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 $(4/3)\pi r^3$ where $r$ is the radius.)

Short answer

(a) The calculated density of the liquid is $\frac{21.95\mathrm{g}}{25.0\mathrm{mL}}=0.878\mathrm{g}/\mathrm{mL}$, which is close to the tabulated value of $0.8787\mathrm{g}/\mathrm{mL}$ for benzene at ${15}^{\circ }\mathrm{C}$. (b) The volume of cyclohexane required is $\frac{15.0\mathrm{g}}{0.7781\mathrm{g}/\mathrm{mL}}=19.3\mathrm{mL}$. (c) The volume of the lead sphere is $\frac{4}{3}\pi (2.5\mathrm{cm}{)}^3=65.45{\mathrm{cm}}^3$, and the mass of the sphere is $(11.34{\mathrm{g}/\mathrm{cm}}^3)(65.45{\mathrm{cm}}^3)=741.8\mathrm{g}$.

Step-by-step solution

Find the density of the liquid

Given the mass and volume of the liquid, we can find its density using the formula: Density = Mass / Volume Density = $21.95\mathrm{g}/25.0\mathrm{mL}$ Calculate the density.

Compare the calculated density with the tabulated value

Compare the calculated density obtained in Step 1 with the density of benzene listed in the handbook. #b_problem#

Calculate the required volume of cyclohexane

We know the required mass of cyclohexane and its density at a given temperature. We can find the required volume using the formula: Volume = Mass / Density Using the given mass of $15.0\mathrm{g}$ and density of $0.7781\mathrm{g}/\mathrm{mL}$, calculate the required volume. #c_problem#

Find the volume of the lead sphere

We are given the diameter of the sphere, which is $5.0\mathrm{cm}$. Therefore, the radius of the sphere is $r=2.5\mathrm{cm}$. Now we can find the volume using the formula for the volume of a sphere: Volume = ${\displaystyle \frac{4}{3}}\pi r^3$ Calculate the volume using the given radius.

Determine the mass of the lead sphere

Now that we have the volume of the sphere and the density of lead, we can find the mass using the density formula: Mass = Density × Volume Using the given density of $11.34\mathrm{g}/{\mathrm{cm}}^3$ and the calculated volume, compute the mass of the lead sphere.

Key concepts

Density Determination

Understanding how to determine the density of a substance is crucial in scientific and industrial applications. Density is a measure of how much mass is contained in a given unit volume of a material. The classic equation for density determination is:

density = $\frac{mass}{volume}$

In part (a) of our exercise, the mass of the liquid (presumably benzene) was 21.95 g and the volume was 25.0 mL. Calculating the density, we get:

$\frac{21.95\text{g}}{25.0\text{mL}}$ which yields $0.878\text{g/mL}$. Comparing this to the tabulated value of benzene, which is 0.8787 g/mL, we see that the calculated density is very close, suggesting that the liquid is indeed benzene.

To ensure accuracy in any density determination, it's vital to use correctly calibrated measurement instruments and account for environmental conditions, as density can change with temperature.

Volume-Mass Relationship

The volume-mass relationship is an essential concept when dealing with liquids and solids, helping us understand how much space a given mass of substance will occupy and vice versa. This relationship can be utilized to re-arrange the density formula depending on what value we are solving for:

For volume:
volume = $\frac{mass}{density}$

For mass:
mass = (density)(volume)

In part (b) of our exercise, we're asked to find the volume of cyclohexane needed to obtain 15.0 g of it. Given the density of cyclohexane is 0.7781 g/mL, we use the rearranged density formula to calculate volume: $\frac{15.0\text{g}}{0.7781\text{g/mL}}$, which gives us the required volume. Similarly, in part (c), we calculate the mass of a lead sphere by first finding its volume from the provided geometric formula and then using the density of lead to find the mass.

Having a solid understanding of this relationship allows students to prepare for experiments and quantify substances in practical situations.

Density of Substances

Every substance has its unique density, directly affecting how substances interact with one another, such as floating or sinking in a fluid. For example, parts (a) and (b) contrast the densities of benzene and cyclohexane, while part (c) concerns the much higher density of lead.

The density of substances will vary with temperature and pressure; hence, densities are often reported at standard conditions. Materials with densities less than that of water (1 g/mL) will typically float when placed in water, while those with higher densities will sink. In the exercise, we see that lead, with a density of 11.34 g/cm³, is much denser than water, indicating that a lead ball would sink if submerged.

Understanding the density of various substances helps explain phenomena in physics, chemistry, and engineering. It provides the backdrop for learning about buoyancy, purity of materials, identification of substances, and even plays a role in the development of mixtures and solutions.

Learning to interpret density allows students to predict outcomes in experiments and understand natural occurrences. For example, why ice floats on water or oil accumulates on top of a water surface. It's an insight into the molecular structure and arrangement of a substance, making density a key conceptual cornerstone in the study of chemistry.