Problem 16 - Chemistry and Chemical Reactivity

Question

Iron pyrite is often called "fool's gold" because it looks like gold (see page 19 ). Suppose you have a solid that looks like gold, but you believe it to be fool's gold. The sample has a mass of \(23.5 \mathrm{g} .\) When the sample is lowered into the water in a graduated cylinder (see Study Question 15 ), the water level rises from \(47.5 \mathrm{mL}\) to \(52.2 \mathrm{mL}\). Is the sample fool's gold \(\left(d=5.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) or "real" gold \(\left(d=19.3 \mathrm{g} / \mathrm{cm}^{3}\right) ?\)

Step-by-Step Solution

Answer

The sample is likely fool's gold, as its density is approximately 5.00 g/cm³.

Step 1: Determine the Volume of the Solid

The volume of the solid can be calculated by subtracting the initial water level from the final water level in the graduated cylinder. Given that the initial water level is 47.5 mL and the final water level is 52.2 mL, the volume of the solid is \(52.2 \, \mathrm{mL} - 47.5 \, \mathrm{mL} = 4.7 \, \mathrm{mL}\).

Step 2: Calculate the Density of the Solid

Density is defined as mass divided by volume. Given that the mass of the sample is 23.5 g and the volume is 4.7 mL, the density can be calculated as \( \text{Density} = \frac{23.5 \, \mathrm{g}}{4.7 \, \mathrm{mL}} \approx 5.00 \, \mathrm{g/cm}^3\).

Step 3: Compare the Density to Known Values

Compare the calculated density of the solid with the densities of iron pyrite and gold. The density of iron pyrite is \(5.00 \, \mathrm{g/cm}^3\) and the density of gold is \(19.3 \, \mathrm{g/cm}^3\). Since the calculated density matches the density of iron pyrite, the sample is likely iron pyrite, also known as "fool's gold."

Key Concepts

Density Calculation

Understanding the concept of density is crucial in identifying substances based on their mass and volume. Density is a measure of how much mass is contained in a given unit of volume. It is expressed mathematically as:

• \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)

In our example, the mass of the sample is given as \(23.5 \, \text{g}\) and its volume has been determined to be \(4.7 \, \text{mL}\). By substituting these values into the formula, the density is calculated to be approximately \(5.00 \, \text{g/cm}^3\). This calculation reveals how density can help distinguish between different materials by comparing this result to known density values of other substances.

Volume Measurement

Volume measurement is a vital component in calculating the density of a substance. In our exercise, the volume of the solid was found using water displacement in a graduated cylinder. Here's how it works:

• First, record the initial water level in the graduated cylinder before the solid is added. In this case, it was \(47.5 \, \text{mL}\).
• Then, gently lower the solid into the cylinder and note the new water level, which was \(52.2 \, \text{mL}\) here.
• The volume of the solid is simply the difference between these two levels: \(52.2 \, \text{mL} - 47.5 \, \text{mL} = 4.7 \, \text{mL}\).

This method works because the volume of water displaced by the solid is equal to the volume of the solid itself, assuming no water is spilled out due to the immersion.

Material Identification

Once the density of a sample is known, it can be compared to the known densities of various materials to identify what the sample is likely to be. This is an essential step in distinguishing substances that may appear similar in appearance.

• For example, gold has a density of \(19.3 \, \text{g/cm}^3\), which is notably higher than that of iron pyrite, \(5.00 \, \text{g/cm}^3\).
• In our scenario, the calculated density of the sample was approximately \(5.00 \, \text{g/cm}^3\), perfectly matching the density of iron pyrite, commonly known as "fool's gold."

By using density values, one can make informed conclusions about the composition of a material, even when they are visually deceiving.