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

Expert verified
The sample of calcium carbonate contains approximately 6.49 x 10²4 oxygen atoms.

Step by step solution

01

Calculate the cube's volume

We are given that the cube has an edge length of 2.005 in. First, convert this length to centimeters: \(1 \mathrm{~in} = 2.54 \mathrm{~cm}\) (2.005 in.) x (2.54 cm/in.) = 5.0927 cm Now, calculate the volume of the cube: (volume of a cube) = (edge length)³ volume = (5.0927 cm)³ = 132.6316 cm³
02

Find the mass of the sample using its volume and density

Use the formula: mass = (volume) x (density) mass = (132.6316 cm³) x (2.71 g/cm³) = 359.6320 g
03

Calculate the number of moles of calcium carbonate in the sample

Use the molar mass of calcium carbonate (CaCO₃) to convert the mass to moles: molar mass of CaCO₃ = 40.08 g (Ca) + 12.01 g (C) + (3 x 16.00 g (O)) = 100.09 g/mol number of moles of CaCO₃ = (mass of CaCO₃) / (molar mass of CaCO₃) number of moles of CaCO₃ = (359.6320 g) / (100.09 g/mol) = 3.5953 mol
04

Calculate the number of moles of oxygen in the sample

In one mole of calcium carbonate, there are three moles of oxygen (due to the formula CaCO₃). Multiply the moles of CaCO₃ by three to find the moles of oxygen: number of moles of O = (3.5953 mol of CaCO₃) x (3 mol of O per mol of CaCO₃) = 10.7860 mol of O
05

Convert moles of oxygen to the number of oxygen atoms using Avogadro's constant

Use Avogadro's constant (6.022 x 10²³ atoms/mol) to convert moles of oxygen to atoms: number of oxygen atoms = (10.7860 mol of O) x (6.022 x 10²³ atoms/mol) = 6.492 x 10²4 atoms The sample of calcium carbonate contains approximately 6.49 x 10²4 oxygen atoms.

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