Question

A tabulation of data lists the following equation for calculating the densities \((d)\) of solutions of naphthalene in benzene at \(30^{\circ} \mathrm{C}\) as a function of the mass percent of naphthalene. $$d\left(\mathrm{g} / \mathrm{cm}^{3}\right)=\frac{1}{1.153-1.82 \times 10^{-3}(\% \mathrm{N})+1.08 \times 10^{-6}(\% \mathrm{N})^{2}}$$ Use the equation above to calculate (a) the density of pure benzene at \(30^{\circ} \mathrm{C} ;\) (b) the density of pure naphthalene at \(30^{\circ} \mathrm{C} ;\) (c) the density of solution at \(30^{\circ} \mathrm{C}\) that is 1.15\% naphthalene; (d) the mass percent of naphthalene in a solution that has a density of \(0.952 \mathrm{g} / \mathrm{cm}^{3}\) at \(30^{\circ} \mathrm{C} .[\text { Hint: For }(\mathrm{d}),\) you need to use the quadratic formula. See Section A-3 of Appendix A.]

Short answer

The density of pure benzene is found in step 1, the density of pure naphthalene in step 2, the density of a solution of 1.15% naphthalene in step 3, and the %N in a solution with a given density in step 4. The specifics for each solution will depend upon the actual calculations done in each step.

Step-by-step solution

Calculating the density of pure benzene

For pure benzene, there is no naphthalene involved, hence, the mass percent of naphthalene (\(% N\)) is 0. Substituting \%N = 0 into the density equation \(d=(1)/(1.153-1.82 \times 10^{-3}(\%N)+1.08 \times 10^{-6}(\%N)^{2})\), gives the density of pure benzene.

Calculating the density of pure naphthalene

For pure naphthalene, the mass percent of naphthalene (\(% N\)) is 100. Substituting \%N = 100 into the density equation \(d=(1)/(1.153-1.82 \times 10^{-3}(\%N)+1.08 \times 10^{-6}(\%N)^{2})\), gives the density of pure naphthalene.

Calculating the density of the solution with 1.15% naphthalene

Here, the mass percent of naphthalene (\(% N\)) is 1.15. Substituting \%N = 1.15 into the density equation \(d=(1)/(1.153-1.82 \times 10^{-3}(\%N)+1.08 \times 10^{-6}(\%N)^{2})\), gives the density of the solution.

Finding the mass percent of naphthalene in a solution with a given density

Here, the density of the solution (\(d\)) is given and the mass percent of naphthalene (\(% N\)) needs to be determined. Thus, the equation needs to be rewritten as \(%N = (1.153d -1)/(1.82 \times 10^{-3} + 1.08 \times 10^{-6}d)\) and solved using the quadratic formula to find \%N.

Key concepts

Naphthalene

Naphthalene is an organic compound best known for its strong smell and is primarily used in mothballs and some pharmaceuticals. It is composed of fused benzene rings that make it a polycyclic aromatic hydrocarbon (PAH). Naphthalene has distinct properties such as being solid at room temperature. When determining the density of naphthalene in solutions like benzene, it involves calculating how much naphthalene is present. This helps chemists understand the behavior of naphthalene in various solvents and at different conditions.

Understanding how naphthalene interacts in solutions is crucial for industrial applications where the concentration can affect not just the density but also other physical and chemical properties of the mixture. Calculating its contribution to the density of a solution gives insights into its behavior and potential reactivity.

Benzene

Benzene is a widely used solvent in the chemical industry. It is a simple aromatic hydrocarbon with the formula \( C_6H_6 \). At 30°C, benzene is in liquid form and is often used as a base to dissolve other substances, such as naphthalene, for density calculations. Its unique structure, known as a resonance-stabilized ring, lends benzene certain qualities like stability and lack of reactivity compared to other unsaturated hydrocarbons.

To find the density of benzene or its solutions, we consider the absence or presence of other solutes — in this case, naphthalene. The task involves using a derived density function where the mass percent of additives like naphthalene is accounted for to determine the overall density of the solution. Knowing the density can aid in understanding the solution's concentration and behavior.

Quadratic Formula

The quadratic formula is a fundamental tool used to solve equations of the form \( ax^2 + bx + c = 0 \). When performing density calculations related to naphthalene and benzene, the quadratic formula comes into play when you need to solve for the mass percent of naphthalene given the density of the solution.

The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using the quadratic formula helps solve for unknowns, especially when the relationship between variables is not linear. In this exercise, it's used to determine the mass percent, which is one of the variables in the density equation.

Mass Percent

Mass percent is a way of expressing concentration in a solution, specifically the amount of solute relative to the total mass of the solution. When dealing with naphthalene in benzene, mass percent reflects the weight of naphthalene compared to the entire solution weight.

This calculation involves converting the mass of the solute (naphthalene) into a percentage of the total mass mixture (naphthalene plus benzene). Understanding mass percent is essential when you're required to adjust or predict the physical properties of a solution, such as density. It provides a clear picture of the solute’s proportion contributing to the overall mixture.

Step-by-Step Solution

The step-by-step approach makes complex calculations manageable by breaking them down into smaller parts. In the context of the naphthalene-benzene density problem, it involves:

• Calculating densities given no naphthalene (pure benzene) and full naphthalene (pure naphthalene) using the formula provided.
• This method helps visualize how adding varying amounts of naphthalene affects the solution.
• Applying basic substitution to find unknown values based on given percentages or densities.
• Utilizing the quadratic formula to solve for unknown mass percent when the density is known.

This structured process allows students to comprehend how individual components of the equation contribute to the overall solution, reinforcing mathematical and chemical concepts simultaneously.