Question

Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 g$/$mol, and its atomic number is 13.) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 $\times$ 10${}^4$ N (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Short answer

(a) Each sphere contains $7.25\times {10}^{24}$ electrons. (b) $3.41\times {10}^{14}$ electrons must be transferred. (c) This is $4.70\times {10}^{-11}$ of the total electrons in each sphere.

Step-by-step solution

Calculate moles of aluminum in each sphere

Each sphere has a mass of 0.0250 kg. First, convert this to grams: 0.0250 kg = 25.0 g. Using the atomic mass of aluminum, calculate the moles: $$\text{moles}=\frac{25.0\text{ g}}{26.982\text{ g/mol}}\approx 0.926\text{ mol}$$

Determine number of atoms in each sphere

Use Avogadro's number to find the number of atoms in the sphere. Avogadro's number is $6.022\times {10}^{23}\text{atoms/mol}$: $$0.926\text{ mol}\times 6.022\times {10}^{23}\text{ atoms/mol}\approx 5.58\times {10}^{23}\text{ atoms}$$

Calculate number of electrons in each sphere

Each aluminum atom has 13 electrons. Therefore, the number of electrons per sphere is: $$5.58\times {10}^{23}\text{ atoms}\times 13\text{ electrons/atom}\approx 7.25\times {10}^{24}\text{ electrons}$$

Use Coulomb's law to determine electrons to transfer

We need a force of $1.00\times {10}^4\text{ N}$ between the spheres. Coulomb's law is $F=\frac{k\cdot |q_1\cdot q_2|}{r^2}$, where $k=8.99\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$. Assume the number of electrons transferred is $x$. Thus, charge on each sphere is $q=e\times x$, where $e=1.602\times {10}^{-19}\text{ C}$. Solve $$1.00\times {10}^4=\frac{8.99\times {10}^9\times (1.602\times {10}^{-19}\times x{)}^2}{(0.8{)}^2}$$ to find $$x\approx 3.41\times {10}^{14}\text{ electrons}$$

Calculate fraction of electrons

Now, determine the fraction of electrons transferred relative to the total electrons in the sphere: $$\frac{3.41\times {10}^{14}}{7.25\times {10}^{24}}\approx 4.70\times {10}^{-11}$$

Key concepts

aluminum atomic structure

Aluminum is an element found in group 13 of the periodic table, known as the boron group. It has an atomic number of 13, which means each aluminum atom has 13 protons in its nucleus. These protons are positively charged and are balanced by 13 negatively charged electrons orbiting around the nucleus. These electrons are distributed across different energy levels or shells around the nucleus.

Aluminum is notable for having three electrons in its outermost shell, making these electrons available for chemical reactions, usually resulting in aluminum forming a trivalent cation, Al³⁺. The arrangement of electrons in an aluminum atom can be represented in its electronic configuration:

• First shell: 2 electrons
• Second shell: 8 electrons
• Third shell: 3 electrons

Understanding its electron configuration helps explain many of aluminum's chemical properties and behaviors, such as its ability to conduct electricity and participate in bonds.

Avogadro's number

Avogadro's number, approximately equal to 6.022 x 10^23, is a fundamental constant used to count atoms, ions, or molecules in a given amount of substance. Named after the scientist Amedeo Avogadro, this number represents the quantity of atoms or molecules in one mole of any substance.

In practical terms, Avogadro's number allows chemists to convert between the mass of a sample and the number of constituent particles. This is crucial for calculations such as determining how many atoms, and eventually electrons, exist in a sample like our aluminum spheres. By multiplying the number of moles in a sample by Avogadro's number, we can find out the total number of atoms present, enabling further calculations about the sample's properties.

electron transfer calculation

Electron transfer plays a critical role in understanding how objects can exert electrical forces on one another. In the context of the aluminum spheres, calculating how many electrons must be moved from one sphere to the other helps us understand how substantial electric forces can be achieved using Coulomb's Law.

Coulomb's Law, expressed as $$F=\frac{k\cdot |q_1\cdot q_2|}{r^2}$$is used to calculate the force between two point charges based on their charge magnitudes and the distance between them. Here, $F$ depicts the force, $k$ is Coulomb's constant $8.99\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$, $q_1$ and $q_2$ are charges, and $r$ is the separation between them.

By transferring electrons, we change the charge on each sphere. The number of electrons transferred, denoted as $x$, thus modifies the charge magnitude $q=e\times x$, where $e$ is the elementary charge ($1.602\times {10}^{-19}\text{ C}$). Solving for $x$ involves rearranging Coulomb's law to achieve a specific force, here $1.00\times {10}^4\text{ N}$, confirming the profound impact small-scale electron movements have on macroscale forces.