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 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 $\left(\frac{4}{3}\right)\pi r^3$ where $r$ is the radius.)

Short answer

(a) The calculated density is $\rho =\frac{21.95\mathrm{g}}{25.0\mathrm{mL}}=0.878\mathrm{g}/\mathrm{mL}$, which is in agreement with the tabulated value of $0.8787\mathrm{g}/\mathrm{mL}$ at ${15}^{\circ }\mathrm{C}$. (b) The volume of cyclohexane required is $V=\frac{15.0\mathrm{g}}{0.7781\mathrm{g}/\mathrm{mL}}=19.3\mathrm{mL}$. (c) The volume of the lead sphere is $V=\frac{4}{3}\pi (2.5\mathrm{cm}{)}^3=65.4{\mathrm{cm}}^3$, and thus, the mass of the lead sphere is $m=11.34\mathrm{g}/{\mathrm{cm}}^3\times 65.4{\mathrm{cm}}^3=741.7\mathrm{g}$.

Step-by-step solution

Part (a) - Calculating the density and verifying the identity of benzene

First, we calculate the density using the mass and volume given. We are given: - Mass ($m$): $21.95\mathrm{g}$ - Volume ($V$): $25.0\mathrm{mL}$ Using the density formula, we can calculate the density ($\rho$): $\rho =\frac{m}{V}=\frac{21.95\mathrm{g}}{25.0\mathrm{mL}}$ Now, check if the calculated density is in agreement with the tabulated value $0.8787\mathrm{g}/\mathrm{mL}$ at ${15}^{\circ }\mathrm{C}$.

Part (b) - Finding the volume needed for a given mass of cyclohexane

We need to find the volume of cyclohexane required for a given mass and known density. We are given: - Mass ($m$): $15.0\mathrm{g}$ - Density ($\rho$): $0.7781\mathrm{g}/\mathrm{mL}$ Rearranging the density formula to find the volume ($V$): $V=\frac{m}{\rho }=\frac{15.0\mathrm{g}}{0.7781\mathrm{g}/\mathrm{mL}}$ Calculate the volume of cyclohexane needed.

Part (c) - Calculating the mass of the lead sphere

We need to find the mass of the lead sphere given its diameter and density. We are given: - Diameter: $5.0\mathrm{cm}$ - Density ($\rho$): $11.34\mathrm{g}/{\mathrm{cm}}^3$ First, we must calculate the volume of the sphere using its diameter. We know that the volume of a sphere is given by the formula: $V=\frac{4}{3}\pi r^3$ where $r$ is the radius. The diameter is given, so the radius of the sphere is half the diameter, i.e., $2.5\mathrm{cm}$. Now calculate the volume: $V=\frac{4}{3}\pi (2.5\mathrm{cm}{)}^3$ Now we have the volume and density, we can use the density formula to determine the mass ($m$): $m=\rho V=11.34\mathrm{g}/{\mathrm{cm}}^3\times V$ Calculate the mass of the lead sphere.

Key concepts

Benzene Identification

When identifying a liquid as benzene through its density, a comparison between experimental and known values is crucial. In this exercise, a chemist measures the density of an unknown liquid to verify if it is benzene. The measured density found from the ratio of mass to volume is compared with benzene's tabulated density at a specific temperature. Density is defined as mass divided by volume. For instance, with a mass of 21.95 g and a volume of 25.0 mL, we find the density using the formula:

• $\rho =\frac{m}{V}=\frac{21.95\text{g}}{25.0\text{mL}}$

To confirm if the liquid is benzene, compare this calculated density with the known value of 0.8787 g/mL. Agreement between these values indicates the liquid may indeed be benzene.

Cyclohexane Volume

The volume needed for a given mass of cyclohexane can be calculated using its density. Density allows conversion between mass and volume, given the formula:

• $\rho =\frac{m}{V}$

Rearranging this formula helps find the volume given mass and density:

• $V=\frac{m}{\rho }$

For 15.0 g of cyclohexane with a density of 0.7781 g/mL, the volume needed is:

• $V=\frac{15.0\text{g}}{0.7781\text{g/mL}}$

Calculate this to get the exact volume to utilize in your experiment.

Lead Sphere Mass

To find the mass of a lead sphere, we need both its density and volume. Density is provided, but first, calculate the sphere's volume using its diameter. For a sphere, the formula for volume is:

• $V=\frac{4}{3}\pi r^3$

Given diameter (5.0 cm), the radius is half: 2.5 cm. Compute the volume:

• $V=\frac{4}{3}\pi (2.5\text{cm}{)}^3$

With volume calculated, use the density formula to determine mass, given density of lead as 11.34 g/cm³:

• $m=\rho V=11.34{\text{g/cm}}^3\times V$

This calculation provides the mass of the spherical ball of lead.

Density Formula

Density is a fundamental concept in chemistry, characterizing how much mass is found in a given volume. The mathematical expression for density is:

• $\rho =\frac{m}{V}$

Here, $\rho$ denotes density, $m$ is mass, and $V$ represents volume. This relationship enables easy conversion between mass and volume if the density is known. Hence, understanding and applying this formula is crucial for correctly identifying substances and predicting properties in chemical reactions. Whether determining volumes needed for a reaction or validating a material's identity, density serves as a fundamental pillar of chemical calculations.

Volume of a Sphere

Understanding the volume of a sphere is vital in numerous scientific applications. The formula for calculating a sphere's volume is:

• $V=\frac{4}{3}\pi r^3$

Here, $V$ stands for volume, $r$ is the sphere's radius, and $\pi$ is a constant approximately equal to 3.14159. To apply this formula, always ensure the radius is derived accurately from any given diameter (by taking half of the diameter). This concept culminates in real-world applications such as calculating the mass of spherical objects, like in the case of lead spheres, where once the volume is known, further calculations such as determining mass become straightforward.