Question

Calculate the average density of a single Al-27 atom by assuming that it is a sphere with a radius of \(0.143 \mathrm{nm}\). The masses of a proton, electron, and neutron are \(1.6726 \times 10^{-24} \mathrm{~g}, 9.1094 \times 10^{-28} \mathrm{~g}\), and \(1.6749 \times 10^{-24} \mathrm{~g}\) respectively. The volume of a sphere is \(4 \pi r^{3} / 3,\) where \(r\) is its radius. Express the answer in grams per cubic centimeter. The density of aluminum is found experimentally to be \(2.70 \mathrm{~g} / \mathrm{cm}^{3} .\) What does that suggest about the packing of aluminum atoms in the metal?

Short answer

Answer: The higher calculated average density of a single Al-27 atom compared to the experimentally determined density of aluminum suggests that the packing of aluminum atoms in the metal is not close-packed and there is some empty space between the atoms.

Step-by-step solution

Determine the Atomic Composition of Al-27

An Al-27 atom consists of 13 protons, 13 electrons, and 14 neutrons. This information is crucial for calculating the atom's mass.

Calculate the Mass of Al-27 Atom

We will now calculate the mass of a single Al-27 atom by adding up the masses of its protons, electrons, and neutrons. Mass of Al-27 atom = (Mass of 13 protons) + (Mass of 13 electrons) + (Mass of 14 neutrons) Mass of Al-27 atom = (13 × 1.6726 × 10^{-24} g) + (13 × 9.1094 × 10^{-28} g) + (14 × 1.6749 × 10^{-24} g) After calculating, we get: Mass of Al-27 atom ≈ 4.4803 × 10^{-23} g

Calculate the Volume of the Sphere

We are given the radius of the sphere, r = 0.143 nm. We will first need to convert the radius to centimeters and then use the volume formula to compute the volume of the sphere. 1 nm = 10^{-9} m = 10^{-7} cm So, r = 0.143 nm = 0.143 × 10^{-7} cm Volume of the sphere = (4 * pi * r^3) / 3 Volume ≈ 1.2312 × 10^{-23} cm³

Calculate the Average Density

We will now compute the average density by dividing the mass by the volume. Average Density = Mass / Volume Average Density ≈ (4.4803 × 10^{-23} g) / (1.2312 × 10^{-23} cm³) Average Density ≈ 3.64 g/cm³

Compare the Calculated Density to the Experimentally Determined Value

The average density of a single Al-27 atom, calculated as 3.64 g/cm³, is higher than the experimentally determined density of aluminum (2.70 g/cm³). This suggests that the packing of aluminum atoms in the metal is not close-packed, and there is some empty space between the atoms.

Key concepts

Aluminum Atom Mass

Understanding the mass of an aluminum atom requires a closer look at its atomic composition. Aluminum has the atomic number 13, which means each atom has 13 protons and typically, 13 electrons. However, the isotope Al-27 also contains 14 neutrons. The mass of an atom is primarily from its protons and neutrons, as the mass of electrons is negligible in comparison.

To compute the mass, we consider the known masses of these particles: protons (\(1.6726 \times 10^{-24} \text{g}\)), electrons (\(9.1094 \times 10^{-28} \text{g}\)), and neutrons (\(1.6749 \times 10^{-24} \text{g}\)). By multiplying the number of each type of particle by their respective masses and adding the sums together, we find the mass of the Al-27 atom. Such calculations are fundamental to understanding the atomic weight and its significance in the physical properties of elements.

Volume of a Sphere

Calculating the volume of a sphere is a key concept in geometry, and it is especially pertinent when considering atoms on a microscopic scale. The formula for the volume of a sphere is \(\frac{4}{3} \times \pi r^3\) where \(r\) is the radius. For an Al-27 atom, assuming it's a perfect sphere with a radius of \(0.143 \text{nm}\) or \(0.143 \times 10^{-7} \text{cm}\), we apply the formula to find the spherical volume, which is a factor in determining atomic density. Remember, when dealing with atoms, we convert the measurements to appropriate units, like centimeters, to match up with conventional units used in density calculations.

Through this example, students can appreciate how geometric principles apply to atomic-level structures and why mastery of these mathematical concepts is essential in the field of chemistry and physics.

Atomic Structure

The atomic structure is a core concept in chemistry that describes the arrangement of particles within an atom. In our case, the aluminum atom consists of a central nucleus made of protons and neutrons, surrounded by electrons in orbitals. This structure is pivotal in determining the physical and chemical properties of the element. For example, the arrangement of electrons determines the atom's reactivity, while the mass of the nucleus influences its weight and, consequently, the atomic mass.

In analyzing the atomic structure, we can also infer the atomic volume and thereby gauge the density of the atom which relates to how tightly the atoms are packed within a substance. Therefore, a comprehensive understanding of atomic structure is invaluable for interpreting various chemical reactions and material characteristics.

Density Comparison

The concept of density comparison is vital when analyzing substances at the atomic or bulk scale. Density is defined as mass per unit volume. In the context of the exercise, we calculated the theoretical density of a single Al-27 atom by treating it as a sphere and compared it to the experimentally determined bulk density of aluminum.

This comparison reveals that the calculated atomic density is higher than the aluminum's bulk density, which suggests two things: firstly, that atoms are not packed as closely in the solid metal as they are in our model; secondly, it hints at the presence of empty spaces or what we refer to as voids in the metal's crystal lattice. Such insights are crucial for fields like material science where understanding the packing and spacing of atoms affects how we interpret the material's properties, including strength, ductility, and conductivity.

When it comes to materials' properties, knowing how to calculate and compare densities at different scales is a fundamental skill for students and professionals alike.