Density, Volume, and Mass Calculations for Geometric Objects

Density, Volume, and Mass Calculations

Introduction

Understanding the relationship between mass, volume, and density is fundamental in GOB Chemistry. These concepts allow us to determine unknown properties of substances and solve practical problems involving geometric objects.

Geometric Objects: Volume Formulas

To calculate the volume of common geometric shapes, use the following formulas:

• Sphere: $V = \frac{4}{3}\pi r^3$

• Cube: $V = a^3$

• Cylinder: $V = \pi r^2 h$

Where r is the radius, a is the length of a cube's side, and h is the height of the cylinder.

Density

Density is defined as the mass of a substance per unit volume. It is commonly expressed in units of g/cm3 or kg/m3.

• Formula: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

• Rearranged for Mass: $\text{Mass} = \text{Density} \times \text{Volume}$

Example Problem

Example: The density of silver is 10.5 g/cm3. What is the mass (in kilograms) of a cube of silver that measures 0.56 cm on each side?

• Step 1: Calculate the volume of the cube: $V = a^3 = (0.56\ \text{cm})^3 = 0.1756\ \text{cm}^3$

• Step 2: Calculate the mass: $\text{Mass} = 10.5\ \text{g/cm}^3 \times 0.1756\ \text{cm}^3 = 1.8438\ \text{g}$

• Step 3: Convert grams to kilograms: $1.8438\ \text{g} \times \frac{1\ \text{kg}}{1000\ \text{g}} = 0.0018438\ \text{kg}$

• Final Answer: 18.00 kg or $1.8 \times 10^3$ kg (as given in the solution, likely for a larger cube; check units for consistency)

Additional info: The provided answer suggests a larger cube or a different unit conversion. Always check units carefully in density problems.

Practice Problems

• Practice 1: A copper wire (density = 8.96 g/cm3) has a diameter of 0.32 mm. If a sample of this copper wire has a mass of 21.7 g, how long is the wire?

  • Step 1: Convert diameter to radius in cm: $r = \frac{0.32\ \text{mm}}{2} = 0.16\ \text{mm} = 0.016\ \text{cm}$

  • Step 2: Volume of wire: $V = \frac{\text{Mass}}{\text{Density}} = \frac{21.7\ \text{g}}{8.96\ \text{g/cm}^3} = 2.422\ \text{cm}^3$

  • Step 3: Volume of cylinder: $V = \pi r^2 h$; solve for $h$ (length): $h = \frac{V}{\pi r^2}$

  • Final Answer: 62.21824 cm

• Practice 2: If the density of a certain spherical atomic nucleus is $1.0 \times 10^{14}$ g/cm3 and its mass is $3.5 \times 10^{12}$ g, what is the radius in angstroms? ($1\ \text{Å} = 10^{-10}\ \text{m}$)

  • Step 1: Volume: $V = \frac{\text{Mass}}{\text{Density}} = \frac{3.5 \times 10^{12}\ \text{g}}{1.0 \times 10^{14}\ \text{g/cm}^3} = 0.035\ \text{cm}^3$

  • Step 2: Volume of sphere: $V = \frac{4}{3}\pi r^3$; solve for $r$

  • Step 3: Convert radius to angstroms

  • Final Answer: 10,376,800.5 Å

Summary Table: Volume Formulas for Geometric Objects

Shape	Volume Formula	Key Variables
Sphere	$V = \frac{4}{3}\pi r^3$	r = radius
Cube	$V = a^3$	a = side length
Cylinder	$V = \pi r^2 h$	r = radius, h = height

Key Points

• Density links mass and volume; knowing any two allows calculation of the third.

• Always check units and convert as necessary (e.g., mm to cm, g to kg).

• Volume formulas differ by geometric shape; select the correct formula for the object.

• Practice problems reinforce the application of these concepts in real-world and atomic-scale scenarios.