Problem 105 - Introductory Chemistry Atoms First

Question

Gold has a density of \(19.3 \mathrm{~g} / \mathrm{mL}\). Suppose you have \(100.0\) glonkins of gold. What volume in liters will the gold occupy? Here are some conversion factors to help you: \(0.911\) ounce per glonkin and \(28.35 \mathrm{~g}\) per ounce. Use unit analysis to calculate your answer, and show your work. Treat both conversion factors as exact.

Step-by-Step Solution

Answer

The gold will occupy approximately \(0.137\) liters of volume.

Step 1: Use the conversion factor of \(0.911\) ounce per glonkin to convert the given gold mass of \(100.0\) glonkins to ounces: \(100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}}\) #Step 2: Convert ounces to grams#:

Use the conversion factor of \(28.35\) grams per ounce to convert the mass in ounces to grams: \(100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}\) #Step 3: Find the volume in milliliters#:

Step 2: Divide the mass in grams by the density of gold (19.3 g/mL) to get the volume in milliliters: \(\frac{100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}}{19.3\frac{\text{g}}{\text{mL}}}\) #Step 4: Convert volume in milliliters to liters#:

Use the conversion factor of \(1\) liter \(=1000\) milliliters to convert volume in milliliters to liters: \(\frac{100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}}{19.3\frac{\text{g}}{\text{mL}}} \times \frac{1\,\text{L}}{1000\,\text{mL}}\) Now, calculate the volume of gold in liters: \(\frac{100.0\,\text{glonkins} \times 0.911\,\text{ounce} \times 28.35\,\text{g}}{19.3\,\text{g/mL} \times 1000\,\text{mL}} \approx 0.137\,\text{L}\) So, the gold will occupy approximately \(0.137\) liters of volume.

Key Concepts

Unit Conversion

Unit conversion is a useful mathematical technique that allows us to change one unit of measurement to another, using specific conversion factors. This skill is particularly handy in chemistry and physics where different measurement systems are often used. For this exercise, we started with glonkins — an imaginary unit.
Here’s how we converted it:

• Convert glonkins to ounces using the factor 0.911 ounce per glonkin. This step translates the imaginary mass measurement (glonkins) into a more standard unit (ounces).


• Then, convert ounces to grams using the factor 28.35 grams per ounce. Gram is the preferred metric unit for mass, which simplifies further calculations.

Each conversion factor is considered to be exact, meaning no additional sources of error are introduced during the conversion process. By multiplying these conversion factors sequentially with the initial value, we effectively alter the unit of the initial quantity.

Mass to Volume Conversion

Mass to volume conversion involves taking the mass of a substance and translating it into the volume that substance occupies, using its density. This type of conversion is essential in sciences where understanding how much space a material takes up can be just as important as how much of it there is.
In our problem, we needed to find the volume in milliliters of gold given its mass in grams.

• First, determine the mass of gold in grams after unit conversion from glonkins.


• Then, use the density of gold, which is 19.3 grams per milliliter, as the conversion factor. This density tells us how much 1 mL of gold weighs.

By dividing the mass by the density, we get the volume in milliliters. Finally, to convert the volume into liters, which is a more convenient unit for larger quantities, we use the conversion factor that 1 liter equals 1000 milliliters.

Density of Gold

The density of gold is a fundamental property that describes how tightly packed gold atoms are. It is given as 19.3 grams per milliliter (g/mL). Understanding density helps explain why gold feels heavier than many other materials, even if they're the same size.
Density is calculated from the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In essence, it tells us how much mass exists in a particular volume of a substance.

• High density implies that the material has a lot of mass in a small volume, like gold.


• A lower density indicates that the material is lighter and less compact.

In contexts like this exercise, knowing the density allows us to convert between mass and volume efficiently. The 19.3 g/mL value is specific enough to help make accurate conversions, helping us determine just how much space a given mass of gold will occupy. This calculation plays a crucial role in industries that require precious metal handling.