Problem 94 - Chemistry: The Central Science in SI Units, Global Edition

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 (b) Calculate the surface area for \(10.0-\mathrm{mg}\) sample of this material. a \(10.0-\mathrm{mg}\) sample of this material. \((\mathbf{c})\) 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 (d) What is the final mass mercury was removed from the water? of the spongy material after the exposure to mercury?

Step-by-Step Solution

Answer

The volume of a 10.0-mg sample of this spongy material is 0.05 cm³, and its surface area is 12.42 m². After treating a 10.0-mL contaminated water sample with this material, 99.99% of mercury was removed. The final mass of the spongy material after exposure to mercury is 17.747 mg.

Step 1: Calculate the volume of a 10.0-mg sample

First, we need to convert the mass of the sample from milligrams (mg) to grams (g): \(10.0 \mathrm{~mg} = 0.010 \mathrm{~g}\) We are given the density of the material: \(0.20 \mathrm{~g/cm^3}\) Now we can calculate the volume using the formula: \(V = \frac{mass}{density}\) \(V = \frac{0.010 \mathrm{~g}}{0.20 \mathrm{~g/cm^3}}\) \(V = 0.05 \mathrm{~cm^3}\)

Step 2: Calculate the surface area of a 10.0-mg sample

We are given the surface area per gram of the material: \(1242 \mathrm{~m^2/g}\) Now we can calculate the surface area using the formula: \(A = m \frac{1242 \mathrm{~m^2}}{1 \mathrm{~g}}\) \(A = 0.010 \mathrm{~g} \times \frac{1242 \mathrm{~m^2}}{1 \mathrm{~g}}\) \(A = 12.42 \mathrm{~m^2}\)

Step 3: Calculate the percentage of mercury removed from the water

We are given the initial and final amounts of mercury in the water sample: Initial amount = 7.748 mg Final amount = 0.001 mg Amount of mercury removed = Initial amount - Final amount Amount of mercury removed = 7.748 mg - 0.001 mg = 7.747 mg Now we can calculate the percentage of mercury removed: \(\% = \frac{7.747 \mathrm{~mg}}{7.748 \mathrm{~mg}} \times 100\% \) \(\% = 99.99 \% \)

Step 4: Calculate the final mass of the spongy material after exposure to mercury

Initial mass of spongy material = 10.0 mg Amount of mercury removed = 7.747 mg Final mass of spongy material = Initial mass + Amount of mercury removed Final mass of spongy material = 10.0 mg + 7.747 mg = 17.747 mg

Key Concepts

Density

Density is like a measure of how tightly matter is packed together in a space. For the new spongy material discussed, it has a density of \(0.20 \mathrm{~g/cm}^3\). This means that each cubic centimeter of this material has a mass of 0.20 grams. To understand this, let’s break down the formula for density: - Density \(= \frac{Mass}{Volume}\)With this formula, we see that if you know the density and the mass of a material, you can rearrange it to find out the volume:- Volume \(= \frac{Mass}{Density}\)In the exercise, we calculated the volume of a 10.0 mg sample of spongy material. First, we need to convert milligrams to grams, because density is given in grams per cubic centimeter. After converting 10.0 mg to 0.010 grams, we found the material takes up 0.05 \(\mathrm{cm^3}\). This illustrates how density can help us see how much space a material will occupy based on its mass.

Surface Area

Surface area is about how much area is on the outside of a material. Imagine wrapping a gift; the surface area is like the amount of wrapping paper you need. For the spongy material, the surface area is exceptionally large, \(1242 \mathrm{~m^2}\) per gram. Why does surface area matter? Especially in material science, the more surface area there is, the more of the material is in contact with its surroundings. This means it can interact more with things it touches, like water molecules in this case. - To calculate surface area: - Multiply the surface area per gram by the mass of the sample.For our 10.0 mg sample:- Convert the mass to grams (0.010 grams),- Then multiply: \(0.010 \mathrm{~g} \times 1242 \mathrm{~m^2/g} = 12.42 \mathrm{~m^2}\)Having this understanding explains why the spongy material can interact so effectively with substances like mercury in water—it has a large surface area!

Mercury Removal

Removing mercury from water is important because mercury is a hazardous pollutant. The spongy material excels at this job. It acts like a sponge, soaking up the mercury. Here’s how it removes mercury from water effectively:- It has a huge surface area, allowing it to capture more mercury.- In the exercise, we know that a 10.0 mL water sample originally had 7.748 mg of mercury. After treatment with 10.0 mg of the spongy material, only 0.001 mg of mercury was left.To calculate the efficiency:- Subtract the final amount of mercury from the initial amount to find how much was removed.- Calculate the removal percentage: \[\% = \frac{7.747 \mathrm{mg}}{7.748 \mathrm{mg}} \times 100\% = 99.99\%\]The material removed 99.99% of mercury! This high efficiency is part of why this substance is a breakthrough in mercury removal.

Chemical Material Properties

Chemical material properties tell us about how a substance behaves or interacts with other materials and conditions. For this spongy material, understanding its chemical properties helps us grasp why it's effective in particular uses, like removing mercury from water. Some of the important properties include: - **Density:** Affects how the material interacts in a solution. - **Surface Area:** Affects the amount of interaction the material can have with mercury particles. - **Composition:** Made of nickel, molybdenum, and sulfur, which may aid in its reactivity with mercury. Also, from the exercise, we learned the material doesn't dissolve or lose much of itself in the process, maintaining its mass after absorbing mercury: - Initial mass before interaction: 10.0 mg. - After exposure, it had effectively gained mercury, showing a total related mass increase, reflecting its good binding properties. Understanding these properties tells us why this material is so good at its job of mercury removal.