The volume of a unit cell of diamond is 0.0454 nm³, and the density of diamond is 3.52 g/cm³. Find the number of carbon atoms in a unit cell of diamond.

Name: Chemistry: A Molecular Approach
Author: Tro
ISBN: 9780137832217

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Convert the volume of the unit cell from nm³ to cm³ by using the conversion factor: 1 nm = 10⁻⁷ cm.

Calculate the mass of the unit cell using the formula: mass = density × volume.

Use the molar mass of carbon (12.01 g/mol) to convert the mass of the unit cell to moles of carbon atoms.

Use Avogadro's number (6.022 × 10²³ atoms/mol) to convert moles of carbon atoms to the number of carbon atoms.

Determine the number of carbon atoms in the unit cell by considering the structure of diamond, which is a face-centered cubic lattice with additional atoms inside the cell.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Unit Cell

A unit cell is the smallest repeating unit in a crystal lattice that reflects the symmetry and structure of the entire crystal. In the case of diamond, which has a face-centered cubic (FCC) structure, the unit cell contains a specific arrangement of atoms that defines its properties. Understanding the unit cell is crucial for calculating the number of atoms it contains.

Density and Volume Relationship

Density is defined as mass per unit volume, typically expressed in grams per cubic centimeter (g/cm³). To find the mass of the unit cell, the formula mass = density × volume can be applied. This relationship is essential for determining how many atoms are present in the unit cell based on the mass calculated from the given density and volume.

Avogadro's Number

Avogadro's number, approximately 6.022 × 10²³, is the number of atoms or molecules in one mole of a substance. This concept is vital for converting the mass of carbon in the unit cell to the number of carbon atoms. By using the molar mass of carbon, one can determine how many atoms correspond to the mass calculated from the density and volume of the unit cell.

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Textbook Question

Copper iodide crystallizes in the zinc blende structure. The sep- aration between nearest neighbor cations and anions is approximately 311 pm, and the melting point is 606 °C. Potassium chloride, by contrast, crystallizes in the rock salt structure. Even though the separation between nearest-neighbor cations and anions is greater (319 pm), the melting point of potassium chlo- ride is higher (776 °C). Explain.

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