂O)}}_{n}$ or ${C}_{n} {H}_{2 n} {O}_{n}$, 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.

$n = \left(180.18 \text{ u")/(30.03" u}\right)$ = 6.000 ≈ 6

∴ The molecular formula = ${C}_{n} {H}_{2 n} {O}_{n}$ = C₆H₁₂O₆.

Hope this helps.