Question

An artist's statue has a surface area of \(14.6 \mathrm{ft}^{2}\). The artist plans to apply gold leaf to the statue and wants the coating to be \(2.50 \mu \mathrm{m}\) thick. If the price of gold were \(\$ 1,774.10\) per troy ounce, how much would it cost to give the statue its gold coating? \((1\) troy ounce \(=31.1035 \mathrm{~g}\); the density of gold is \(19.3 \mathrm{~g} / \mathrm{mL}\).)

Short answer

The total cost to cover the statue with gold is calculated by finding the volume of the gold needed, converting it to mass, then finding the equivalent weight in troy ounces and finally multiplying by the cost per troy ounce.

Step-by-step solution

Calculate the Volume of Gold Needed

To calculate the volume of the gold leaf that will cover the statue, use the formula for the volume of a rectangular prism: volume (V) is equal to the surface area (A) times the thickness (t). The thickness must be converted from micrometers to feet: \(2.50 \mu m = 2.50 \times 10^{-6} m\) and then to feet by multiplying by 3.28084 (since 1 meter is equal to 3.28084 feet). The volume in cubic feet is then \(V = A \times t\).

Convert Volume to Cubic Meters

Convert the volume from cubic feet to cubic meters for easier computation with the density, which is given in grams per milliliter. Use the conversion factor \(1 \mathrm{ft}^3 = 0.0283168 \mathrm{m}^3\).

Find the Mass of Gold Needed

To find the mass of gold needed, multiply the volume by the density of gold. The volume should be in cubic meters and the density of gold is given as \(19.3 \mathrm{g}/\mathrm{mL}\), with \(1 \mathrm{mL} = 1\times10^{-3} \mathrm{L} = 1\times10^{-3} \mathrm{m}^3\). The result will give you the mass in grams.

Convert Mass to Troy Ounces

Convert the mass of gold from grams to troy ounces by using the conversion factor given in the problem: \(1 \mathrm{troy \ ounce} = 31.1035 \mathrm{g}\).

Calculate the Cost

To find the total cost to cover the statue, multiply the mass of the gold in troy ounces by the cost per troy ounce. The final result will be the cost in US dollars.

Key concepts

Understanding Surface Area to Volume Ratio

The surface area to volume ratio is a fundamental concept in various scientific fields, including chemistry. It is especially important when dealing with coatings, such as in the example of applying gold leaf to a statue. The surface area represents the total area that is exposed to the external environment, while the volume measures the space that a substance occupies.

To calculate the gold leaf needed, understanding the relationship between the surface area of the object and the volume of coating is crucial. This is because the amount of gold required will directly depend on the volume that needs to be covered, and this volume can be found by multiplying the surface area by the thickness of the coating. A higher surface area to a given volume means a thinner layer of material is necessary to cover the object, which in the case of gold leaf application, can significantly affect the cost.

For the artist's statue with a specific surface area, the gold leaf needed to cover it will have the same surface area but with a very small thickness, making the volume comparatively small. Hence, in cost calculations, we always strive to minimize the volume of expensive materials like gold to reduce overall expenses.

Importance of Unit Conversion

Unit conversion is a crucial skill in chemistry and everyday life. It allows us to translate measurements into different units for accurate comparisons and calculations. When working with units from different measurement systems, like micrometers (µm) and feet, or grams and troy ounces, conversions ensure that computations are correct and meaningful.

For example, in our exercise, the thickness of the gold leaf is given in micrometers, but the surface area of the statue is in square feet. To calculate the volume of the gold leaf needed, we need to convert the thickness to feet before multiplying by the surface area. Misinterpreting units or omitting conversion steps can lead to significant errors, potentially resulting in a costly miscalculation of the amount of gold needed.

Always pay close attention to the units presented and employ the correct conversion factors. Common conversions, such as meters to feet, grams to troy ounces, and cubic centimeters to cubic meters, are essential tools in the problem-solving toolkit of anyone working within the realm of science and mathematics.

Density Calculations in Context

Density calculations are a cornerstone in understanding how much space a certain mass of a substance will occupy and, conversely, the mass of a substance that will fill a particular volume. Density is defined as the mass per unit volume and is expressed in various units like grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³).

In the case of the gold leaf, knowing the density of gold allows us to determine how much mass corresponds to the calculated volume that will cover the statue. By multiplying the volume by the density of gold, we obtain the mass in grams. To proceed with the cost calculation, we convert this mass to trestroy ounces since the price of gold is given per troy ounce.

Density plays a pivotal role in a wide range of applications from industrial to artistic, as it allows us to estimate the quantities of materials needed and their costs. Always remember that when dealing with density, the consistency of units throughout the calculation is just as important as applying the correct values for density itself.