Problem 88 - Chemistry The Science in Context

Question

In Section 20.2 we discussed the rotation of a beam of polarized light when it passes through a solution containing an optically active molecule. Equimolar solutions of \(\alpha\) -glucose and \(\beta\) -glucose rotate plane-polarized light by \(+112^{\circ}\) and \(+18.7^{\circ},\) respectively. If these two solutions are then mixed and allowed to reach equilibrium, the solution then rotates the polarized light by \(+53.4 "\) Calculate the percent glucose in the \(\alpha\) and \(\beta\) forms in this solution.

Step-by-Step Solution

Answer

Answer: The solution contains approximately 46.7% α-glucose and 53.3% β-glucose.

Step 1: Assign variables for mole fractions

Let x be the mole fraction of \(\alpha\)-glucose, and y be the mole fraction of \(\beta\)-glucose. Since they are the only two components in the mixture, their sum is equal to 1: $$ x + y = 1 $$

Step 2: Write the equation for the total rotation

As mentioned earlier, the total rotation is a weighted sum of the rotations caused by the \(\alpha\) and \(\beta\) forms, considering their mole fractions. Therefore, we have $$ 53.4 = 112x + 18.7y $$

Step 3: Solve for the mole fractions of the \(\alpha\) and \(\beta\) forms

Now we have a system of two linear equations with two variables: $$ x + y = 1 \\ 53.4 = 112x + 18.7y $$ From the first equation, we can solve for y: $$ y = 1 - x $$ Substitute this expression for y into the second equation: $$ 53.4 = 112x + 18.7(1-x) $$ Now solve for x: $$ x \approx 0.467 $$ Substitute x back into the equation for y to find its value: $$ y \approx 0.533 $$ These are the mole fractions of the \(\alpha\) and \(\beta\) forms, respectively.

Step 4: Calculate the percentages of the \(\alpha\) and \(\beta\) forms

To find the percentage of glucose in the \(\alpha\) and \(\beta\) forms, simply multiply the mole fractions by 100: $$ \alpha \text{-glucose \%} = 46.7\% \\ \beta \text{-glucose \%} = 53.3\% $$ The solution contains approximately 46.7% \(\alpha\)-glucose and 53.3% \(\beta\)-glucose.

Key Concepts

Polarized Light

Polarized light is a fascinating and important concept in the study of chemistry and physics. When regular light is emitted, it vibrates in multiple planes. However, when light passes through a polarizer, it becomes polarized light. This means that the light is now vibrating only in one plane.

This kind of light becomes especially useful when studying optically active substances. Optically active substances rotate the plane of polarized light. The direction and extent of this rotation depend on the nature of the substance, the concentration of its molecules, and the path length it travels through.

• In our context, \(\alpha\)-glucose and \(\beta\)-glucose each have a distinct effect on polarized light.
• When a solution of each is separately prepared, \(\alpha\)-glucose rotates light by \(+112^{\circ}\), while \(\beta\)-glucose rotates it by \(+18.7^{\circ}\).

By understanding how these molecules interact with polarized light, we can also understand the proportion of each molecule present in a mixture. For example, when the solution's overall rotation is a combination of the rotations induced by both \(\alpha\) and \(\beta\) forms, it indicates their relative abundances.

Mole Fractions

Mole fraction is a key concept in the solutions and mixtures domain of chemistry. It is a way to express the concentration of each component in a mixture. Specifically, the mole fraction (\(x\) or \(y\)) is the ratio of the number of moles of a specific component to the total number of moles in a solution. This simple but powerful tool enables chemists to calculate many properties of mixtures.

In the given exercise, the mole fractions (\(x\) for \(\alpha\)-glucose and \(y\) for \(\beta\)-glucose) were used to determine the compound percentages in a mixture.

• The sum of the mole fractions must equal 1, i.e., \(x + y = 1\).
• You solved for one mole fraction using the second equation, blending their weighted contributions to polar rotation: \(112x + 18.7y = 53.4\).
• This led to finding \(x\) and \(y\) values, demonstrating the proportion of each glucose form in the mixture.

The exercise elegantly shows that mole fractions are not just about numbers but are essential for understanding composition and behavior in mixtures.

Equilibrium in Solutions

Equilibrium in solutions is a vital and broad concept in chemistry that applies to various systems. It refers to the state where the concentrations of all reactants and products remain constant over time. This happens once the forward and backward processes within the solution occur at an equal rate.

For glucose mixtures, equilibrium refers to the balance achieved in the mixture concentrations of \(\alpha\) and \(\beta\) forms.

• Initially, these glucose forms have distinct initial proportions.
• Upon mixing, they interact and change until they reach a stable ratio.
• At equilibrium, the rotation effect on polarized light reflects this balanced state because their proportions relative to each other yield a specific combined rotation value.

Understanding equilibrium in solutions helps chemists predict and control reactions and mixtures. It provides in-depth insights into reaction dynamics, much needed for effective solution and reaction management.