The cubic volume is defined to be 0.478 nm along each edge.

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                                                                        Andy Somody

                                                                        97300-6222

                                                                        ENSC 330

Assignment #3

1).

The cubic volume is defined to be 0.478 nm along each edge. Since the unit cell has a cubic configuration, the volume of the unit cell can be determined by the following formula:

Since each of the sides of the unit cell are defined to be 0.478 nm in length, the volume of the unit cell of CaO can be calculated to be:

with significant figures applied

The question also states that the cubic volume of CaO contains four Ca2+ ions and four O2- ions. The mass of the unit cell contributed by the ions is given by the following formula:

Therefore, the contribution to the mass of the unit cell from each of the Ca2+ and O2- ions is given by modifying the above formula in the following manner.

The mass of one mole of Ca2+ ions is equal to the atomic mass of calcium. The atomic mass of calcium is given as 40.08 g/mol. Similarly, the mass of one mole of O2- ions is equal to the atomic mass of oxygen. The atomic mass of oxygen is given as 15.999 g/mol. The number of atoms or ions in one mole of any element (or the number of molecules in one mole of any compound) is defined to be equal to Avogadro’s number (Van Vlack, 1989). Avogadro’s number states that a mole of any element possesses 6.02 x 1023 atoms or ions of that element (Van Vlack, 1989). Substituting these values into the above equations, we can calculate the contribution to the mass of the CaO unit cell from the Ca2+ and O2- ions.

with significant figures applied

with significant figures applied

Therefore, the total mass of the unit cell of CaO can be calculated by summing the mass component of the Ca2+ ions and the mass component of the O2- ions. This is determined in the following manner:

with significant figures applied

The mass/volume of a material is defined to be equal to its density. We can state this relationship in the following manner.

When considering the mass within a unit cell of a material and the volume within a unit cell of a material, a more specific definition of density can be stated.

Since we have obtained the values for the mass of a unit cell of CaO and the volume of a unit cell of CaO, we can substitute these values into the above equation to solve for the density of CaO.

with significant figures applied

2).

a).

The mass of a unit cell can be calculated by the following formula.

The mass of one mole of tin atoms is equal to the atomic mass of tin. The atomic mass of tin is given as 118.69 g/mol. The question states that tin possesses 4 atoms/unit cell. As explained in question #1, the number of atoms or ions in one mole of any element (or the number of molecules in one mole of any compound) is equal to Avogadro’s number (Van Vlack, 1989). Avogadro’s number states that a mole of any element possesses 6.02 x 1023 atoms or ions of that element (Van Vlack, 1989). Using these values in the above equation, we can calculate the mass of a unit cell of tin:

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with significant figures applied

The volume of a unit cell can be calculated by the following formula.

Since the mass/volume of a material is defined to be its density, the above formula can be restated in the following manner.

The density of tin is defined to be equal to 7.17 g/cm3. Thus, using our previous result for the mass in a unit cell of tin, we can use the above formula to calculate the volume of a unit cell of tin.

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