Problem 61 - Chemistry: Principles and Reactions

Question

Lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) and oxygen has a density of \(1.31 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^{3}\) at room temperature. How many \(\mathrm{cm}^{3}\) are occupied by \(1 \mathrm{~g}\) of lead? By \(1 \mathrm{~g}\) of oxygen? Comment on the difference in volume for the two elements.

Step-by-Step Solution

Answer

Question: Compare and comment on the difference in volume occupied by 1 gram of lead and 1 gram of oxygen based on their densities. Answer: 1 gram of lead occupies a volume of approximately 0.0882 cm³, while 1 gram of oxygen occupies a much larger volume of approximately 763.4 cm³. This significant difference in volume is due to the much lower density of oxygen compared to lead. In other words, oxygen is much less dense than lead, which makes it occupy a larger volume for the same mass.

Step 1: Understand the definition of density

The density formula is given by \(\rho = \frac{m}{V}\), where \(\rho\) is density, \(m\) is mass, and \(V\) is volume. We'll use this formula to find the volume of each element.

Step 2: Calculate volume of 1 gram of lead

We are given the density of lead (\(\rho_{lead} = 11.34 \mathrm{~g}/\mathrm{cm}^{3}\)) and are asked to find the volume (\(V_{lead}\)) occupied by 1 gram of lead. Using the density formula, we can write the equation: $$\rho_{lead} = \frac{1 \mathrm{~g}}{V_{lead}}$$Solving for \(V_{lead}\), we get:$$V_{lead} = \frac{1 \mathrm{~g}}{11.34 \mathrm{~g}/\mathrm{cm}^{3}}$$Now, calculate the value:$$V_{lead} \approx 0.0882 \mathrm{~cm}^{3}$$

Step 3: Calculate volume of 1 gram of oxygen

Similarly, we are given the density of oxygen (\(\rho_{oxygen} = 1.31 \times 10^{-3} \mathrm{~g}/\mathrm{cm}^{3}\)) and are asked to find the volume (\(V_{oxygen}\)) occupied by 1 gram of oxygen. Using the density formula, write the equation:$$\rho_{oxygen} = \frac{1 \mathrm{~g}}{V_{oxygen}}$$Solving for \(V_{oxygen}\), we get:$$V_{oxygen} = \frac{1 \mathrm{~g}}{1.31 \times 10^{-3} \mathrm{~g}/\mathrm{cm}^{3}}$$Now, calculate the value:$$V_{oxygen} \approx 763.4 \mathrm{~cm}^{3}$$

Step 4: Compare the volumes and comment on the difference

We have calculated the volumes occupied by 1 gram of lead and 1 gram of oxygen:$$V_{lead} \approx 0.0882 \mathrm{~cm}^{3}$$$$V_{oxygen} \approx 763.4 \mathrm{~cm}^{3}$$Clearly, the volume occupied by 1 gram of oxygen is significantly larger than that occupied by 1 gram of lead. This difference in volume is due to the much lower density of oxygen compared to lead. In other words, oxygen is much less dense than lead, which makes it occupy a larger volume for the same mass.

Key Concepts

Volume Calculation

Volume calculation is crucial in understanding how much space an object or substance occupies. To determine the volume (\(V\)) of an object when its mass (\(m\)) and density (\(\rho\)) are known, you can rearrange the density formula to solve for volume:

• Density formula: \(\rho = \frac{m}{V}\)
• Rearranged to find volume: \(V = \frac{m}{\rho}\)

For example, if you have 1 gram of lead and its density is 11.34 g/cm³, you can calculate the volume it occupies by
plugging the values into the formula:
\[V_{lead} = \frac{1 \mathrm{~g}}{11.34 \mathrm{~g}/\mathrm{cm}^3} \approx 0.0882 \mathrm{~cm}^3\]This calculation tells you that 1 gram of lead doesn't take up much space, due to its high density.

Density Formula

The density formula is a fundamental concept in science that relates mass and volume. Density (\(\rho\)) is defined
as the mass (\(m\)) of a substance divided by its volume (\(V\)):
\[ \rho = \frac{m}{V} \]This tells us that density is a measure of how tightly matter is packed in a given space. The unit of density usually combines
the units of mass (such as grams) with units of volume (such as cubic centimeters or cm³).
Using the density formula, if you know any two of these three quantities (mass, volume, and density), you can easily find the third.
This makes the density formula extremely useful when conducting scientific calculations. For instance, in the original problem,
we were able to determine how much space 1 gram of oxygen occupies by applying this formula:\[V_{oxygen} = \frac{1 \mathrm{~g}}{1.31 \times 10^{-3} \mathrm{~g}/\mathrm{cm}^{3}}\]

Comparison of Densities

Comparing densities helps us understand why substances behave differently in terms of weight and volume. For example,
in the exercise above, the density of lead is much greater than that of oxygen. Lead has a density of 11.34 g/cm³,
while oxygen's density is a mere 1.31 x 10⁻³ g/cm³.

• Due to lead's higher density, it occupies a smaller volume (0.0882 cm³), even if it weighs the same as 1 gram of oxygen.
• Conversely, oxygen, with its lower density, occupies a significantly larger volume of 763.4 cm³.

This vast difference highlights how different densities affect how much space the same mass of various materials will take up.
Understanding these differences is crucial in fields like material science and engineering, where the physical properties
of materials must be considered. The density comparison clearly shows why heavier, denser materials like lead will
often sink, while lighter, less dense gases like oxygen will seem to "float."