Question

To determine the approximate mass of a small spherical shot of copper, the following experiment is performed. When 125 pieces of the shot are counted out and added to \(8.4 \mathrm{mL}\) of water in a graduated cylinder, the total volume becomes \(8.9 \mathrm{mL}\). The density of copper is \(8.92 \mathrm{g} / \mathrm{cm}^{3} .\) Determine the approximate mass of a single piece of shot, assuming that all of the pieces are of the same dimensions.

Short answer

The approximate mass of a single piece of shot is \(0.03568 \mathrm{g}\).

Step-by-step solution

Determine the Volume of the Shot Pieces

The volume of the shot pieces can be determined using volume displacement. When the shot pieces are added to the water, they displace a volume of water equal to their own volume. The initial volume of water was \(8.4 \mathrm{mL}\), and the final volume is \(8.9 \mathrm{mL}\). Therefore, the volume of the shot pieces is the difference, which is \((8.9-8.4) \mathrm{mL}=0.5 \mathrm{mL}\) or \(0.5 \mathrm{cm}^{3}\).

Calculate the Collective Mass of the Shot Pieces

The mass of the shot pieces can be calculated using the formula mass = volume × density, where the density of copper is given as \(8.92 \mathrm{g} / \mathrm{cm}^{3}\). So, the mass = \(0.5 \mathrm{cm}^3\) × \(8.92 \mathrm{g} / \mathrm{cm}^3\)= 4.46 g.

Calculate the Mass of a Single Shot Piece

To calculate the mass of a single shot piece, divide the total mass by the number of pieces. Hence, the mass of a single shot piece is \(4.46 \mathrm{g} / 125 = 0.03568 \mathrm{g}\).

Key concepts

Volume displacement

When we say "volume displacement," we're talking about a very useful method to measure the volume of irregularly shaped objects. This method is particularly handy when you don't have all the dimensions of the object needed for traditional volume calculations. Imagine you have some tiny copper spheres. You could just submerge them in water and observe how much the water level rises. The amount the water level increases by is equal to the volume of the objects submerged.

• Initial volume of water: 8.4 mL
• Final volume with shot added: 8.9 mL
• Displaced volume: 8.9 mL - 8.4 mL = 0.5 mL

In this case, the volume displaced by the copper shot is 0.5 mL which is equivalent to 0.5 cm³ because 1 mL equals 1 cm³. This is the volume of the 125 copper pieces together.

Mass calculation

Once you have the volume of the shot pieces from the displacement method, the next step is to figure out their mass. This involves using the formula for density, which is mass divided by volume. We can rearrange this to find the mass if we know the density and volume:

\[\text{Mass} = \text{Volume} \times \text{Density}\]

For copper, the density is given as 8.92 g/cm³. Using the volume calculated earlier (0.5 cm³), you can multiply the two to get the mass of all 125 pieces of copper.

$\backslash text\{Total\; Mass\}\; =\; 0.5\; \backslash text\{\; cm\}^3\; \backslash times\; 8.92\; \backslash text\{\; g/cm\}^3\; =\; 4.46\; \backslash text\{\; g\}$
This shows that the combined mass of all the copper shot is 4.46 grams. To find the mass of a single piece of shot, you divide the total mass by the number of pieces: 4.46 g ÷ 125 = 0.03568 g per piece. This calculation helps understand the individual mass of each copper shot efficiently.

Copper properties

Copper is a fascinating and highly useful metal with many properties that make it perfect for various applications. Understanding these properties helps explain why objects made from copper behave the way they do in experiments like our volume and mass calculations.

Some important properties of copper include:

• **Excellent Conductivity**: Copper is an excellent conductor of electricity, which is why it's used in electrical wiring.
• **High Density**: With a density of 8.92 g/cm³, copper is relatively heavy, which is evident when calculating the mass of small copper objects.
• **Malleability and Ductility**: Copper can be easily shaped and drawn into wires, which contributes to its widespread use in industries.

The density property is particularly relevant to our exercise. Knowing the density helps predict how much an object will weigh without needing to directly weigh it, which is crucial for precise calculations.