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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 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

Expert verified
The empirical formula of valproic acid is C鈧凥鈧圤, and its molecular formula is C鈧圚鈧佲倖O鈧�.

Step by step solution

01

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 H鈧侽 produced = 0.166 g. (Each water molecule contains 2 Hydrogens). So, the mass of Hydrogen = (2 * mass of hydrogen in 1 H鈧侽 molecule) / mass of 1 H鈧侽 molecule * mass of H鈧侽 produced = \( \frac{2 * 1}{18} * 0.166 = 0.0184 g \) ii. The mass of CO鈧� produced = 0.403 g. (Each CO鈧� molecule contains 1 carbon). So, the mass of Carbon = (1 * mass of carbon in 1 CO鈧� molecule) / mass of 1 CO鈧� molecule * mass of CO鈧� 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
02

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 鈮� 0.00916 mol Moles of H = mass of H / molar mass of H = 0.0184 g / 1.008 g/mol 鈮� 0.0183 mol Moles of O = mass of O / molar mass of O = 0.0366 g / 16.00 g/mol 鈮� 0.00229 mol
03

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 C鈧凥鈧圤.
04

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 鈮� 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 鈮� 2 Finally, multiply the empirical formula by 2 to get the molecular formula: Molecular formula = (C鈧凥鈧圤) 脳 2 = C鈧圚鈧佲倖O鈧� Therefore, the molecular formula of valproic acid is C鈧圚鈧佲倖O鈧�.

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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 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 (CO鈧�) and water (H鈧侽).
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 CO鈧� and hydrogen in H鈧侽 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.

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Most popular questions from this chapter

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a 2.558-g sample of washing soda is heated at \(125^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). What is the value of \(x\) ?

(a) What is the mass, in grams, of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate? (b) How many moles of chloride ions are in \(0.2550 \mathrm{~g}\) 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 \(0.406 \mathrm{~g}\) ?

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.
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