Problem 57 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

Empirical formula: \( \text{C}_4\text{H}_8\text{O} \). Molecular formula: \( \text{C}_8\text{H}_{16}\text{O}_2 \).

Step 1: Determine Moles of Carbon from CO₂

The mass of CO₂ produced gives us the amount of carbon. Use the molar mass of CO₂ (44.01 g/mol) to find moles of CO₂: \[ \text{moles of CO}_2 = \frac{0.403\, \text{g}}{44.01\, \text{g/mol}} = 0.00916\, \text{mol} \]Each mole of CO₂ contains one mole of carbon, so: \[ \text{moles of C} = 0.00916\, \text{mol} \]

Step 2: Determine Moles of Hydrogen from H₂O

The mass of water gives us the amount of hydrogen. Use the molar mass of H₂O (18.02 g/mol) to find moles of water:\[ \text{moles of H}_2\text{O} = \frac{0.166\, \text{g}}{18.02\, \text{g/mol}} = 0.00921\, \text{mol} \]Each mole of H₂O contains two moles of hydrogen atoms, so:\[ \text{moles of H} = 2 \times 0.00921\, \text{mol} = 0.01842\, \text{mol} \]

Step 3: Determine Mass of Oxygen in the Compound

The total mass of the valproic acid sample is 0.165 g. You already know the mass of carbon and hydrogen from the CO₂ and H₂O:- Mass of C: \(0.00916\, \text{mol} \times 12.01\, \text{g/mol} = 0.110\, \text{g} \)- Mass of H: \(0.01842\, \text{mol} \times 1.008\, \text{g/mol} = 0.01857\, \text{g} \)Now, subtract the combined mass of C and H from the total mass to find the mass of oxygen:\[ \text{mass of O} = 0.165\, \text{g} - (0.110\, \text{g} + 0.01857\, \text{g}) = 0.03643\, \text{g} \]

Step 4: Determine Moles of Oxygen

Using the molar mass of oxygen (16.00 g/mol), calculate the moles of oxygen:\[ \text{moles of O} = \frac{0.03643\, \text{g}}{16.00\, \text{g/mol}} = 0.00228\, \text{mol} \]

Step 5: Determine Empirical Formula

Use the moles calculated for C, H, and O to determine the ratios:- C: 0.00916 mol- H: 0.01842 mol- O: 0.00228 molDivide each number of moles by the smallest value to get the simplest ratio:\[ \begin{align*}C &: \frac{0.00916}{0.00228} = 4.02 \ H &: \frac{0.01842}{0.00228} = 8.08 \ O &: \frac{0.00228}{0.00228} = 1.00 \end{align*} \]Thus, the empirical formula is \( \text{C}_4\text{H}_8\text{O} \).

Step 6: Determine Molecular Formula

The empirical formula weight of \(\text{C}_4\text{H}_8\text{O}\) is:\[ (4 \times 12.01 + 8 \times 1.008 + 16.00)\, \text{g/mol} = 72.12\, \text{g/mol} \] The molar mass of valproic acid is given as 144 g/mol. Find the multiple:\[ n = \frac{144}{72.12} \approx 2 \]Therefore, the molecular formula is \( \text{C}_8\text{H}_{16}\text{O}_2 \).

Key Concepts

Molecular Formula

Understanding molecular formulas is key to identifying the actual number of atoms of each element present in a compound. While the empirical formula gives the simplest whole-number ratio of the elements in a compound, the molecular formula reveals the precise number of each type of atom in a molecule. This is crucial for distinguishing between compounds that may have the same empirical formula but differ in their structure and properties.
For instance, if the empirical formula of a compound is \( \text{C}_4\text{H}_8\text{O} \) and its molar mass is 144 g/mol. By calculating the empirical formula mass, which is approximately 72 g/mol, and dividing the molar mass by the empirical formula mass, we find that \( n = 2 \). This number indicates that the molecular formula is \( \text{C}_8\text{H}_{16}\text{O}_2 \), showing that this compound contains twice the number of atoms compared to its empirical formula.

Combustion Analysis

Combustion analysis is a widely used method in chemistry to determine the mass percentages of carbon and hydrogen in a compound. It involves burning the compound in excess oxygen, resulting in the formation of carbon dioxide (CO₂) and water (H₂O).
This allows us to trace back to the original amounts of carbon and hydrogen in the compound. In our given problem, we obtained \( 0.403 \, \text{g} \) of CO₂ and \( 0.166 \, \text{g} \) of H₂O. From these, we calculated the moles of carbon and hydrogen because each molecule of CO₂ contains one carbon atom and each molecule of H₂O contains two hydrogen atoms.
This is a critical procedure in organic chemistry as it gives us insights into a compound's composition based solely on its combustion products.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationship between reactants and products in a chemical reaction.
Through stoichiometry, we interpreted the amounts of CO₂ and H₂O generated from combustion to determine the moles of carbon and hydrogen in the original compound.
The stoichiometry of combustion ensures that all measurements and derived calculations are consistent with the law of conservation of mass, allowing us to accurately compute the empirical formula and further deduce the molecular formula.

• We started by determining moles of CO₂, which directly told us about the moles of C.
• We then considered moles of water, allowing us to identify moles of H.
• Finally, by accounting for the total mass and remaining elements, the moles of oxygen were determined.

This process exemplifies stoichiometric calculations in chemical analysis.

Chemical Composition

The chemical composition of a substance refers to the arrangement, type, and ratio of atoms in molecules. It is crucial in understanding both the properties and the behavior of a substance during reactions.
In the given exercise, chemical composition is highlighted through the decision of the empirical formula \(\text{C}_4\text{H}_8\text{O}\) which represents the ratio of carbon, hydrogen, and oxygen in valproic acid.
These analyses are invaluable in organic chemistry, where recognizing the makeup of complex molecules aids in predicting reaction outcomes. Chemical compositions guide precise formulation of compounds for intended applications such as pharmaceuticals.

Organic Chemistry

Organic chemistry is the study of carbon-containing compounds and their properties. Vacating exclusively in carbon, hydrogen, and oxygen, organic compounds are diverse in structure and abundant in applications.
Valproic acid, for example, is a well-known compound in organic chemistry due mainly to its medicinal use in treating bipolar disorder and seizures. Understanding its structure, from empirical to molecular formula, unveils facets vital for its synthesis and application in medicine.
Organic chemistry encourages the determination of chemical structures that have real-world benefits, ensuring innovations transform into effective solutions.

Moles Calculation

The concept of moles is essential in chemical quantification, allowing chemists to translate grams into an understandable quantity of molecules or atoms.
In our exercise, calculating moles from grams was fundamental. For CO₂, \( \text{moles of CO}_2 = \frac{0.403\, \text{g}}{44.01\, \text{g/mol}} = 0.00916\, \text{mol} \). Similarly, for H₂O, \( \text{moles of H}_2\text{O} = \frac{0.166\, \text{g}}{18.02\, \text{g/mol}} = 0.00921\, \text{mol} \). These calculations allowed us to derive the moles of carbon and hydrogen present in the original compound.
This transformation is vital in carrying out balanced reactions and achieving accurate compositions in chemical processes. Understanding moles and applying them effectively is central to successful chemistry problem-solving.