Problem 75 - General Chemistry: Principles and Modern Applications

Question

Thiophene \(\left(\mathrm{fp}=-38.3 ; \mathrm{bp}=84.4^{\circ} \mathrm{C}\right)\) is a sulfur containing hydrocarbon sometimes used as a solvent in place of benzene. Combustion of a \(2.348 \mathrm{g}\) sample of thiophene produces \(4.913 \mathrm{g} \mathrm{CO}_{2}, 1.005 \mathrm{g} \mathrm{H}_{2} \mathrm{O},\) and \(1.788 \mathrm{g} \mathrm{SO}_{2} .\) When a \(0.867 \mathrm{g}\) sample of thiophene is dissolved in \(44.56 \mathrm{g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),\) the freezing point is lowered by \(1.183^{\circ} \mathrm{C} .\) What is the molecular formula of thiophene?

Step-by-Step Solution

Answer

The molecular formula of thiophene is \( C_{4}H_{4}S \).

Step 1: Determine empirical formula from combustion products

Write the combustion reaction: \(C_{n}H_{m}S + aO_{2} \rightarrow bCO_{2} + cH_{2}O + dSO_{2}\). We know that combustion produces water, carbon dioxide, and sulfur dioxide. From stoichiometric coefficients, we can find that \(n=b\), \(m=2c\), and \(1=d\). Now, find the amount (mol) of each product by dividing weight by the molecular weight. You obtain: \(b = 4.913g/44.01g/mol = 0.1116 mol\), \(c = 1.005g/18.02g/mol = 0.0557 mol\), and \(d = 1.788g/64.07g/mol = 0.0279 mol\). Now, calculate \(n, m\), and the empirical formula.

Step 2: Find molar mass from freezing point depression and find molecular formula

According to the formula \(\Delta T = K_f * m\), where \(\Delta T = 1.183^{\circ} C\) is the freezing point depression, \(K_f = 5.12 C*mol/g\) is the molal freezing point depression constant for benzene, and \(m=n_2m_2/n_1\) is the molality of the solution, solve for \(m\), where \(n_2 = mass_2/M_2\), and \(M_2\) is what you're looking for. Given that \(mass_2 = 0.867g\), and \(n_1 = 44.56g/78.12g/mol = 0.570 mol\), \(M_2 = mass_2/(m*n_1)\). Then find the molecular formula knowing that the ratio of molar mass to the mass corresponding to empirical formula equals to the ratio of numbers of atoms.

Step 3: Calculate empirical formula mass

Given the empirical formula obtained in step 1, calculate the empirical formula mass by adding up the atomic masses of the elements. For \( C_{4}H_{4}S \), the empirical formula mass is \(4*12.00g/mol + 4*1.01g/mol + 32.07g/mol = 68.15g/mol\).

Step 4: Calculate molar mass and molecular formula

Using the formula obtained in step 2, calculate \(M_2\) . Then, find the ratio of the \(M_2\) to empirical formula mass, which should be an integer. If it's not equal to one, multiply the empirical formula by this number to get the molecular formula.

Key Concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements within that compound. It's like finding the basic ingredients for a recipe without any excess amounts. For thiophene, a sulfur-containing hydrocarbon, we engage with this concept during its combustion analysis.
The combustion reaction is as follows:

• Thiophene reacts with oxygen to produce carbon dioxide, water, and sulfur dioxide.
• By analyzing the amounts of these products, we can measure how much carbon, hydrogen, and sulfur are in thiophene.

This involves converting the given masses of combustion products to moles by dividing each by its respective molecular weight.
Once the number of moles is determined, we translate these into whole-number ratios to find the empirical formula. In thiophene's case, assuming you initially didn't know its formula, you'd notice a ratio falling around one sulfur per formula unit. The empirical formula helps us see the basic chemical architecture of the molecule.

Combustion Analysis

Combustion analysis is a massively useful technique in determining the elemental composition of a compound. It works by burning the compound and carefully measuring the resulting amounts of various products.
For thiophene:

• The combustion products are carbon dioxide ( CO_2 ), water ( H_2O ), and sulfur dioxide ( SO_2 ).
• Each product can be directly correlated back to one of the original elements in the compound: carbon, hydrogen, and sulfur, respectively.

By calculating the moles of each combustion product, we deduce the amount of its parent element present in the original sample of thiophene. This provides an essential piece of information for building up the empirical formula.
Overall, combustion analysis reveals not just the presence, but the proportions of elements within the compound, allowing in-depth understanding of the substance's makeup.

Freezing Point Depression

A fascinating application of colligative properties is the concept of freezing point depression. This occurs when a solute is added to a solvent, lowering the solvent's freezing point.
In the context of thiophene:

• Thiophene is dissolved in benzene, resulting in a lower freezing point than pure benzene would exhibit.
• This change in freezing point enables the calculation of thiophene's molar mass when using benzene's winter properties as a reference.

The formula \( \Delta T = K_f \times m \) connects the change in freezing point (\Delta T) to the solution's molality (m) and benzene's freezing point depression constant (K_f).
Through this relationship, we find the molar mass of thiophene, which is essential for determining its molecular formula when combined with information from the empirical formula.

Stoichiometry

Stoichiometry is the mathematical analysis of chemical reactions and compounds, allowing us to predict quantities of reactants and products. It is vital for understanding molecular interactions.
In the case of thiophene:

• We use stoichiometry to switch between mass and moles of elements and compounds based on balanced equations.
• It also helps in comparing the empirical and molecular formulas by relating their mass relationships.

This calculation extends to understanding freezing point depression and combustion analysis, which both rely on stoichiometry for accurate results.
Ultimately, stoichiometry bridges the gap between qualitative descriptions and quantitative predictions, providing clarity and precision in chemical analyses such as determining the molecular formula of thiophene.