Problem 98 - Chemistry The Central Science

Question

An unknown substance is found to contain only carbon and hydrogen. It is a liquid that boils at \(49^{\circ} \mathrm{C}\) at 1 atm pressure. Upon analysis it is found to contain 85.7\(\%\) carbon and 14.3\(\%\) hydrogen by mass. At \(100^{\circ} \mathrm{C}\) and 735 torr, the vapor of this unknown has a density of 2.21 \(\mathrm{g} / \mathrm{L} .\) When it is dissolved in hexane solution and bromine water is added, no reaction occurs. What is the identity of the unknown compound?

Step-by-Step Solution

Answer

The identity of the unknown compound is cyclopentane (C5H10), which is determined by following these steps: 1) Calculate the empirical formula (CH2) using the given mass percentages, 2) Determine the molar mass (70.16 g/mol) using the density and Ideal Gas Law, 3) Calculate the molecular formula (C5H10) using the molar mass and empirical formula, and 4) Deduce the identity based on the molecular formula, boiling point, and reaction with bromine water.

Step 1: Determine the empirical formula from mass percentages

First, we need to find the empirical formula of the unknown compound using the given mass percentages of carbon and hydrogen. We will assume 100g of the substance, which gives us 85.7g of carbon and 14.3g of hydrogen. Then, convert the mass to moles by dividing by their respective atomic masses: Carbon: \(\frac{85.7\,\text{g}}{12.01\,\text{g/mol}} = 7.14\,\text{mol}\) Hydrogen: \(\frac{14.3\,\text{g}}{1.008\,\text{g/mol}} = 14.2\,\text{mol}\) Now, we will find the mole ratio by dividing each number of moles by the smallest mole value among them: Carbon: \(\frac{7.14}{7.14}=1\) Hydrogen: \(\frac{14.2}{7.14}=1.99 \approx 2\) Hence, the empirical formula of the unknown compound is CH2.

Step 2: Calculate the molar mass of the compound using its density

The density of the vapor of the unknown compound is given at 100°C and 735 torr: 2.21 g/L. We can determine the molar mass (MM) of the compound using the Ideal Gas Law: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. First, we need to convert the temperature and pressure into Kelvin and atm: Temperature: \(100^{\circ}\text{C} + 273.15 = 373.15\,\text{K}\) Pressure: \(\frac{735\,\text{torr}}{760\,\text{torr/atm}} = 0.967\,\text{atm}\) Now we can substitute the values and solve for n: \(0.967\,\text{atm} \times 1\,\text{L} = n \times 0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} \times 373.15\,\text{K}\) \(n = \frac{0.967 \times 1}{0.0821 \times 373.15} = 0.0315\,\text{mol}\) Since we have the vapor density and number of moles, we can now calculate the molar mass: Molar Mass = \(\frac{2.21\,\text{g}}{0.0315\,\text{mol}} = 70.16\,\text{g/mol}\)

Step 3: Calculate the molecular formula

Using the molar mass of the compound (70.16 g/mol) and the empirical formula (CH2), we can determine the molecular formula: Empirical formula mass of CH2: 12.01 g/mol (C) + 2 x 1.008 g/mol (H) = 14.03 g/mol Divide molar mass by empirical formula mass to find the number of empirical units per molecule: \(\frac{70.16\,\text{g/mol}}{14.03\,\text{g/mol}} = 5\) Since there are 5 empirical formula units per molecule, the molecular formula of the unknown compound is: C5H10

Step 4: Deduce the identity of the unknown compound

Since the compound does not react with bromine water, it does not have any carbon-carbon double bonds or triple bonds, meaning it must be an alkane. The molecular formula of the unknown compound is C5H10, which does not correspond to an alkane. Therefore, it must be a cyclic hydrocarbon. Thus, the identity of the unknown compound is cyclopentane (C5H10) as it perfectly matches the molecular formula, boiling point, and reaction information provided in the exercise.

Key Concepts

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry that helps determine the weight of one mole of a substance. It is vital in identifying unknown compounds, as in this exercise, where we found the molar mass using the Ideal Gas Law.
To calculate the molar mass, divide the mass of the substance by the number of moles:

• For example, the unknown compound's vapor density provides a mass of 2.21 g per liter.
• Using the Ideal Gas Law, we calculated the moles as 0.0315 mol.
• Thus, the molar mass is: \[\text{Molar Mass} = \frac{2.21\, \text{g}}{0.0315\, \text{mol}} = 70.16\, \text{g/mol}.\]

This molar mass value assists in further identifying the compound's molecular formula by comparing it to the empirical formula calculated earlier.

Ideal Gas Law

The Ideal Gas Law is a powerful tool that connects the physical properties of gases through an equation. It is expressed as \( PV = nRT \), where each variable stands for:

• \( P \): Pressure in atm,
• \( V \): Volume in L,
• \( n \): Number of moles,
• \( R \): Universal gas constant \( 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \), and
• \( T \): Temperature in K.

In this problem, the Ideal Gas Law helps us find the number of moles from known density, temperature, and pressure:

• The temperature is converted from Celsius to Kelvin: \( 100^\circ \text{C} + 273.15 = 373.15 \text{K} \).
• The pressure is converted from torr to atm: \( \frac{735\, \text{torr}}{760\, \text{torr/atm}} = 0.967\, \text{atm} \).
• By substituting the values, we solve for \( n \): \[0.967\, \text{atm} \times 1\, \text{L} = n \times 0.0821\, \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} \times 373.15\, \text{K}.\]
• The result, \( n = 0.0315 \text{mol} \), enables us to calculate the molar mass, essential for identifying the molecular formula.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a compound. It stems from the empirical formula but provides a full picture using the compound's molar mass. In our exercise:

• The empirical formula, derived from mass percentages, is CH₂.
• The molar mass of the empirical formula: \[ \text{C} = 12.01\, \text{g/mol}, \; \text{H}_2 = 2 \times 1.008 = 2.016\, \text{g/mol}; \quad \text{summed: } 14.03\, \text{g/mol}. \]
• Divide the compound's molar mass (70.16 g/mol) by this empirical mass to find how many empirical units fit: \[ \frac{70.16\, \text{g/mol}}{14.03\, \text{g/mol}} = 5. \]
• The molecular formula is thus five times the empirical formula: C₅H₁₀.

Knowing this provides insights into the compound's properties and structure.

Chemical Analysis

Chemical analysis involves examining substances to determine their composition and molecular structure. In this situation, it identifies the unknown compound.

• We start with elemental analysis, identifying the mass percentages of elements in the substance, which are 85.7% for carbon and 14.3% for hydrogen.
• These percentages led to the empirical formula, which serves as a foundational piece.
• Further analysis through the Ideal Gas Law provided insight into the molar mass, aligning with the molecular formula determination.
• Lastly, chemical behavior, such as the non-reactivity with bromine water, concluded the compound type. Since there is no reaction, it lacks double/triple bonds, indicating it is a saturated hydrocarbon, specifically a cyclic one, known as cyclopentane (C₅H₁₀).

This systematic approach, integrating different pieces of information, successfully reveals the identity of the unknown compound.