Problem 31 - Chemistry: The Central Science

Question

Gold can be hammered into extremely thin sheets called gold leaf. If a 200-mg piece of gold (density \(=19.32 \mathrm{~g} / \mathrm{cm}^{3}\) ) is hammered into a sheet measuring \(2.4 \times 1.0 \mathrm{ft}\), what is the average thickness of the sheet in meters? How might the thickness be expressed without exponential notation, using an appropriate metric prefix?

Step-by-Step Solution

Answer

The average thickness of the gold leaf is approximately 46.5 nanometers.

Step 1: Calculate the volume of the gold piece

To find the volume, we can use the formula \(V = \frac{m}{\rho}\), where \(V\) is volume, \(m\) is mass, and \(\rho\) is density. Given the mass (200 mg) and the density (19.32 g/cm³), we can determine the volume. First, we need to convert the mass into grams: \(m = 200~mg = 0.2~g\) Now, we can find the volume: \(V = \frac{0.2\,\text{g}}{19.32\,\text{g}/\text{cm}^3} = 0.01036~\text{cm}^3\)

Step 2: Determine the surface area of the gold leaf

We are given the dimensions of the gold leaf in feet, so we need to convert them into centimeters. 1 foot = 30.48 cm, so the gold leaf dimensions are \(2.4 \times 1.0~ft = 73.152 \times 30.48~cm\). Next, we calculate the surface area (A) by multiplying the length (L) and width (W) dimensions: \(A = L \times W = 73.152\,\text{cm} \times 30.48\,\text{cm} = 2229.26528\,\text{cm}^2\)

Step 3: Calculate the thickness of the gold leaf

To find the average thickness (t) of the gold leaf, we can use the formula: \(t = \frac{V}{A}\), where V is the volume of the gold piece and A is the surface area of the gold leaf. \(t = \frac{0.01036\,\text{cm}^3}{2229.26528\,\text{cm}^2} = 4.6532 \times 10^{-6}\,\text{cm}\) Now we need to convert the thickness into meters. 1 cm = 0.01 m, so \(t = 4.6532 \times 10^{-6}~\text{cm} \times 0.01 = 4.6532 \times 10^{-8}\,\text{m} \)

Step 4: Express the result using an appropriate metric prefix

To express the thickness without exponential notation, we can convert meters to nanometers. 1 meter = \(10^9\) nanometers, so \(t = 4.6532 \times 10^{-8}\,\text{m} \times 10^9 = 46.532\,\text{nm}\) The average thickness of the gold leaf is approximately 46.5 nanometers.

Key Concepts

Density and Volume Calculations

To solve the gold leaf thickness problem, understanding density and volume calculations is crucial. Density is defined as mass per unit volume and can be expressed using the equation \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume of a substance.In the exercise, we are given the mass of gold as 200 mg. Before using it in calculations, it's essential to convert this mass into grams, since the density is provided in g/cm³. Therefore, 200 mg equals 0.2 grams.Once the mass is in the correct units, we use the formula for volume: \( V = \frac{m}{\rho} \). By plugging in the mass (0.2 grams) and the density (19.32 g/cm³), we calculate the volume of the gold piece as approximately 0.01036 cm³.
With this information, a link between mass, density, and volume is established, facilitating further calculations such as determining thickness.

Metric System Conversions

Metric system conversions are indispensable in science and engineering. They help us easily translate measurements into useful forms. In this exercise, we need to convert several measurements to ensure all calculations are consistent and correctly processed. For instance, we start by converting the gold leaf's dimensions from feet to centimeters. Since 1 foot equals 30.48 cm, converting the dimensions of 2.4 ft by 1 ft yields 73.152 cm by 30.48 cm. After calculating, the surface area in centimeters allows us to continue with other calculations, such as finding thickness. Additionally, understanding the conversion of units of thickness is crucial. The thickness can first be determined in centimeters and later converted to meters and also expressed in nanometers for more practical usage.
This flexibility in conversion makes metric systems highly valuable for accuracy and easy interpretation of results.

Surface Area and Volume Formulas

In calculating the thickness of a gold leaf, knowledge of surface area and volume formulas is essential. The surface area formula for a rectangle is given by \( A = L \times W \), where \( L \) is the length, and \( W \) is the width.For this exercise, after converting the dimensions to centimeters, we calculate the surface area to be 2229.26528 cm². This area is used to find the thickness of the gold leaf by dividing the volume of the gold by its surface area.The formula \( t = \frac{V}{A} \) for thickness indicates that by knowing the volume of the gold piece and the surface area of the leaf, the thickness can be calculated efficiently. This understanding of surface area and volume equations allows these two measurements to be used together, determining not just the thickness, but also illustrating properties about the sheet that might impact its physical applications.
These formulas are fundamental in many areas of physics and engineering, providing concise yet powerful tools for analysis and design.