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Metallic iron crystallizes in a cubic lattice. The unit cell edge length is \(287 \mathrm{pm}\). The density of iron is \(7.87 \mathrm{~g} /\) \(\mathrm{cm}^{3}\). How many iron atoms are within a unit cell?

Short Answer

Expert verified
There are 2 iron atoms in the unit cell.

Step by step solution

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01

Convert Units

First, let's convert the unit cell edge length from picometers to centimeters since density is given in grams per cubic centimeter. 1 pm = \(10^{-12} \text{m} = 10^{-10} \text{cm}\)Therefore, \(287 \mathrm{pm} = 287 \times 10^{-10} \mathrm{cm} = 2.87 \times 10^{-8} \mathrm{cm}\).
02

Calculate Volume of the Unit Cell

Next, calculate the volume of the cubic unit cell. Since the unit cell is a cube, its volume is the cube of its edge length.\[ V = a^3 = (2.87 \times 10^{-8} \mathrm{cm})^3 = 2.36 \times 10^{-23} \mathrm{cm}^3 \]
03

Calculate Mass of the Unit Cell

The mass of a single unit cell can be calculated using its volume and the density of iron.\[ \text{Mass of unit cell} = \text{Density} \times \text{Volume} = 7.87 \mathrm{g/cm}^3 \times 2.36 \times 10^{-23} \mathrm{cm}^3 = 1.85732 \times 10^{-22} \mathrm{g} \]
04

Calculate Moles of Iron in the Unit Cell

We know that the molar mass of iron is approximately 55.85 g/mol. To find the moles of iron in a unit cell, divide the mass of the unit cell by the molar mass of iron using the formula:\[ \text{Moles of Fe} = \frac{\text{Mass of unit cell}}{\text{Molar mass of Fe}} = \frac{1.85732 \times 10^{-22} \mathrm{g}}{55.85 \mathrm{g/mol}} = 3.3253 \times 10^{-24} \mathrm{mol} \]
05

Calculate Number of Atoms in the Unit Cell

Now, convert the moles of iron to atoms using Avogadro's number, which is \(6.022 \times 10^{23} \text{atoms/mol}\).\[ \text{Number of atoms} = (3.3253 \times 10^{-24} \mathrm{mol}) \times (6.022 \times 10^{23} \mathrm{atoms/mol}) = 2 \text{atoms} \]

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Cell
In crystallography, the term 'unit cell' is fundamental. A cubic unit cell is the smallest section of a crystal repeated in three dimensions to create the entire lattice structure. Each point of the lattice is called a lattice point, coinciding with an atom or a group of atoms.
The unit cell is defined by its edge length, denoted 'a'. For a cubic lattice, this means that all the edges are equal. So, the volume of a unit cell is simply the length of one edge cubed. For iron with an edge length of 287 pm, conversion to centimeters and calculating the volume is crucial, as it helps in further computations, such as finding density and mass.
Understanding the structure of the unit cell helps us predict properties like mechanical strength, electrical conductivity, and more. The arrangement also aids in visualizing how atoms interact within the material's framework, which in the case of iron, helps determine the number of atoms within each unit cell.
Density of Iron
Density is a primary physical property representing how much mass is contained in a given volume. It's calculated as mass divided by volume. For iron, the density is given as 7.87 g/cm³. This figure represents the mass of iron per cubic centimeter.
In the context of the exercise, the density provides a bridge between mass and volume data. By knowing the cube's volume (from the unit cell calculations) and using the density, we can derive the mass of the unit cell. This mass calculation is crucial for determining the number of atoms within the cubic structure since the scenario requires us to find out how many atoms occupy the space provided by the unit cell.
It's interesting to note that density is also affected by crystal structure. For iron, which crystallizes in a body-centered cubic structure, its density helps us understand how tightly its atoms are packed. This insight is useful in fields like material science and engineering, where knowing this property can impact product development and materials choice.
Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry, often defined as the number of constituent particles (usually atoms or molecules) in one mole of a substance. It is approximately \(6.022 \times 10^{23}\). This vast number serves as a bridge between the observable macroscopic world and the world of atoms and molecules at the microscopic scale.
In the context of unit cells and lattice structures, Avogadro's Number enables the conversion from moles of a substance to actual numbers of atoms. For this exercise, once we have calculated the moles of iron per unit cell using the unit cell's mass, multiplying by Avogadro’s Number gives the exact number of atoms in the unit cell.
  • This approach is essential because direct counting at the atomic scale is unfeasible.
  • It enables chemists and scientists to communicate about atomic-scale quantities in comprehensible human terms.
Thus, Avogadro's Number is a powerful tool in modern science, linking theoretical chemistry concepts to real-world experimental data and facilitating better comprehension and application of molecular and atomic interactions.

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Most popular questions from this chapter

The boiling point and freezing point of sulfur dioxide are \(-10^{\circ} \mathrm{C}\) and \(-72.7^{\circ} \mathrm{C}\) (at \(\left.1 \mathrm{~atm}\right)\), respectively. The triple point is \(-75.5^{\circ} \mathrm{C}\) and \(1.65 \times 10^{-3} \mathrm{~atm},\) and its critical point is at \(157^{\circ} \mathrm{C}\) and 78 atm. On the basis of this information, draw a rough sketch of the phase diagram of \(\mathrm{SO}_{2}\).

Given the general properties of water and ammonia, comment on the problems that a biological system (as we know it) would have developing in an ammonia medium. $$ \begin{array}{lll} & \mathrm{H}_{2} \mathrm{O} & \mathrm{NH}_{3} \\ \hline \text { Boiling point } & 373.15 \mathrm{~K} & 239.65 \mathrm{~K} \\ \text { Melting point } & 273.15 \mathrm{~K} & 195.3 \mathrm{~K} \\ \text { Molar heat capacity } & 75.3 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} & 8.53 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \\ \text { Molar heat of vaporization } & 40.79 \mathrm{~kJ} / \mathrm{mol} & 23.3 \mathrm{~kJ} / \mathrm{mol} \\ \text { Molar heat of fusion } & 6.0 \mathrm{~kJ} / \mathrm{mol} & 5.9 \mathrm{~kJ} / \mathrm{mol} \\ \text { Viscosity } & 0.001 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2} & 0.0254 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2} \\ & & (\text { at } 240 \mathrm{~K}) \\ \text { Dipole moment } & 1.82 \mathrm{D} & 1.46 \mathrm{D} \\ \text { Phase at } 300 \mathrm{~K} & \text { Liquid } & \text { Gas } \end{array} $$

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