Problem 57 - Chemistry: The Central Science

Question

Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a \(100 \mathrm{ft} \times 82 \mathrm{ft}\) ceiling with gold leaf that is five-millionths of an inch thick. The density of gold is \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\), and gold costs \(\$ 1654\) per troy ounce \((1\) troy ounce \(=31.1034768 \mathrm{~g})\). How much will it cost the architect to buy the necessary gold?

Step-by-Step Solution

Answer

The cost for the architect to buy the necessary gold to cover the ceiling with a gold leaf of five-millionths of an inch thick is \(\$ 99,103.43\).

Step 1: Convert ceiling dimensions to cm

First, we need to convert the ceiling dimensions from feet to centimeters: - \(1 \mathrm{ft} = 30.48 \mathrm{cm}\) So, the ceiling dimensions in cm are: - \(100 \mathrm{ft} \times 30.48 \mathrm{cm/ft} = 3048 \mathrm{cm}\) - \(82 \mathrm{ft} \times 30.48 \mathrm{cm/ft} = 2499.36 \mathrm{cm}\)

Step 2: Calculate the volume of gold leaf

Now that we have the ceiling dimensions in cm, we can find the volume of gold leaf required to cover the ceiling: Volume of gold leaf = Area of ceiling × Thickness of gold leaf We will calculate this in the next step, after converting the thickness of the gold leaf to cm.

Step 3: Convert gold leaf thickness to cm

Next, we need to convert the thickness of the gold leaf from inches to centimeters: - \(5 \times 10^{-6} \mathrm{in} = 5 \times 10^{-6} \mathrm{in} \times 2.54 \mathrm{cm/in} = 1.27 \times 10^{-5} \mathrm{cm}\)

Step 4: Calculate the volume of gold leaf

Now that we have the thickness of the gold leaf in cm, we can calculate the volume of gold leaf: Volume of gold leaf = (\(3048 \mathrm{cm} \times 2499.36 \mathrm{cm}) \times 1.27 \times 10^{-5} \mathrm{cm} = 96.5107 \mathrm{cm^3}\)

Step 5: Calculate the mass of gold required

Next, we will find the mass of gold using its density: Mass of gold required = Volume of gold leaf × Density of gold Mass of gold required = \(96.5107 \mathrm{cm^3} \times 19.32 \mathrm{g/cm^3} = 1864.8740 \mathrm{g}\)

Step 6: Convert mass of gold to troy ounces

Now, we need to convert the mass of gold from grams to troy ounces: - \(1\) troy ounce \(= 31.1034768 \mathrm{g}\) Mass of gold required in troy ounces = \(\frac{1864.8740 \mathrm{g}}{31.1034768 \mathrm{g/troy~ounce}} = 59.9250\) troy ounces

Step 7: Calculate the cost of gold required

Finally, we will find the cost of gold required to cover the ceiling: Cost of gold required = Mass of gold in troy ounces × Cost per troy ounce Cost of gold required = \(59.9250\) troy ounces × \(\$ 1654\) per troy ounce Cost of gold required = \(\$ 99,103.43\) So, it will cost the architect \(\$ 99,103.43\) to buy the necessary gold to cover the ceiling.

Key Concepts

Density Calculations

Understanding the concept of density is crucial for a range of scientific inquiries and applications, particularly in chemistry and material science. Density, a measure of how much mass is contained in a given volume, is often expressed using the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
In the context of our exercise, we know the density of gold is \(19.32 \text{g/cm}^3\). With this information, calculating the mass of the gold required involves finding the volume of gold leaf needed to cover the specified area and then applying the formula for density. To solve such problems, it is essential to ensure all units are in the right system which, in this case, means converting measurements to a consistent unit—centimeters and grams for metric.
Solving density problems often involves a step-by-step approach: identify the required density, calculate the volume of the material, and then compute the mass by multiplying the volume by the density. By understanding this formula and the concept of density as mass per unit volume, you will be able to tackle a myriad of problems involving the calculation of material amounts in both scientific research and practical, everyday situations.

Unit Conversions

Unit conversions are the backbone of many scientific calculations, ensuring accurate and useful results across different measurement systems. For students, gaining proficiency in unit conversions is necessary for success in chemistry and other scientific fields. The process involves multiplying the given value by a conversion factor that cancels out the original unit and introduces the desired unit.

Creating Conversion Factors

A conversion factor is a ratio or fraction which represents the relationship between two different units. For instance, in the problem at hand, we have the following relationships:

• \(1 \text{ft} = 30.48 \text{cm}\)
• \(1 \text{in} = 2.54 \text{cm}\)
• \(1 \text{troy ounce} = 31.1034768 \text{g}\)

Conversions like these are a necessary step in many calculations. They require attention to detail and can often be a source of error if not done carefully. Ensuring that units cancel out properly and that the right conversion factor is used is critical in obtaining the correct result. When working through chemistry problems, always pause to verify that your units match the units required for your final answer. This habit can prevent common mistakes and is a good practice for achieving precision in solving chemistry-related problems.

Volume Determination

Volume determination is essential for solving a variety of problems in chemistry, particularly when dealing with the physical properties of substances. In the context of our exercise, we needed to find the volume of gold leaf necessary to cover a ceiling. The volume is a measure of the three-dimensional space an object occupies, and for a thin sheet like gold leaf, it is calculated by multiplying its area by its thickness.
Area is a measure of a two-dimensional space, determined by the formula:

• \( \text{Area} = \text{Length} \times \text{Width}\)

Once you have the area, volume can be found by incorporating the object's height—or thickness, in the case of a sheet:

• \( \text{Volume} = \text{Area} \times \text{Thickness}\)

The precision of volume determination is critical when calculating material costs or understanding substance quantities. Errors in measuring or calculating volume can lead to incorrect results in the overall problem, such as the quantity of material needed or the final cost. By taking the time to do these calculations with care and double-checking measurements, such issues can be minimized, leading to accurate and reliable outcomes in chemistry and beyond.