Problem 97 - Chemistry The Central Science In Si Units

Question

An organic compound is analyzed and found to contain \(62.0 \%\) carbon, \(10.4 \%\) hydrogen, and \(27.6 \%\) oxygen by mass. The compound boils at \(48.8^{\circ} \mathrm{C}\). At \(25^{\circ} \mathrm{C}\) and 76.5 \(\mathrm{kPa}\), the vapor has a density of \(1.8 \mathrm{~g} / \mathrm{L}\). The compound has a carbonyl group and can be oxidized to a carboxylic acid. Suggest a structure for the compound.

Step-by-Step Solution

Answer

The compound is likely propanal, \( \text{C}_3\text{H}_6\text{O} \).

Step 1: Determine the Empirical Formula

Given percentages: Carbon (C) - 62.0%, Hydrogen (H) - 10.4%, Oxygen (O) - 27.6%. Assume 100 g of the compound, so you have 62 g of C, 10.4 g of H, and 27.6 g of O. Use the atomic masses (C: 12.01 g/mol, H: 1.01 g/mol, O: 16.00 g/mol) to convert masses to moles: \ \( \text{moles of C} = \frac{62}{12.01} \approx 5.16 \), \( \text{moles of H} = \frac{10.4}{1.01} \approx 10.3 \), \( \text{moles of O} = \frac{27.6}{16} \approx 1.73 \). Divide by the smallest number of moles (1.73): \ \( \frac{5.16}{1.73} \approx 2.98 \approx 3 \), \( \frac{10.3}{1.73} \approx 5.95 \approx 6 \), \( \frac{1.73}{1.73} = 1 \). The empirical formula is \( \text{C}_3\text{H}_6\text{O} \).

Step 2: Calculate the Molecular Formula

Find the molar mass from the vapor density data. The molar mass is determined using the formula: \( \text{Molar mass} = \frac{\text{Density} \times \text{RT}}{\text{Pressure}} \). Using 1.8 g/L for density, 76.5 kPa for pressure, and assuming the ideal gas constant \( R = 8.314 \text{ J/mol K} \) with temperature 298 K: \ \( \text{Molar mass} = \frac{1.8 \times 8.314 \times 298}{76.5} \approx 58 \text{ g/mol} \). \ The empirical formula mass of \( \text{C}_3\text{H}_6\text{O} \) is approximately 58 g/mol, suggesting that the empirical formula is the molecular formula.

Step 3: Identify Functional Groups and Structure

The compound contains a carbonyl group and can be oxidized to a carboxylic acid, indicating the presence of an aldehyde (formyl group) which oxidizes to a carboxylic acid. Given the empirical formula \( \text{C}_3\text{H}_6\text{O} \), and common structure for a three-carbon aldehyde: it is likely propanal and its structure is \( \text{CH}_3\text{CH}_2\text{CHO} \), an aldehyde.

Key Concepts

Empirical Formula

When analyzing a chemical compound, determining the empirical formula is a critical first step. The empirical formula provides the simplest whole-number ratio of elements present in a compound. In this context, we're given the percentages of carbon, hydrogen, and oxygen in an organic compound. To convert these percentages to an empirical formula:

• Assume you have 100 grams of the compound. This makes the percentage equivalent to grams.
• Convert the grams of each element to moles using their atomic masses.
• Divide each element's moles by the smallest number of moles to find the simplest ratio.

In our example, this results in the empirical formula of \( \text{C}_3\text{H}_6\text{O} \), meaning there are three carbon atoms, six hydrogen atoms, and one oxygen atom in the simplest ratio.

Vapor Density

Vapor density is key in finding the molar mass of a gaseous compound. It measures how dense a vapor is in comparison to air. By using the ideal gas law (\( PV = nRT \)), you can relate vapor density to molar mass:

• Vapor density is expressed in g/L, a measurement of mass per unit volume.
• The known vapor density, along with the ideal gas constant and specific conditions for temperature and pressure, allows calculation of the compound's molar mass.

For example, in the above exercise, the vapor density given was 1.8 g/L. This value enabled the calculation of a molar mass of 58 g/mol, suggesting that the empirical formula \( \text{C}_3\text{H}_6\text{O} \) matches the molecular formula.

Functional Groups

Functional groups are specific groups of atoms within molecules that determine the characteristic chemical reactions of those molecules. They play a crucial role in determining the properties and reactions of organic compounds.

• They include groups such as hydroxyls, carbonyls, amines, and carboxyls.
• These groups are the "active" regions of molecules and are used to classify many organic molecules.

In the given compound, the presence of a carbonyl group, known for its distinct reactions such as oxidation, is a functional group that hints at the compound being an aldehyde or a ketone. Functional groups allow chemists to predict and explain the behavior of chemical species.

Molecular Structure

Molecular structure refers to the arrangement of atoms within a molecule. Understanding it helps in predicting the chemical behavior and reactivity of the molecule.

• The structure of a molecule is more complex than its empirical formula, which is a simple ratio of elements.
• It determines the shape and size of the molecule, electron distribution, and potential points of chemical reactivity.

In this case, we deduced that the molecular structure of the compound could be represented as propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)), based on its empirical formula and functional groups. This structure gives insight into its chemical properties and reactivity pattern.

Aldehyde

Aldehydes are a type of organic compound characterized by the presence of a carbonyl group (\( C=O \)) bonded to at least one hydrogen atom. They are intermediates in many important biological synthesis processes.

• This group is represented by the suffix 'al'.
• Aldehydes can be oxidized to form carboxylic acids, giving them significant importance in organic synthesis.

In our problem, the compound can be oxidized to a carboxylic acid, indicating that it is likely an aldehyde. Given its empirical formula and reactivity, propanal (a three-carbon aldehyde) is a suitable structure.

Carboxylic Acid

Carboxylic acids are organic acids characterized by the presence of a carboxyl group (\( -COOH \)). This makes them highly reactive and widely used as precursors to various organic compounds.

• The carboxyl group imparts acidic properties to the molecule, allowing it to participate in a wide range of reactions.
• These compounds are typically named with the suffix 'oic acid'.

In the problem, once the aldehyde is oxidized, it forms a carboxylic acid. This transformation is a common chemical reaction pathway, where for instance, propanal could be oxidized to propanoic acid. Understanding this reaction pathway is essential for grasping organic chemical transformations.