Problem 38 - Chemistry The Central Science

Question

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0-\) mL portion of the liquid had a mass of 21.95 \(\mathrm{g} .\) A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density in agreement with the tabulated value? (b) An experiment requires 15.0 \(\mathrm{g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is 0.7781 \(\mathrm{g} / \mathrm{mL}\) . What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of 5.0 \(\mathrm{cm} .\) What is the mass of the sphere if lead has a density of 11.34 \(\mathrm{g} / \mathrm{cm}^{3} ?\) (The volume of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius.)

Step-by-Step Solution

Answer

After calculating their densities, the clear liquid has a density of 0.878 g/mL, which is quite close to the tabulated value of 0.8787 g/mL for benzene at \(15^{\circ} \mathrm{C}\). This suggests that the liquid is likely benzene. In the second part, 19.29 mL of cyclohexane should be used for the given mass of 15.0 g. In the third part, the mass of the lead sphere is approximately 696.26 g.

Step 1: Problem (a) - Calculating the Density

To find the density (d) of the clear liquid, use the formula: \(d = \frac{m}{V}\) where \(m\) is the mass and \(V\) is the volume. The mass of the liquid is 21.95 g, and its volume is 25.0 mL. Now, we can calculate the density: \(d = \frac{21.95 \ \text{g}}{25.0 \ \mathrm{mL}}\)

Step 2: Problem (a) - Comparing the Calculated Density to the Tabulated Value

Now that we've calculated the density, let's compare it to the tabulated value for benzene at \(15^{\circ} \mathrm{C}\): \(d_{\text{calculated}} = \frac{21.95 \ \text{g}}{25.0 \ \mathrm{mL}} = 0.878\ \text{g/mL}\) \(d_{\text{tabulated}} = 0.8787\ \text{g/mL}\) These values are very close, suggesting that the clear liquid is likely to be benzene.

Step 3: Problem (b) - Finding the Volume of Cyclohexane

To find the volume (V) of cyclohexane required for a given mass (m), use the formula: \(V = \frac{m}{d}\) where \(d\) is the density of cyclohexane at \(25^{\circ} \mathrm{C} (0.7781\ \text{g/mL})\), and \(m\) is the required mass (15.0 g). Now, we can calculate the volume: \(V = \frac{15.0 \ \text{g}}{0.7781 \ \mathrm{g/mL}}\)

Step 4: Problem (b) - Calculating the Volume of Cyclohexane

Calculate the volume using the given values: \(V = \frac{15.0 \ \text{g}}{0.7781 \ \mathrm{g/mL}} = 19.29\ \mathrm{mL}\) Thus, 19.29 mL of cyclohexane should be used.

Step 5: Problem (c) - Finding the Volume of the Lead Sphere

To find the volume (V) of the lead sphere, use the formula for the volume of a sphere: \(V = \frac{4}{3} \pi r^{3}\) where r is the radius, which is half of the diameter (2.5 cm). Calculate the volume: \(V = \frac{4}{3} \pi (2.5\ \text{cm})^{3}\)

Step 6: Problem (c) -Calculating the Mass of the Lead Sphere

With the volume of the lead sphere calculated, we can find the mass (m) using the density (d) formula: \(m = Vd\) where \(d\) is the density of lead (11.34 g/cm³). Calculate the mass: \(m = (\frac{4}{3} \pi (2.5\ \text{cm})^{3})(11.34 \ \mathrm{g/cm}^{3})\) Thus, the mass of the lead sphere is approximately 696.26 g.

Key Concepts

Chemistry Problem Solving

Understanding chemistry problems often involves a systematic approach to dissect the problem, identify the relevant concepts, and apply mathematical relationships. To tackle chemistry exercises, one must first carefully read the problem and determine what is being asked. Next, the known quantities are identified, and the correct formula is applied to find the unknown. For instance, in the given exercise, the problems require knowledge of the density formula, volume calculations for spheres, and the ability to compare calculated values with standard tabulated ones. By following a stepwise approach, such as in the provided solutions, students can incrementally solve each part. This cultivates not only a strong grasp of individual concepts but also hones overall problem-solving skills.

Density Formula

When working with the concept of density in chemistry, it's vital to familiarize oneself with the density formula: \(d = \frac{m}{V}\), where \(d\) represents density, \(m\) is mass, and \(V\) is volume. Density is a measure of how much mass is contained in a given volume and is a fundamental property in identifying substances, as seen in the original exercise where a chemist must verify the identity of a liquid by its density. Recognizing the importance of units is also crucial in density calculations—consistency across mass and volume units ensures accurate calculations.

Volume of Spheres

In geometry and applied mathematics, the volume of a sphere can be calculated using the formula \(V = \frac{4}{3} \[pi\] r^{3}\), with \(r\) representing the sphere's radius. The exercise underscores the need to understand geometric formulas within chemistry problems, especially when considering the properties of substances in different shapes. Converting the diameter to radius (by halving it) and then applying the volume formula allows us to find the quantity of material required for other calculations, such as determining the mass of a spherical object when given its diameter and density.

Comparing Calculated and Tabulated Values

Comparing experimental or calculated values to known tabulated ones is a common practice in chemistry to verify substances or to check the accuracy of experimental results. When the calculated density of a liquid nearly matches that of a tabulated reference, it often confirms the substance's identity. This comparative process also helps in assessing the reliability of measurements and identifying any potential errors in experimentation. The importance of precision and accuracy is highlighted as small differences might lead to incorrect identifications or conclusions, making careful calculations and comparisons essential.