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

Step-by-step solution

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

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

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

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

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.

Key concepts

Moles and Avogadro's Number

Understanding moles is crucial in chemistry as it helps quantify atoms, molecules, and other particles. It's like a bridge between the atomic and macroscopic worlds. A mole represents a unit of measurement that contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\). This number is incredibly large because it reflects the tiny scale at which atoms and molecules operate.

When solving stoichiometry problems, chemists often need to calculate how many particles are in a sample. For example, how many oxygen atoms are in calcium carbonate.
Here, the number of moles of oxygen is multiplied by Avogadro's number to find the total number of oxygen atoms.

• Moles: A measure of quantity used in chemistry.
• Avogadro’s Number: \(6.022 \times 10^{23}\), the number of particles in one mole.

Density Calculations

Density is a measure of how compact the mass in a substance is within a given volume. It is often expressed in units like \( \text{g/cm}^3 \). For example, with a sample of calcium carbonate, understanding its density helps determine its mass from its volume in space.

By calculating density, chemists can deduce several properties about a material such as how it will behave under specific conditions or how much mass a particular volume contains. The formula for density is:
\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]Knowing this relationship is particularly useful in converting between volumes and masses, especially when solving problems related to mass-to-mole conversions.

Unit Conversion

Unit conversion is a fundamental skill in chemistry and science as a whole. It involves translating one unit of measurement into another to ensure consistency, particularly between metric and imperial systems.

Converting inches to centimeters is a common example. Each inch is equal to 2.54 centimeters. When solving problems involving volumes and dimensions, proper unit conversion is essential.
For instance, the cube's edge length in inches needs to be converted to centimeters for subsequent calculations involving density, which is given in \( \text{g/cm}^3 \).

• Inches to Centimeters: 1 inch = 2.54 centimeters.
• Volume Calculations: Converting length units before calculating volume is crucial.

Chemical Composition

Chemical composition refers to the makeup of a substance, specifically the types, number, and arrangement of atoms. It is central to understanding both physical and chemical properties.

In case of calcium carbonate (\( \text{CaCO}_3 \)), the composition is by definition one calcium atom, one carbon atom, and three oxygen atoms per formula unit. Knowing the chemical makeup allows for precise calculations such as deriving mass from given moles using molar mass.
The molar mass for \( \text{CaCO}_3 \) adds up from:

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol each, three needed, totaling 48.00 g/mol

The sum, 100.09 g/mol, is the molar mass used to convert grams to moles. Calculating using the chemical composition ensures accuracy when determining the quantity of each component element in a sample.