Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted in a combustion apparatus. The gain in mass of the H2O absorber is 0.166 g, whereas that of the CO2 absorber is 0.403 g.What empirical formula for valproic acid do these results indicate?

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09-09-2019
Chemistry

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Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted in a combustion apparatus. The gain in mass of the H2O absorber is 0.166 g, whereas that of the CO2 absorber is 0.403 g.What empirical formula for valproic acid do these results indicate?

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Аноним Аноним · Answer 1 · 2019-09-09T17:18:43+02:00

Answer:

C4H8O1.

Explanation:

The amount of carbon in the sample comes from the 0.403 g of CO2.

This = (12/ 44) * 0.403 = 0.1099 g C.

The amount of hydrogen is (2.016/ (16+2.016) * 0.166 = 0.01858 g H.

Thus the mass of oxygen in the valproic acid = 0.165 - 0.1099 - 0.011858

= 0.03652 g O.

Dividing by the relative atomic masses of the 3 elements the ratio of C : H : O is

0.00916 : 0.01843 : 0.002285

= 4 : 8 : 1.

So the empirical formula is C4H8O1.

Eduard22sly Eduard22sly · Answer 2 · 2021-10-14T14:39:01+02:00

The empirical formula for the valproic acid that the result indicate is C₄H₈O

We'll begin by calculating the mass of Hydrogen (H), Carbon (C) and Oxygen (O) in the sample. This can be obtained as follow:

For Hydrogen, H:

Mass of H₂O = 0.166 g

Molar mass of H₂O = 18 g/mol

Molar mass of H₂ = 2 g/mol

Mass of H =?

Mass of H = 2/18 × 0.166

Mass of H = 0.0184 g

For carbon, C:

Mass of CO₂ = 0.403 g

Molar mass of CO₂ = 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = 12/44 × 0.403

Mass of C = 0.1099 g

For oxygen, O

Mass of H = 0.0184 g

Mass of C = 0.1099 g

Mass of compound = 0.165 g

Mass of O =?

Mass of O = mass of compound – (mass of H + mass of C)

Mass of O = 0.165 – (0.0184 + 0.1099)

Mass of O = 0.165 – 0.1283

Mass of O = 0.0367 g

Finally, we shall determine the empirical formula of the compound.

Mass of C = 0.1099 g

Mass of H = 0.0184 g

Mass of O = 0.0367 g

Empirical formula =?

Divide by their molar mass

C = 0.1099 / 12 = 0.0092

H = 0.0184 / 1 = 0.0184

O = 0.0367 / 16 = 0.0023

Divide by the smallest

C = 0.0092 / 0.0023 = 4

H = 0.0184 / 0.0023 = 8

O = 0.0023 / 0.0023 = 1

Thus, the empirical formula of the valproic acid is C₄H₈O

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