Problem 57 - Chemistry The Central Science

Question

Valproic acid, used to treat seizures and bipolar disorder, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} . \mathrm{A}\) . \(165-\mathrm{g}\) 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

The empirical formula of valproic acid is \(\color{blue}{C_1\:H_2\:O_{1122}}\). However, this is not a reasonable empirical formula. There seems to be a mistake in the calculation method for moles of Oxygen which led to the large ratio for oxygen. Please consult alternative sources and methods for finding the correct empirical and molecular formulas.

Step 1: Find moles of each element produced

We're given the mass of valproic acid, including masses of water and carbon dioxide produced after combustion. Let's first convert these masses into moles of each element produced. The molecular weight of carbon dioxide (CO2) is \(44.01 \frac{g}{mol}\), and the molecular weight of water (H2O) is \(18.02 \frac{g}{mol}\). Moles of carbon dioxide: \[\frac{0.403\: g}{44.01\: g/mol} = 0.00916\: mol\] Moles of water: \[\frac{0.166\: g}{18.02\: g/mol} = 0.00921\: mol\]

Step 2: Find moles of Carbon, Hydrogen, and Oxygen

Now that we have the moles of carbon dioxide and water, we can find the moles of each individual element in the combusted sample. Moles of Carbon (C) from CO2: \(0.00916\: mol × 1\: C\: in\: CO2 = 0.00916\: mol\: C\) Moles of Hydrogen (H) from H2O: \(0.00921\: mol × 2\: H\: in\: H2O = 0.01842\: mol\: H\) Now we need to find moles of Oxygen (O) in the sample. We know that the sample contains \(C, H,\) and \(O\), so the mass of the valproic acid sample is equal to the mass of these elements combined. We're given the sample mass: \(165\: g\). Mass of Carbon: \(0.00916\: mol\: C × 12.01\: g/mol\: C = 0.110\: g\: C\) Mass of Hydrogen: \(0.01842\: mol\: H × 1.008\: g/mol\: H = 0.0186\: g\: H\) We can now find the mass of Oxygen. Mass of Oxygen: \(165\: g - (0.110\: g + 0.0186\: g) = 164.87\: g\) Moles of Oxygen: \[\frac{164.87\: g}{16.00\: g/mol} = 10.304\: mol\]

Step 3: Determine the empirical formula

We now have the moles of each element in the sample. To find the empirical formula, we'll divide each of the moles by the lowest number of moles (which is 0.00916) and round to the nearest whole number. Ratio of Carbon: \(\frac{0.00916\: mol\: C}{0.00916\: mol} = 1\: C\) Ratio of Hydrogen: \(\frac{0.01842\: mol\: H}{0.00916\: mol} \approx 2\: H\) Ratio of Oxygen: \(\frac{10.304\: mol\: O}{0.00916\: mol} \approx 1126\: O\) However, 1126 is not the correct ratio for an empirical formula. We made an error in our calculation for moles of Oxygen, so let's quickly correct that. Moles of Oxygen: \[\frac{165\: g- (0.403\: g + 0.166\: g)}{16.00\: g/mol}=10.331\: mol\] Now divide by the lowest number of moles again (which is still 0.00916) and round to the nearest whole number. Ratio of Oxygen: \(\frac{10.331\: mol\: O}{0.00916\: mol} \approx 1128\: O\) The ratio is still too large for an empirical formula, which indicates an error. In Step 1, we should have used the moles of Carbon and Hydrogen to determine the moles of Oxygen instead of using the mass of valproic acid. Apologies for the confusion, and let's fix that now. Moles of Oxygen: \(10.304\: mol - (0.00916\: mol\: C + 0.01842\: mol\: H) = 10.276\: mol\: O\) Now divide by the lowest number of moles (which is 0.00916) and round to the nearest whole number. Ratio of Oxygen: \(\frac{10.276\: mol\: O}{0.00916\: mol} \approx 1122\: O\) This is still a very large ratio for an empirical formula, which indicates the solution method contains errors. Please disregard this solution, and consult alternative sources for a correct method.

Key Concepts

Combustion Analysis

Combustion analysis is an invaluable technique in chemistry that allows you to decipher the elemental composition of a substance by burning it and analyzing the resulting products. This method is particularly useful when dealing with organic compounds containing elements like carbon and hydrogen. During combustion, these elements react with oxygen to form carbon dioxide and water, which can be measured to determine the original amounts of carbon and hydrogen in the sample.

In the case of valproic acid, a known mass was combusted, producing specific amounts of water and carbon dioxide. To correct an error in the analysis, a precise procedure should involve determining the moles of carbon from carbon dioxide and the moles of hydrogen from water, taking care to account for all atoms present in the molecules. The moles of oxygen can then be derived by considering the remaining mass of the sample after carbon and hydrogen have been accounted for.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In this context, stoichiometry helps us understand the proportions of each element in the empirical formula of a substance. To apply stoichiometry to combustion analysis, we use the mole concept, which is a fundamental unit in chemistry representing a specific number of particles.

It's important to identify and correct stoichiometric errors in calculations, such as mistakes in determining the moles of oxygen, which can arise from incorrect subtraction of the masses of carbon and hydrogen from the original sample. Understanding stoichiometry can prevent these errors and ensure accurate empirical formulas are derived.

Molecular Weight

The molecular weight (or molecular mass) of a compound is the sum of the atomic weights of all atoms in its molecules, expressed in atomic mass units (amu) or grams per mole (g/mol). This measurement is crucial when determining the number of moles of each substance involved in a chemical reaction.

In the exercise, molecular weights of carbon dioxide and water were used to convert the mass of these compounds into moles, which is the first step of combustion analysis. It's essential to use accurate molecular weights to ensure the subsequent calculation of moles is correct. Molecular weight is also the key to transitioning from an empirical formula to a molecular formula, as seen when the molar mass of valproic acid was provided to deduce its molecular formula.

Chemical Composition

Chemical composition refers to the identity and quantity of the elements within a compound. The empirical formula represents the simplest whole-number ratio of these elements, while the molecular formula represents the actual numbers of atoms in a molecule. Missteps in calculations can significantly affect the determination of chemical composition.

In resolving errors within the exercise related to valproic acid, it's imperative to accurately calculate the moles of each element present. The empirical formula is deduced by finding the simplest integer ratio of the moles of elements, with the molecular formula providing further insight into the specific number of atoms in the compound. To ensure accuracy, it's critical to verify each calculation step, particularly when dealing with the mass and moles of oxygen, as incorrectly deduced amounts can lead to the wrong chemical composition.