Problem 151 - Chemistry and Chemical Reactivity

Question

Polystyrene can be prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. A sample prepared by this method has the empirical formula \(\mathrm{Br}_{3} \mathrm{C}_{6} \mathrm{H}_{3}\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)_{n},\) where the value of \(n\) can vary from sample to sample. If one sample has \(0.105 \% \mathrm{Br},\) what is the value of \(n ?\)

Step-by-Step Solution

Answer

The value of \(n\) is approximately 2194.

Step 1: Determine Molecular Weight of Empirical Formula

First, we need to find the molecular weight of the empirical formula \(\mathrm{Br}_3\mathrm{C}_6\mathrm{H}_3(\mathrm{C}_8\mathrm{H}_8)_n\).- The molecular weight of \(\mathrm{Br}_3\mathrm{C}_6\mathrm{H}_3\) is calculated as: - \(3 \times 79.90\) g/mol (for Br) = 239.7 g/mol - \(6 \times 12.01\) g/mol (for C) = 72.06 g/mol - \(3 \times 1.008\) g/mol (for H) = 3.024 g/mol - Total: 314.784 g/mol- The molecular weight of \(\mathrm{C}_8\mathrm{H}_8\) is: - \(8 \times 12.01\) g/mol (for C) = 96.08 g/mol - \(8 \times 1.008\) g/mol (for H) = 8.064 g/mol - Total: 104.144 g/mol.

Step 2: Calculate the Total Molecular Weight

Next, calculate the molecular weight of the complete empirical formula, including the \(n\) units of \(\mathrm{C}_8\mathrm{H}_8\):- Molecular weight of the empirical formula = \(314.784 + 104.144n\).

Step 3: Calculate the Mass Percentage of Bromine

The mass of bromine in the compound is the mass from three bromine atoms (239.7 g/mol).The formula used for the mass percentage of bromine is: \[\% ext{Br} = \left(\frac{239.7}{314.784 + 104.144n}\right) \times 100\%\]Given that the percentage by mass of Br is \(0.105\%\).

Step 4: Set Up the Equation

Set up an equation using the percentage information:\[\frac{239.7}{314.784 + 104.144n} \times 100 = 0.105\]

Step 5: Solve for n

Solve the equation for \(n\):1. Simplify: \[\frac{239.7}{314.784 + 104.144n} = 0.00105\]2. Multiply both sides by \(314.784 + 104.144n\): \[239.7 = 0.00105 \times (314.784 + 104.144n)\]3. Distribute: \[239.7 = 0.3304232 + 0.00105 \times 104.144n\]4. Isolate \(n\): \[239.3695768 = 0.00105 \times 104.144n\]5. Simplify: \[n = \frac{239.3695768}{0.00105 \times 104.144}\]6. Solve for \(n\): \[n \approx 2194.24\]

Step 6: Round to Nearest Whole Number

Since \(n\) represents a number of repeating units, it makes sense to round to the nearest whole number.\(n \approx 2194\).

Key Concepts

Molecular Weight Calculation

The concept of molecular weight is a fundamental aspect of chemistry that refers to the weight of a molecule of a chemical compound. It is usually expressed in atomic mass units (amu) or grams per mole (g/mol). Understanding how to calculate this value is essential for solving many chemistry problems, including determining the composition of complex compounds.

For our empirical formula \(\mathrm{Br}_3\mathrm{C}_6\mathrm{H}_3(\mathrm{C}_8\mathrm{H}_8)_n \), we calculate the molecular weight by summing the weights of all atoms present. The fixed part of this empirical formula \(\mathrm{Br}_3\mathrm{C}_6\mathrm{H}_3 \) consists of three bromine, six carbon, and three hydrogen atoms. Bromine contributes significantly with each Br atom having an atomic mass of approximately 79.9 amu. In total, bromine atoms contribute 239.7 g/mol to the molecular weight.

Carbon and hydrogen are lighter, with six carbon atoms and three hydrogen atoms adding 72.06 g/mol and 3.024 g/mol respectively. When you add these together, you find the total fixed part of the formula weighs 314.784 g/mol. For each unit of \(\mathrm{C}_8\mathrm{H}_8 \) added, you need to calculate it separately, which contributes 104.144 g/mol. Thus, the total molecular weight depends on the value of \( n \) which indicates how many repeating units of \(\mathrm{C}_8\mathrm{H}_8 \) there are.

Mass Percentage

Calculating the mass percentage of an element in a compound is another crucial task in chemistry. It allows us to determine the composition of a chemical compound and is essential in the context of empirical and molecular formulas.

Mass percentage is calculated by dividing the mass of the element of interest by the total molecular weight of the compound, then multiplying by 100 to express it as a percentage. In our example, the mass percentage of bromine is given as 0.105%.

To find this value, you need to know the total molecular weight of the compound and the exact contribution from bromine. With bromine contributing 239.7 g/mol to the compound, you use the formula:\[ \% \text{Br} = \left( \frac{239.7}{314.784 + 104.144n} \right) \times 100 \% \] This relationship provides insight into the proportion of bromine compared to the overall compound. When the percentage of bromine is known, this equation can be rearranged to solve for the variable of interest, which in this problem is the number of repeating units, \( n \).

Polymer Chemistry

Polymer chemistry is a vast field that deals with the synthesis and properties of polymers, which are long, repeating structures formed by monomers. Polystyrene is one such polymer, widely used in products ranging from packaging materials to disposable food containers.

In our example, polystyrene is synthesized from styrene monomers, and its structure is symbolized by the empirical formula \( \mathrm{Br}_3\mathrm{C}_6\mathrm{H}_3(\mathrm{C}_8\mathrm{H}_8)_n \). The value of \( n \) indicates the number of \( \mathrm{C}_8\mathrm{H}_8 \) units in the polymer chain. This variation in \( n \) can affect the polymer's physical properties such as its strength, flexibility, and melting point.

Understanding polymer chemistry not only involves grasping the synthesis process but also requires knowledge of how these variations impact the polymer's end-use. The presence of bromine, as seen in our exercise, suggests the polymer has altered characteristics, potentially improving flame resistance or changing its mechanical properties.

Chemical Problem Solving

Solving chemical problems often involves a systematic approach where you take complex questions and break them into smaller, manageable parts. This method is essential to tackle detailed queries like determining the value of \( n \) in our polymer chemistry example.

Initially, identifying what is given and what needs to be calculated is crucial. For instance, we knew the percentage of bromine by mass and moved on to figure out the larger unknown—how many repeating units of \( \mathrm{C}_8\mathrm{H}_8 \) are present.

By setting up the equation \( \frac{239.7}{314.784 + 104.144n} \times 100 = 0.105 \), we analyzed the relationship between the elements of the equation to isolate \( n \). Solving this highlighted the importance of algebra in deciphering chemical compositions.

• Begin with known values and rearrange equations to solve for what you need.
• Check units for consistency to ensure accurate calculations.
• Always question if the outcome makes logical sense based on initial conditions.

Approaching chemical questions with clear, logical thinking reinforces foundational concepts, preparing students for handling more advanced chemistry topics with confidence.