Question

Consider a sample of calcium carbonate in the form of a cube measuring 2.005 in. on each edge. If the sample has a density of \(2.71 \mathrm{~g} / \mathrm{cm}^{3},\) how many oxygen atoms does it contain?

Short answer

The calcium carbonate cube contains approximately 6.46 × 10²⁴ oxygen atoms.

Step-by-step solution

Calculate the volume of the cube in cm³

To find the volume, use the formula for the volume of a cube: \( Volume = edge^3 \). First, convert the edge length from inches to centimeters (1 inch = 2.54 cm) Edge length (in cm) = 2.005 in × 2.54 cm/in ≈ 5.0927 cm Now, compute the volume: Volume = (5.0927 cm)³ ≈ 132.1424 cm³

Calculate the mass of the cube

Next, calculate the mass of the sample with the given density: Density = Mass / Volume Rearrange to find mass: Mass = Density × Volume Mass ≈ 2.71 g/cm³ × 132.1424 cm³ ≈ 358.0658 g

Calculate the number of moles of calcium carbonate

To convert from mass to moles of calcium carbonate, use its molar mass of CaCO₃ = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 16.00 g/mol × 3 (O) ≈ 100.09 g/mol: Moles of CaCO₃ = Mass of CaCO₃ / Molar Mass Moles of CaCO₃ ≈ 358.0658 g / 100.09 g/mol ≈ 3.5764 mol

Calculate the number of moles of oxygen

In one mole of calcium carbonate, there are 3 moles of oxygen atoms: Moles of O = 3 × Moles of CaCO₃ Moles of O ≈ 3 × 3.5764 mol ≈ 10.7292 mol

Calculate the number of oxygen atoms

Use Avogadro's number to find the number of oxygen atoms: Number of Oxygen atoms = Moles of oxygen × Avogadro's number Number of Oxygen atoms ≈ 10.7292 mol × 6.022 × 10²³ atoms/mol ≈ 6.46 × 10²⁴ oxygen atoms Therefore, the calcium carbonate cube contains approximately 6.46 × 10²⁴ oxygen atoms.