Problem 160 - General Chemistry

Question

Prospectors are considering searching for gold on a plot of land that contains \(1.31 \mathrm{~g}\) of gold per bucket of soil. If the volume of the bucket is \(4.67 \mathrm{~L}\), how many grams of gold are there likely to be in \(2.38 \times 10^{3}\) cubic feet of soil?

Step-by-Step Solution

Answer

There are approximately 18,909.3 grams of gold in the soil.

Step 1: Convert cubic feet to liters

Since the volume of soil available is given in cubic feet and the bucket's volume is in liters, we first convert cubic feet to liters. We know that 1 cubic foot is approximately 28.3168 liters. Thus, the total liters of soil is calculated as: \[2.38 \times 10^3 \text{ cubic feet} \times 28.3168 \frac{\text{liters}}{\text{cubic foot}} = 67,404.064 \text{ liters}\]

Step 2: Calculate the number of buckets in the soil

We need to find out how many buckets of soil there would be in 67,404.064 liters. Since each bucket holds 4.67 liters, the number of buckets is: \[ \frac{67,404.064 \text{ liters}}{4.67 \text{ liters per bucket}} \approx 14,430 \text{ buckets}\]

Step 3: Calculate grams of gold in the total soil

Now, knowing that there are approximately 1.31 grams of gold per bucket, we calculate the total gold in the total soil. The total amount of gold is given by:\[14,430 \text{ buckets} \times 1.31 \frac{\text{grams of gold}}{\text{bucket}} \approx 18,909.3 \text{ grams of gold}\]

Key Concepts

Unit Conversion

Understanding unit conversion is essential in stoichiometry because measurements may initially be provided in various units.
To solve problems efficiently, we must convert these measurements to compatible units.
Let's break this down into a simple process:

• Identify the units you are converting from and to.
• Find the conversion factor between these units.
• Apply the conversion factor to change the units.

For example, in the original problem, you convert cubic feet to liters because one unit expresses volume in cubic feet, while another uses liters.
Knowing that 1 cubic foot equals approximately 28.3168 liters allows you to perform a straightforward multiplication to achieve this conversion!

Volume Calculation

Volume calculation helps determine the space occupied by a substance or item.
In stoichiometry, interpreting or measuring volumes accurately is crucial.
This task typically involves computing how much space a particular object occupies, which can vary depending on the units.

• Determine the volume of the substance.
• Understand the appropriate unit for measurement (liters, cubic feet, etc.).
• Use relevant formulas and conversion factors if needed.

In the exercise, you calculate the volume of soil you have in liters by converting it from cubic feet.
You then find how many buckets this volume equals by dividing the entire volume in liters by the volume of one bucket, clearly showcasing the importance of precise volume calculation.

Mass Calculation

Mass calculation is a core element when handling chemical quantities in stoichiometry.
It determines how much of a substance you have, usually in grams or kilograms.
To calculate mass:

• Obtain an accurate mass per unit of volume or item.
• Multiply this value by the number of units or total volume.

In our example, the bucket contains 1.31 grams of gold.
By knowing how many buckets fit into the total volume of soil, multiplying by the mass of gold in each bucket results in calculating the entire mass of gold available.

Density

Density relates mass to volume and signifies how compact a substance is.
It is expressed as mass per unit volume, such as grams per liter or kilograms per cubic meter.
It's important because:

• It helps identify substances.
• It's used in converting between mass and volume.
• It aids in comparing material properties.

In stoichiometry problems, understanding density assists in translating between known values, like converting volume to mass when the density is known.
Though density isn't directly involved in the provided exercise, knowing it would facilitate more advanced calculations if the gold's distinct mass-to-volume relationship had been a factor.