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 6.50 × 10²4 oxygen atoms.

Step-by-step solution

Calculate the volume of the cube

We are given that each edge of the cube is 2.005 in. We first need to convert this length from inches to centimeters (1 inch = 2.54 cm). Then, we will calculate the volume of the cube by raising the length of one edge to the power of 3. Length in cm = \(2.005 \times 2.54\) cm = 5.0927 cm Volume of the cube = edge length³ = \(5.0927^3\) = 132.65 \(cm^3\)

Determine the mass of the cube using its density

The sample has a density of 2.71 g/ \(cm^3\). Density is defined as mass/volume, so we can calculate the mass by multiplying the density by the volume: Mass of the cube = density × volume = 2.71 g/ \(cm^3 \times 132.65\) \(cm^3\) = 359.63 g

Calculate the moles of calcium carbonate (CaCO3) present in the cube

First, we need to determine the molar mass of CaCO3. We know that the molar mass of Ca is 40.08 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol. Therefore, the molar mass of CaCO3 is: Molar mass of CaCO₃ = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol Now, we can calculate the moles of CaCO3 in the cube by dividing the mass by the molar mass: Moles of CaCO₃ = \( \frac{359.63 \, \text{g}}{100.09 \frac{\text{g}}{\text{mol}}} \) = 3.596 mol

Calculate the number of oxygen atoms using Avogadro's number

In a single molecule of CaCO₃, there are 3 oxygen atoms. So, to find the total number of oxygen atoms in the cube, we multiply the number of moles by Avogadro's number (6.022 × 10²³ atoms/mol) and then multiply by the 3 oxygen atoms: Number of oxygen atoms = moles of CaCO₃ × Avogadro's number × 3 = 3.596 mol × 6.022 × 10²³ atoms/mol × 3 = 6.50 × 10²4 oxygen atoms So, the sample of calcium carbonate contains 6.50 × 10²4 oxygen atoms.

Key concepts

Molar Mass Calculation

Calculating molar mass is essential in stoichiometry, which involves chemical equations and reactions. For any compound, the molar mass is the combined mass of each element's atoms within the compound. Understanding how to calculate it can bridge the gap between mass and moles in chemical calculations. For calcium carbonate (CaCO₃), you start by noting down the atomic masses:

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol

The formula for the compound is crucial. In CaCO₃, you have:

• 1 Calcium atom
• 1 Carbon atom
• 3 Oxygen atoms

To compute the molar mass of CaCO₃, sum up these values:\[40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol}\] This molar mass tells you how much one mole of calcium carbonate weighs, which is fundamental for further calculations involving moles and mass.

Avogadro's Number

Avogadro's number is a cornerstone of stoichiometry. It provides the link between macroscopic quantities of substances and the number of particles they contain. Avogadro's number is \(6.022 \times 10^{23}\), representing the number of atoms, ions, or molecules in one mole of a substance. This makes it possible to calculate the number of particles from a given amount of moles.For example, with CaCO₃:If you know you have 3.596 moles of CaCO₃, you can find the number of oxygen atoms because each mole contains:

• 6.022 × 10²³ molecules of CaCO₃
• 3 oxygen atoms per molecule

So, the total calculation would be:\[3.596 \, \text{mol} \times 6.022 \times 10^{23} \frac{\text{atoms}}{\text{mol}} \times 3 = 6.50 \times 10^{24} \, \text{oxygen atoms}\]Hence, Avogadro's number helps determine the immense quantities of atoms or molecules in bulk substances.

Density and Volume Conversion

Density is one of the key terms in chemistry, used to convert between mass and volume. It's typically in the form of grams per cubic centimeter (g/cm³), signifying how much mass is contained in a specific volume. Here, to solve the problem, you start by converting the length of the cube's edges into centimeters, since the density of the substance is given in these units. Converting from inches to centimeters (since 1 inch = 2.54 cm) is essential before calculating volume:

• For a cube with each side equal to 2.005 inches, the length in centimeters is:
• \[2.005 \, \text{in} \times 2.54 \, \text{cm/in} = 5.0927 \, \text{cm}\]

The volume of the cube:\[(5.0927 \, \text{cm})^3 = 132.65 \, \text{cm}^3\]After obtaining the volume, you can find the mass using density:

• Density of CaCO₃ is 2.71 g/cm³.
• Mass \(= \text{Density} \times \text{Volume} = 2.71 \, \text{g/cm}^3 \times 132.65 \, \text{cm}^3 = 359.63 \, \text{g}\)

Understanding these conversions is critical as it forms the basis for making accurate measurements and calculations in chemistry.