Question

In 2009 , a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is \(0.20 \mathrm{~g} / \mathrm{cm}^{3}\), and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0 -mg sample of this material. (b) Calculate the surface area for a 10.0 \(\mathrm{mg}\) sample of this material. (c) A \(10.0-\mathrm{mL}\) sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

Short answer

(a) The volume of a 10.0 mg sample of this material is 0.050 cm³. (b) The surface area for a 10.0 mg sample of this material is 12.42 m². (c) 99.99% of the mercury was removed from the contaminated water. (d) The final mass of the spongy material after exposure to mercury is 17.747 mg.

Step-by-step solution

(a) Calculate the volume of a 10.0 mg sample of the material

We are given the density of the material, which is defined as \(\rho = \frac{m}{V}\), where \(\rho\) is the density, m is the mass, and V is the volume. We want to find the volume of a 10.0 mg sample of the material. Rearranging the formula, we get \(V = \frac{m}{\rho}\). Given: Density: \(\rho = 0.20 \mathrm{~g}/\mathrm{cm}^{3}\) Mass: \(m = 10.0 \mathrm{~mg}\) (we must convert this to grams) Converting the mass to grams: \(m = 10.0 \mathrm{~mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}} = 0.010 \mathrm{~g}\) Now we can plug the values into the formula to find the volume: \(V = \frac{0.010 \mathrm{~g}}{0.20 \mathrm{~g}/\mathrm{cm}^{3}} = 0.050 \mathrm{~cm}^{3}\) Hence, the volume of a 10.0 mg sample of this material is 0.050 cm³.

(b) Calculate the surface area for a 10.0 mg sample

We are given the surface area per gram of material, which is 1242 m²/g. To find the surface area for a 10.0 mg sample, we need to multiply the surface area per gram by the mass of the sample in grams. We have the mass from part (a): \(m = 0.010 \mathrm{~g}\) Now, we can find the surface area of the 10.0 mg sample: Surface Area = (Surface Area per Gram) x (Mass) Surface Area = \(1242 \mathrm{~m}^{2}/\mathrm{g} \times 0.010 \mathrm{~g} = 12.42 \mathrm{~m}^{2}\) Hence, the surface area for a 10.0 mg sample of this material is 12.42 m².

(c) Calculate the percentage of mercury removed from contaminated water

Initially, the contaminated water sample has 7.748 mg of mercury in it. After treatment with the spongy material, only 0.001 mg of mercury remains in the water. To find the percentage of mercury removed, we need to compute the difference between the initial and final amount of mercury, then divide by the initial amount of mercury, and multiply by 100%. Percentage Removed = \( \frac{\text{Initial Amount} - \text{Final Amount}}{\text{Initial Amount}} \times 100\%\) Percentage Removed = \( \frac{7.748 \mathrm{~mg} - 0.001 \mathrm{~mg}}{7.748 \mathrm{~mg}} \times 100\% \) Percentage Removed = 99.99% Hence, 99.99% of the mercury was removed from the contaminated water.

(d) Calculate the final mass of the spongy material after exposure to mercury

Initially, the spongy material has a mass of 10.0 mg. The amount of mercury removed from the water sample was 7.748 mg - 0.001 mg = 7.747 mg. To find the final mass of the spongy material after exposure to mercury, we simply add the mass of the removed mercury to the initial mass of the spongy material. Final Mass of Spongy Material = Initial Mass + Mass of Removed Mercury Final Mass of Spongy Material = \(10.0 \mathrm{~mg} + 7.747 \mathrm{~mg} = 17.747 \mathrm{~mg}\) Hence, the final mass of the spongy material after exposure to mercury is 17.747 mg.

Key concepts

Density

Density is a key concept in material science and it refers to how much mass is contained in a given volume. It is usually expressed in units of grams per cubic centimeter (g/cm³) in these types of problems. For the spongy material introduced in the exercise, its density is noted as 0.20 g/cm³. Understanding how to manipulate the formula for density—\(\rho = \frac{m}{V}\), where \(m\) is mass and \(V\) is volume—is crucial. By rearranging this formula, students can calculate unknown variables such as the volume of a material when its mass and density are known. This enables engineers and scientists to predict how a material will behave in various environments, or how much space it will occupy when deployed. Calculating the density of materials aids in determining their applications, especially when dealing with things like structural supports or any scientific equipment that requires precise material usage. Additionally, density is an intrinsic property, meaning it doesn't change with the size of the sample, helping in identifying substances and checking the purity of materials.

Surface Area

Surface area is an important concept when working with materials at the micro or nanoscale. It essentially measures the total area that the surface of the material exposes. For porous or powdery materials, like the spongy nickel-molybdenum-sulfur composition discussed, surface area can significantly influence reactions both physically and chemically. The exercise specifies a surface area of 1242 m²/g for the material. This large surface area means more sites are available for interactions, which include adsorption of contaminants like mercury. Calculation is straightforward when the surface area per gram is given; you simply multiply this value by the mass of the material. This large area enhances the effectiveness of the material in operations like filtration or catalysis, as more of the contaminant can be captured or reacted at once. Understanding surface area is essential, particularly in fields that utilize catalysis, drug delivery, and environmental cleanup, where maximizing interaction points can lead to more efficient and effective processes.

Contamination Removal

Removing contaminants from environments, such as mercury in water, is one of the practical applications of using materials with high surface areas and specific chemical compositions. In our problem, the spongy material is highly effective at removing mercury due to its high surface area, which allows it to adsorb more mercury per unit of material.To evaluate contamination removal, we calculate the percentage of mercury removed, which tells us how effective the treatment is: \[\text{Percentage Removed} = \frac{\text{Initial mercury} - \text{Remaining mercury}}{\text{Initial mercury}} \times 100\%\]This calculation provides insight into the material’s efficiency. In practical terms, a high percentage indicates that the material can serve well in environmental cleanup efforts. The exercise's spongy material reaches a 99.99% removal efficiency, an excellent figure for applications requiring stringent environmental safety, such as drinking water purification.

Chemical Composition

The chemical composition of a material defines what elements and in what ratio they make up the compound. This is crucial as it influences the material’s physical and chemical properties, including reactivity, strength, and suitability for specific applications. For the spongy material, the combination of nickel, molybdenum, and sulfur has been specially designed to interact with mercury. Each element brings unique properties:

• Nickel can provide structural support and has catalytic properties.
• Molybdenum can enhance the hardness and durability of the material while also contributing to catalytic activities.
• Sulfur often interacts with metal ions, making it ideal for capturing and binding with mercury.

This combination creates a robust material that performs well in removing mercury from contaminated water, as shown in the exercise. Such precise control over chemical composition allows for materials to be engineered for specific roles, be it filtering contaminants, catalyzing reactions, or being part of more complex systems.