Problem 38 - Chemistry The Central Science In Si Units

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 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 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

(a) Yes, the densities match closely. (b) Use 19.27 mL of cyclohexane. (c) The mass of the lead sphere is 742.83 g.

Step 1: Calculate the Density of the Liquid

To determine the density, we use the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). The mass of the liquid is given as 21.95 g, and its volume is 25.0 mL. Therefore, the density is \( \frac{21.95 \text{ g}}{25.0 \text{ mL}} = 0.878 \text{ g/mL} \).

Step 2: Compare Densities to Verify Identity

Now that we have the calculated density \(0.878 \text{ g/mL}\), we compare it to the tabulated value of benzene, which is \(0.8787 \text{ g/mL}\). The values are very close, suggesting that the liquid is likely benzene.

Step 3: Calculate Volume of Cyclohexane

Using the formula \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \), calculate the volume needed for 15.0 g of cyclohexane. With the density provided as \(0.7781 \text{ g/mL}\), we find \( \text{Volume} = \frac{15.0 \text{ g}}{0.7781 \text{ g/mL}} = 19.27 \text{ mL} \).

Step 4: Calculate the Volume of the Lead Sphere

First, we find the radius of the sphere. The diameter is 5.0 cm, so the radius \( r \) is \( \frac{5.0}{2} = 2.5 \text{ cm} \). The volume \( V \) of a sphere is given by \( \frac{4}{3} \pi r^3 \). Substituting \( r = 2.5 \text{ cm} \), we compute \( V = \frac{4}{3} \pi (2.5)^3 \approx 65.45 \text{ cm}^3 \).

Step 5: Calculate the Mass of the Lead Sphere

Using the density of lead, \( 11.34 \text{ g/cm}^3 \), and the volume calculated in the previous step, we find the mass by \( \text{Mass} = \text{Density} \times \text{Volume} \). Thus, \( \text{Mass} = 11.34 \text{ g/cm}^3 \times 65.45 \text{ cm}^3 \approx 742.83 \text{ g} \).

Key Concepts

Chemical Identification

Identifying a chemical substance involves comparing its measured properties, such as density, to known values. If a liquid, like the one from this exercise, is thought to be benzene, its density needs to be checked against the standard or tabulated density of benzene at a specific temperature. Here is how it works:

The density of any substance is calculated by dividing its mass by its volume:

• Given a mass of 21.95 g and a volume of 25.0 mL, the calculation shows a density of \(0.878\, \text{g/mL}\).
• This is very close to the standard density of benzene, \(0.8787\, \text{g/mL}\), indicating that the liquid is likely benzene.

Therefore, if the calculated and standard densities are nearly identical, the identity of the substance can be confirmed with confidence. Chemical identification is crucial in verifying and utilizing substances in various chemical applications.

Volume Calculation

Volume calculation is key when determining how much of a liquid you need for a given mass, especially in chemical experiments. Cyclohexane provides a perfect example:

The formula used to calculate volume is:

• \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)

To find out how much cyclohexane is needed for 15.0 g, knowing its density is \(0.7781 \text{g/mL}\):

• Calculate: \( \text{Volume} = \frac{15.0 \text{ g}}{0.7781 \text{ g/mL}} \approx 19.27 \text{ mL} \)

This calculation tells you that you need approximately 19.27 mL of cyclohexane. Volume calculations ensure that chemical reactions or processes are executed accurately, using appropriate quantities of substances.

Mass Calculation

Mass calculation is essential when you need to know the weight of a certain volume of a substance, such as a solid sphere. Lead provides a solid, real-world example:First, determine the sphere's volume:

• With a diameter of 5.0 cm, the radius \( r \) is half of that: \(2.5\, \text{cm}\).
• Using the sphere's volume formula, \( \frac{4}{3} \pi r^3 \), the volume is calculated as \( \approx 65.45\, \text{cm}^3 \).

Next, compute the mass using the known density of lead, \( 11.34\, \text{g/cm}^3 \):

• Mass calculation: \( \text{Mass} = 11.34\, \text{g/cm}^3 \times 65.45\, \text{cm}^3 \approx 742.83 \text{g} \)

This shows the sphere weighs approximately 742.83 g. Understanding mass calculation helps in assessing and handling substances in a variety of scientific and real-world contexts.