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 g>cm3, how many oxygen atoms does it contain?

Key Concepts

Density and Volume Calculation

Density is defined as mass per unit volume, typically expressed in grams per cubic centimeter (g/cm³). To find the mass of the calcium carbonate cube, we first calculate its volume using the formula for the volume of a cube (V = side³). Given the edge length of 2.005 inches, we convert this measurement to centimeters and then apply the density to find the mass.

Molar Mass and Chemical Composition

Calcium carbonate (CaCO₃) consists of calcium (Ca), carbon (C), and oxygen (O) atoms. To determine the number of oxygen atoms in the sample, we need to know the molar mass of CaCO₃, which is approximately 100.09 g/mol. This allows us to convert the mass of the sample into moles of CaCO₃, from which we can calculate the total number of oxygen atoms, considering that each molecule contains three oxygen atoms.

Avogadro's Number

Avogadro's number (6.022 x 10²³) is the number of particles, typically atoms or molecules, in one mole of a substance. After calculating the moles of calcium carbonate in the sample, we can use Avogadro's number to find the total number of molecules present. Multiplying the number of moles of CaCO₃ by three gives the total number of oxygen atoms in the sample.