Problem 28 - General Chemistry: Principles and Modern Applications

Question

For the ion \(^{228} \mathrm{Ra}^{2+}\) with a mass of 228.030 u, determine (a) the numbers of protons, neutrons, and electrons in the ion; (b) the ratio of the mass of this ion to that of an atom of \(^{16} \mathrm{O}\) (refer to page 47 ).

Step-by-Step Solution

Answer

The ion \(^{228} \mathrm{Ra}^{2+}\) has 88 protons, 140 neutrons, and 86 electrons. The ratio of its mass to that of an Oxygen atom is approximately 14.25.

Step 1: Numbers of protons, neutrons, and electrons

In \(^{228} \mathrm{Ra}^{2+}\), 'Ra' stands for Radium and its atomic number is 88. This means it has 88 protons. Since it's a \(^{2+}\) ion, it has lost two electrons, so it has 88 - 2 = 86 electrons. The atomic mass (A) is 228, so the number of neutrons can be calculated by A - atomic number (Z), which is 228 - 88 = 140 neutrons.

Step 2: Mass ratio calculation

The ratio of the mass of the ion to that of an atom of \(^{16} \mathrm{O}\) is simply the mass of \(^{228} \mathrm{Ra}^{2+}\) divided by the mass of \(^{16} \mathrm{O}\). Given that the atomic mass unit (u) of Oxygen is approximately 16 u, the ratio would be 228.030 / 16 = 14.251875.

Key Concepts

Atomic Structure

Atoms consist of three fundamental particles: protons, neutrons, and electrons. The protons and neutrons form the nucleus at the center of the atom. Protons have a positive charge, while neutrons are neutral. Electrons, which are negatively charged, orbit around the nucleus in electron clouds.

For the ion \(^{228} \mathrm{Ra}^{2+}\), 'Ra' signifies Radium, which has an atomic number of 88. This atomic number represents the number of protons in the nucleus. As protons and electrons are equal in a neutral atom, an uncharged Radium atom also has 88 electrons. However, since the ion is \(2+\), it has lost two electrons, leaving it with 86 electrons. This ionization changes the electron structure but not the atomic structure.

The atomic mass "A" of Radium in this ion is given as 228. This indicates the total mass for protons and neutrons, as electrons contribute negligibly to atomic mass. To find the number of neutrons, subtract the number of protons from the atomic mass: \(A - Z = 228 - 88 = 140\) neutrons.

Mass Ratio

The concept of mass ratio is important in comparing the mass of different atoms or ions, especially when dealing with isotopes. When we talk about mass ratio, we are comparing the mass of one entity to another. Here, we compare the mass of the \(^{228} \mathrm{Ra}^{2+}\) ion to the mass of a \(^{16} \mathrm{O}\) atom.

The mass of the \(^{228} \mathrm{Ra}^{2+}\) ion is 228.030 u, while the atomic mass of an oxygen atom \(^{16} \mathrm{O}\) is approximately 16 u. To determine the mass ratio, divide the mass of the Radium ion by the mass of the oxygen atom:

• Mass of Radium ion: 228.030 u
• Mass of Oxygen atom: 16 u
• Mass ratio: \(\frac{228.030}{16} = 14.251875\)

Understanding mass ratio helps in discussions related to isotope calculations, molecular formulas, and stoichiometry.

Ionization

Ionization involves removing or adding electrons to an atom, resulting in a charged ion. This process can significantly alter the chemical and physical properties of a substance.

For the \(^{228} \mathrm{Ra}^{2+}\) ion, ionization has occurred, meaning it has lost two electrons compared to a neutral Radium atom. This causes a net charge of \(+2\), hence the \(^{2+}\) notation. Ionized atoms tend to behave differently in chemical reactions compared to their neutral counterparts due to their charge.

Key points about ionization include:

• It affects the atomic structure by changing the electron arrangement.
• Charged ions often have different reactivity and properties.
• In physics and chemistry, ionization is crucial for processes like conductivity and reactivity.

Understanding ionization is essential for comprehending how atoms interact, form compounds, and conduct electricity in gases and solutions.