What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen?

Jun 20, 2014

The empirical formula is C₃H₃O.

Assume we have 100 g of the compound.

Then we have 65.5 g of C, 5.5 g of H, and 29.0 g of O.

Moles of C = 65.5 g C × $\left(1 \text{ mol C")/(12.01"g C}\right)$ = 5.45 mol C

Moles of H = 5.5 g H × $\left(1 \text{ mol H")/(1.008"g H}\right)$ = 5.5 mol H

Moles of O = 29.0 g O × $\left(1 \text{ mol O")/(16.00"g O}\right)$ = 1.81 mol H

Moles of C:Moles of H:Moles of O = 5.45:5.5:1.81 = 3.01:3.0:1 ≈ 3:3:1

The empirical formula is C₃H₃O.

Hope this helps.