Problem 165 - General Chemistry

Question

A steel sphere has a radius of 1.58 in. If this steel has a density of \(7.88 \mathrm{~g} / \mathrm{cm}^{3},\) what is the mass of the sphere in grams?

Step-by-Step Solution

Answer

The mass of the steel sphere is approximately 2133.48 grams.

Step 1: Calculate the Volume of the Sphere

The formula to calculate the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). First, convert the radius from inches to centimeters. Since 1 inch is 2.54 cm, we have: \( r = 1.58 \text{ in} \times 2.54 \text{ cm/in} = 4.0132 \text{ cm} \). Now use this radius to find the volume: \( V = \frac{4}{3} \pi (4.0132)^3 \approx 270.782 \text{ cm}^3 \).

Step 2: Apply the Density Formula to Find Mass

Density \( (\rho) \) is defined as mass \( (m) \) per unit volume \( (V) \), expressed as \( \rho = \frac{m}{V} \). To find the mass, rearrange this formula to \( m = \rho \times V \). Using the previously calculated volume \( V = 270.782 \text{ cm}^3 \) and given density \( \rho = 7.88 \text{ g/cm}^3 \), calculate the mass: \( m = 7.88 \times 270.782 \approx 2133.48 \text{ grams} \).

Key Concepts

Volume of a Sphere

The volume of a sphere is the amount of space it occupies in three dimensions. To find the volume, we use the formula:

• \( V = \frac{4}{3} \pi r^3 \)

This formula lets you calculate the volume when you know the radius of the sphere.
Here, \( \pi \) (pi) is a constant, approximately equal to 3.14159, and \( r \) is the radius of the sphere.
The \( \frac{4}{3} \) multiplied by \( \pi \) and \( r^3 \) calculates the cubic (three-dimensional) volume.
To use this formula effectively, first ensure that the radius \( r \) is in the same unit as the volume you want, such as centimeters or inches.

Density Formula

Density is a measure of how much mass is contained in a unit volume of a substance. It is defined by the formula:

• \( \rho = \frac{m}{V} \)

In this formula, \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.
Solving this formula for mass (\( m \)) gives us \( m = \rho \times V \).
This equation shows that if a material has a high density, it has more mass packed in each unit of volume.
We use this formula when we know the density of a material and the volume it occupies, allowing us to calculate its mass easily.

Unit Conversion

Unit conversion involves changing measurements from one unit to another. This process is necessary when units used in the problem differ from the units needed in the solution.
For example, converting the radius of a sphere from inches to centimeters is crucial when calculating volume in cubic centimeters.

• 1 inch equals 2.54 centimeters.

Knowing the conversion factor between inches and centimeters allows for accurate calculations.
In many physics and engineering problems, proper unit conversion ensures all calculations are in compatible units, avoiding errors and simplifying the problem-solving process.

Radius Calculation

Calculating the radius is an initial step in solving problems related to spheres, especially when you already have a measurement in a different unit.
Given a radius in inches, you may need to convert it to centimeters for consistency in calculations.
For our example, if the radius is originally given as 1.58 inches:

• Convert it to centimeters: \( r = 1.58 \times 2.54 = 4.0132 \text{ cm} \)

This conversion ensures that when you apply the volume formula, each measurement fits the required units, delivering an accurate and precise solution.