Question

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is -3.20 x 10\(^{-9}\)C. (a) Find the number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.

Short answer

(a) There are \(2.00 \times 10^{10}\) excess electrons; (b) \(8.62 \times 10^{-13}\) excess electrons per lead atom.

Step-by-step solution

Calculate the Charge of Single Electron

The charge of a single electron is \( e = -1.60 \times 10^{-19} \, \text{C} \). We will use this to find the number of excess electrons.

Determine the Number of Excess Electrons

To find the number of excess electrons \( n \) on the sphere, use the equation:\[ n = \frac{q_{\text{net}}}{e} \]where \( q_{\text{net}} = -3.20 \times 10^{-9} \, \text{C} \).Thus, \[ n = \frac{-3.20 \times 10^{-9}}{-1.60 \times 10^{-19}} = 2.00 \times 10^{10} \] This means there are \( 2.00 \times 10^{10} \) excess electrons.

Calculate Moles of Lead

Determine the moles of lead using its mass and atomic mass. Use the formula:\[ \text{moles of lead} = \frac{\text{mass}}{\text{atomic mass}} = \frac{8.00 \, \text{g}}{207 \, \text{g/mol}} = 0.0386 \, \text{mol} \]

Calculate the Number of Lead Atoms

The number of atoms can be calculated using Avogadro's number \( 6.022 \times 10^{23} \text{ atoms/mol} \):\[ \text{Number of atoms} = 0.0386 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 2.32 \times 10^{22} \]

Calculate Excess Electrons Per Lead Atom

Using the total number of excess electrons from Step 2 and the number of lead atoms from Step 4:\[\text{Excess electrons per lead atom} = \frac{2.00 \times 10^{10}}{2.32 \times 10^{22}} = 8.62 \times 10^{-13} \]

Key concepts

Charge of Electron: The Fundamental Unit

Every electron carries a charge, which is one of its most crucial characteristics. The charge of an electron is a fundamental physical constant, denoted by the letter \( e \). It is measured as \(-1.60 \times 10^{-19} \; \text{C}\) (Coulombs). This charge is negative, reflecting the electron's role in electricity and charge interactions.
Understanding the charge of an electron is key to comprehending how electric charge is quantified. For example, in a situation where excess electrons accumulate on an object, the net charge can be computed by multiplying the number of excess electrons by the charge of a single electron.
To find the number of excess electrons, one simply divides the total net charge \( (q_{\text{net}}) \) by the charge of a single electron \( e \). This relationship is expressed in the equation:

• \( n = \frac{q_{\text{net}}}{e} \)

This approach is often used in physics and chemistry to examine how charged particles behave and influence their surroundings.

Avogadro's Number: Linking Microscopic and Macroscopic Worlds

Avogadro's number \( (6.022 \times 10^{23} \text{ atoms/mol}) \) is a cornerstone of chemistry. It links the microscopic world of atoms with the macroscopic quantities we observe in the laboratory.
Named after Amedeo Avogadro, this number is the quantity of atoms, molecules, or particles in a mole of substance, making it fundamental for molecular counting in chemical reactions and compositions.
When calculating the number of atoms in a given amount of a substance, we first determine the number of moles. For example, if we have \( 0.0386 \; \text{mol} \) of lead, we multiply it by Avogadro's number to derive the total number of atoms:

• \( \text{Number of atoms} = 0.0386 \; \text{mol} \times 6.022 \times 10^{23} \; \text{atoms/mol} \)

Here, Avogadro's number transforms the conventional measurement of molar mass into a detailed quantification of individual atoms, enhancing our understanding of atomic-scale processes.

Atomic Mass and Its Role in Chemistry

Atomic mass is the mass of a single atom, typically expressed in atomic mass units (amu) or grams per mole (g/mol). For lead (Pb), the atomic mass is approximately \(207 \; \text{g/mol}\).
Atomic mass is crucial when converting between the mass of a sample and the quantity of atoms. By employing the formula \( \text{moles} = \frac{\text{mass}}{\text{atomic mass}} \), scientists can determine how many moles a sample contains.
This step is essential for calculations involving chemical reactions, where knowing the proportion of reactants and products is necessary. In our exercise, we determined that \( 8.00 \; \text{g} \) of lead corresponds to \( 0.0386 \; \text{mol} \), by dividing the mass by the atomic mass. This empowered us to ascertain the total number of lead atoms present.
This computation is vital in chemical stoichiometry, where balancing equations and quantifying substances depends on precise atomic mass calculations.