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.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

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