Problem 39 - Chemistry and Chemical Reactivity

Question

Cyclopentane is a simple hydrocarbon. If 0.0956 g of the compound is burned in oxygen, \(0.300 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.123 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are isolated. (a) What is the empirical formula of cyclopentane? (b) If a separate experiment gave \(70.1 \mathrm{g} / \mathrm{mol}\) as the molar mass of the compound, what is its molecular formula?

Step-by-Step Solution

Answer

The empirical formula and the molecular formula of the compound are both C5H10.

Step 1: Determine the Moles of CO2 Formed

First, calculate the moles of carbon dioxide (\( CO_2 \)) produced using its molar mass \( 44.01\, \text{g/mol} \): \[ \text{moles of } CO_2 = \frac{0.300\, \text{g}}{44.01\, \text{g/mol}} \approx 0.00682\, \text{mol} \] Since each \(CO_2\) has one carbon atom, this is also the moles of carbon.

Step 2: Determine the Moles of H2O Formed

Calculate the moles of water (\( H_2O \)) produced using its molar mass \( 18.02\, \text{g/mol} \): \[ \text{moles of } H_2O = \frac{0.123\, \text{g}}{18.02\, \text{g/mol}} \approx 0.00683\, \text{mol} \] Each mole of \(H_2O\) has two moles of hydrogen atoms, so \[ \text{moles of H} = 2 \times 0.00683\, \text{mol} \approx 0.01366\, \text{mol} \].

Step 3: Calculate the Mass of Carbon and Hydrogen

Using the moles, find the mass of carbon and hydrogen: - Mass of carbon using \( 12.01\, \text{g/mol} \): \[ 0.00682\, \text{mol} \times 12.01\, \text{g/mol} \approx 0.0819\, \text{g} \] - Mass of hydrogen using \( 1.008\, \text{g/mol} \): \[ 0.01366\, \text{mol} \times 1.008\, \text{g/mol} \approx 0.0138\, \text{g} \]

Step 4: Determine the Mass of Oxygen in the Compound

Subtract the mass of carbon and hydrogen from the total mass of the compound to find the mass of oxygen: \[ 0.0956\, \text{g} - (0.0819\, \text{g} + 0.0138\, \text{g}) \approx 0.0000\, \text{g} \] This shows there is negligible oxygen in the compound, typical of hydrocarbons.

Step 5: Determine the Empirical Formula

Assume the empirical formula is \( C_xH_yO_z \). Using the mole ratio between carbon and hydrogen: \[ \text{Ratio of } C:H = \frac{0.00682}{0.01366/2} = 1:1 \approx 1:2 \] This suggests \( C:H = 5:10 \) or \( C:H = \text{5:10} \) becomes \( C_5H_{10} \).

Step 6: Calculate the Empirical Formula Mass

Calculate the mass of the empirical formula \( C_5H_{10} \) to confirm it matches the molecular formula. The empirical mass is: \[ 5(12.01) + 10(1.008) = 70.13\, \text{g/mol} \] which confirms it as the molecular formula.

Key Concepts

Cyclopentane

Cyclopentane is a member of the cycloalkanes, a class of hydrocarbons characterized by their ring-like structure. The name 'cyclopentane' indicates that it consists of a ring structure with five carbon atoms. Each carbon atom is bonded to two other carbons and two hydrogens, creating a stable, closed-loop formation. This structure provides cyclopentane with unique chemical properties, often used in industrial and laboratory settings.

• Cyclopentane is a colorless liquid at room temperature, known for its distinct gasoline-like odor.
• It is used in various applications, including as a solvent and in the production of other chemicals.
• The ring structure of cyclopentane results in a particular set of chemical behaviors, different from those of linear hydrocarbons, such as n-pentane.

Understanding cyclopentane is essential in organic chemistry, especially when exploring the properties and reactions of cyclic compounds.

Molecular Formula Determination

Molecular formula determination is a fundamental process in chemistry to find out the actual number of each type of atom present in a molecule. This differs from the empirical formula, which gives the simplest whole-number ratio of the atoms. Determining the molecular formula involves several key steps, particularly when starting with combustion analysis.

• The empirical formula provides a basic ratio that can serve as a stepping stone to identify the molecular formula.
• From the empirical formula, the molecular formula is determined by comparing the calculated empirical formula mass with the given molar mass from experimental data.
• Use the relationship: if the empirical formula mass multiplied by a whole number (n) equals the molecular mass, then the empirical formula multiplied by n provides the molecular formula.

In our example, cyclopentane’s empirical formula is found as \( C_5H_{10} \), which is the same as the molecular formula because its empirical formula mass matches the experimentally determined molar mass.

Combustion Analysis

Combustion analysis is a technique used to determine the composition of hydrocarbons by burning the compound and analyzing the combustion products, typically carbon dioxide and water. This method is especially useful for finding the empirical formula of organic compounds like cyclopentane.

• During combustion, the carbon in cyclopentane forms carbon dioxide, while hydrogen turns into water.
• The mass of these products helps calculate the amount of carbon and hydrogen in the original compound.
• For instance, from the mass of carbon dioxide produced, the number of moles of carbon atoms is calculated, offering insight into cyclopentane's composition.

It is key to note that combustion analysis assumes no other elements like oxygen are present in the original hydrocarbon, consistent with the results here where the compound was determined not to contain oxygen.

Hydrocarbons

Hydrocarbons are organic compounds consisting solely of hydrogen and carbon atoms. They form the basis of many organic substances and are primarily found in fuels like natural gas, oil, and coal. Cyclopentane is a classic example of a hydrocarbon, part of the cycloalkane family.

• Hydrocarbons are classified into different types based on their structure. They include alkanes, alkenes, and alkynes, with cyclopentane belonging to the alkane (saturated) group.
• These compounds are key in energy production due to their ability to release significant energy when combusted.
• Their structure and properties vary greatly, influencing their reactivity and function in various chemical applications.

Understanding hydrocarbons like cyclopentane offers insights into a wide array of chemical processes, essential for fields ranging from energy to synthetic materials.