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Chapter 3: Problem 57

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?

Short Answer

Expert verified
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 by step solution

01

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: \[\frac{0.166\: g}{18.02\: g/mol} = 0.00921\: mol\]
02

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: \[\frac{164.87\: g}{16.00\: g/mol} = 10.304\: mol\]
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) What is the mass, in grams, of \(2.50 \times 10^{-3}\) mol of ammonium phosphate? (b) How many moles of chloride ions are in 0.2550 g of aluminum chloride? (c) What is the mass, in grams, of \(7.70 \times 10^{20}\) molecules of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2} ?\) (d) What is the molar mass of cholesterol if 0.00105 \(\mathrm{mol}\) has a mass of 0.406 \(\mathrm{g}\) ?

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately 300 \(\mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(12 \times 15 \times 8.0 \mathrm{ft}\) . The density of air at \(26^{\circ} \mathrm{C}\) is 0.00118 \(\mathrm{g} / \mathrm{cm}^{3} .\) (b) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon or Acrilan burn. Acrilan has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is 50.9\(\%\) of the formula by mass. A rug measures \(12 \times 15 \mathrm{ft}\) and contains 30 oz of Acrilan fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is 20\(\%\) and that the carpet is 50\(\%\) consumed.

(a) What is the mass, in grams, of one mole of \(^{12} \mathrm{C} ?\) (b) How many carbon atoms are present in one mole of \(^{12} \mathrm{C} ?\)

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when 0.500 mol \(\mathrm{Al}(\mathrm{OH})_{3}\) and 0.500 \(\mathrm{mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

An element \(X\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right) .\) The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If 0.5000 \(\mathrm{g}\) of \(\mathrm{XI}_{3}\) is treated with chlorine, 0.2360 \(\mathrm{g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.
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