Question

Valproic acid, used to treat seizures and bipolar disorder, is composed of \(\mathrm{C}, \mathrm{H}\), and O. A 0.165-g sample is combusted in an apparatus such as that shown in Figure 3.14. The gain in mass of the \(\mathrm{H}_{2} \mathrm{O}\) absorber is \(0.166 \mathrm{~g}\), whereas that of the \(\mathrm{CO}_{2}\) absorber is \(0.403 \mathrm{~g}\). What is the empirical formula for valproic acid? If the molar mass is \(144 \mathrm{~g} / \mathrm{mol}\) what is the molecular formula?

Short answer

The empirical formula of valproic acid is C₄H₈O, and its molecular formula is C₈H₁₆O₂.

Step-by-step solution

Find the mass of each element in the sample

: We need to find the mass of C, H, and O in the 0.165g sample. We can determine the mass of C and H from the mass of the products after combustion. i. The mass of H₂O produced = 0.166 g. (Each water molecule contains 2 Hydrogens). So, the mass of Hydrogen = (2 * mass of hydrogen in 1 H₂O molecule) / mass of 1 H₂O molecule * mass of H₂O produced = \( \frac{2 * 1}{18} * 0.166 = 0.0184 g \) ii. The mass of CO₂ produced = 0.403 g. (Each CO₂ molecule contains 1 carbon). So, the mass of Carbon = (1 * mass of carbon in 1 CO₂ molecule) / mass of 1 CO₂ molecule * mass of CO₂ produced = \( \frac{12}{44} * 0.403 = 0.110 g \) iii. Now we need to find the mass of oxygen. The oxygen must come from the remaining non-carbon or non-hydrogen part of the original sample. Mass of O = Total mass of the sample - (mass of C + mass of H) = 0.165 - (0.110 + 0.0184) = 0.0366 g

Calculate the moles of each element

: Next, we'll find the moles of each element by dividing the mass by their respective molar mass. Moles of C = mass of C / molar mass of C = 0.110 g / 12.01 g/mol ≈ 0.00916 mol Moles of H = mass of H / molar mass of H = 0.0184 g / 1.008 g/mol ≈ 0.0183 mol Moles of O = mass of O / molar mass of O = 0.0366 g / 16.00 g/mol ≈ 0.00229 mol

Determine the empirical formula

: To find the empirical formula, we need to find the ratio of moles for these elements. To do that, it's helpful to divide each value by the smallest one among them, and round them to the nearest whole number. C : H : O = 0.00916 : 0.0183 : 0.00229 Divide each element by the smallest number (0.00229): C : H : O = 4 : 8 : 1 The empirical formula is C₄H₈O.

Determine the molecular formula

: The empirical formula is just a simplified version of the molecular formula, and we can determine the molecular formula by multiplying that empirical formula. First, we need to find the molar mass of the empirical formula: Empirical formula molar mass = 4(12.01) + 8(1.008) + 1(16.00) = 48.04 + 8.064 + 16.00 ≈ 72.104 g/mol Now we will compare this to the given molar mass (144 g/mol) by dividing the given molar mass by the empirical formula molar mass: Molecular formula multiplier = given molar mass / empirical formula molar mass = 144 g/mol / 72.104 g/mol ≈ 2 Finally, multiply the empirical formula by 2 to get the molecular formula: Molecular formula = (C₄H₈O) × 2 = C₈H₁₆O₂ Therefore, the molecular formula of valproic acid is C₈H₁₆O₂.

Key concepts

Combustion Analysis

Combustion analysis is a common method used in chemistry to ascertain the composition of unknown compounds. Through this process, an organic substance is completely burned, and the resulting gaseous products are measured. In the case of valproic acid, the combustion products were carbon dioxide (CO₂) and water (H₂O).
This process helps us determine the quantity of carbon (C) and hydrogen (H) within the compound.

• For carbon, we measure the amount of carbon dioxide produced.
• For hydrogen, the amount of water provides the necessary data.

To derive the mass of elements, the mass gain in absorbers after combustion is measured. The carbon in CO₂ and hydrogen in H₂O are calculated using mole ratios and atomic or molecular weights. The remaining mass of the original sample accounts for oxygen. Once these masses are known, they help compute the empirical formula.

Molar Mass Calculations

Molar mass calculations are the foundation for transforming the empirical formula into the molecular formula. For any compound, its empirical formula provides the simplest whole number ratio of its constituent elements. In our exercise, the empirical formula of valproic acid was determined to be \( \text{C}_4\text{H}_8\text{O} \).
Calculating the empirical formula mass involves:

• Multiplying the number of atoms in the empirical formula by their atomic masses.
• Adding these values together to get the total empirical mass.

Once calculated, we compare this mass to the given molar mass. The ratio of the molar mass to the empirical mass determines how many times the empirical unit fits into the molecular form of the compound. For valproic acid, our multiplier of 2 indicated that the molecular formula is double the empirical formula, leading to \( \text{C}_8\text{H}_{16}\text{O}_2 \).

Valproic Acid

Valproic acid is a well-known medication used to manage seizures and bipolar disorder, amongst other conditions. Its chemical significance can be appreciated further when understanding its empirical and molecular structures.

• Empirical formula: Represents the simplest ratio of elements in a compound. For valproic acid, it is \( \text{C}_4\text{H}_8\text{O} \).
• Molecular formula: Reflects the actual number of atoms of each element in a molecule. For this compound, it is \( \text{C}_8\text{H}_{16}\text{O}_2 \).

Knowing both formulas allows chemists to understand the substance's behavior and reactivity better, influencing everything from therapeutic applications to synthesis methods. This interplay between empirical and molecular formulas explains why such calculations are pivotal for chemical analysis and the development of new medications.