Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165 g sample is combusted to produce 0.166 g of water and 0.403 g of carbon dioxide. What is the empirical formula for valproic acid? If the molar mass is 144 g/mol, what is the molecular formula?

The empirical formula is C₄H₈O;

The molecular formula is C₈H₁₆O₂.

Explanation:

The empiric formula is CxHyOz, and the combustion reaction occurs between the fuel and oxygen gas, so it will be:

CxHyOz + O₂ → CO₂ + H₂O

So, all the carbon will form CO₂ and all hydrogen will form H₂O. So, let's calculated the number of moles of carbon in CO₂ and the number of moles of hydrogen in H₂O. They must be the same number f the moles in the valproic acid.

The molar masses are: C = 12 g/mol, O = 156 g/mol and H = 1 g/mol, so CO₂ = 12 + 2x16 = 44 g/mol, and H₂O = 2x1 + 16 = 18 g/mol.

nCO₂ = 0.403/44 = 9.16x10⁻³ mol

In 1 mol of CO₂ there is 1 mol of C, so nC = 9.16x10⁻³ mol.

nH₂O = 0.166/18 = 9.22x10⁻³ mol.

In 1 mol of H₂O there are 2 moles of H, so nH = 0.018 mol.

The mass of Carbon and the mass of H in the compound must be:

mC = nxM = 9.22x10⁻³ x12 = 0.1106 g

mH = nxM = 0.018x1 = 0.018 g

The mass of oxygen must be then:

mC + mH + mO = 0.165

mO = 0.165 - 0.1106 - 0.018

mO = 0.0364g

And its number of moles:

nO = 0.0364/16 = 2.275x10⁻³ mol

To have the empirical formula, the coefficients must be the smallest, so let's divide the number of moles for the small one: 2.275x10⁻³

nC = (9.16x10⁻³) / (2.275x10⁻³) = 4 mol

nH = 0.018 / (2.275x10⁻³) = 8 mol

nO = (2.275x10⁻³) / (2.275x10⁻³) = 1 mol

The empirical formula is C₄H₈O.

The molecular formula must be a multiple of the empirical, so it will be n (C₄H₈O). Knowing the molar mass, we can calculate n:

4xnx12 + 8xnx1 + 1xnx16 = 144

48n + 8n + 16n = 144

72n = 144

n = 2

The molecular formula is 2 (C₄H₈O) = C₈H₁₆O₂.