Question

Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a 100 \(\mathrm{ft} \times 82\) 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?

Short answer

The total cost of the gold leaf required to cover the 100 ft × 82 ft ceiling with a thickness of five-millionths of an inch will be approximately \$835,794.

Step-by-step solution

Find the area of the ceiling

We need to find the total area of the 100 ft × 82 ft ceiling and convert it to cm². Recall that 1 foot = 30.48 cm. So, the ceiling area is: \(A = 100 \ \text{ft} \times 82 \ \text{ft} \times (30.48 \ \text{cm/ft})^2\) Calculate the area of the ceiling.

Calculate the volume of gold leaf

Now, we need to find the volume of gold leaf needed to cover the ceiling. To do this, we will multiply the area of the ceiling by the thickness of the gold leaf. Note that 1 inch = 2.54 cm, and 5 millionths of an inch = \(5 \times 10^{-6}\) inch. So, the volume of gold leaf is: \(V = A \times 5 \times 10^{-6} \ \text{inch} \times 2.54 \ \text{cm/inch}\) Calculate the volume of gold leaf.

Find the mass of the gold leaf

Now, we can find the mass of gold leaf needed by using the density of gold: \(m = V \times \rho\) where \(\rho = 19.32 \ \text{g/cm}^3\) is the density of gold. Calculate the mass of the gold leaf.

Convert the mass to troy ounces

We need to convert the mass of the gold leaf to troy ounces. Recall that 1 troy ounce = 31.1034768 g. So, the mass of gold leaf in troy ounces is: \(m_{\text{troy}} = m \times \frac{1}{31.1034768} \ \text{troy ounce/g}\) Calculate the mass of the gold leaf in troy ounces.

Calculate the cost of the gold leaf

Finally, we can find the cost of the gold leaf using the price per troy ounce: Cost = \(m_{\text{troy}} \times 1654 \ \text{USD/troy ounce}\) Calculate the total cost of the gold leaf.

Key concepts

Density Calculations

Understanding the concept of density is crucial in chemistry, as it is a measure of how much mass is contained within a certain volume. Density is calculated using the formula \( \text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)} \). In practical terms, this tells us how tightly packed the material's particles are. When dealing with precious metals like gold, knowing its density allows us to calculate the mass of a certain volume, an essential step when estimating material costs for practical applications.

For example, gold's density is given as \( 19.32 \frac{g}{cm^3} \) which means that one cubic centimeter of gold has a mass of 19.32 grams. Applying this to a gold leaf chemistry problem requires a multi-step process: first, find the volume of gold needed to cover a certain area, and then use the density to determine its mass. This mass can then be converted into cost using current market values, which is often given in troy ounces for precious metals.

Unit Conversion

Unit conversion is a basic, yet vital skill in chemistry and many other fields of science and engineering. It allows us to express measurements in different units to facilitate calculations and communicate effectively. Often, we need to convert units to match others in a calculation or to adhere to standard units within a problem.

In our gold leaf problem, unit conversion is a multi-step process. We convert the dimensions of the ceiling from feet to centimeters to match the units for the thickness of the gold leaf, which is given in inches. Understanding the relationship between these units (1 foot = 30.48 centimeters, 1 inch = 2.54 centimeters) is key. Similarly, we must convert the mass of gold from grams to troy ounces, as pricing for gold is typically given per troy ounce. Each conversion must remain consistent and accurate to ensure the final cost calculation is correct.

Mass-Volume Relationship in Chemistry

The mass-volume relationship is an application of density that describes how mass and volume relate to each other. As seen in the gold leaf problem, to calculate the mass, you multiply the volume of the material by its density. When we know two of these values, we can always solve for the third.

This relationship is fundamental in chemistry for tasks like preparing solutions or estimating yields in reactions. For our gold leaf problem, once we have the area of the ceiling and the thickness of the leaf, we obtain the volume of gold needed. Then, by using the density of gold, we calculate the mass. Such practical applications show how theoretical concepts in chemistry are applied in real-world scenarios such as construction, manufacturing, and even in art restoration, where materials must be measured and costs estimated.