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Chapter 2: Problem 1

Cite the difference between atomic mass and atomic weight.

Short Answer

Expert verified
Answer: The key differences between atomic mass and atomic weight are: 1. Atomic mass is the mass of a single atom or isotope, while atomic weight is the weighted average mass of an element considering all its isotopes. 2. Atomic mass is expressed in atomic mass units (amu), whereas atomic weight is expressed in unified atomic mass units (u or Da). 3. Atomic mass deals with individual atoms, whereas atomic weight pertains to all the isotopes of an element collectively.

Step by step solution

01

Define Atomic Mass

Atomic mass refers to the mass of an individual atom or isotope, and is measured in atomic mass units (amu). It indicates the total number of protons and neutrons present in the nucleus of an atom. Step 2:
02

Define Atomic Weight

Atomic weight, also known as relative atomic mass, is the average mass of the atoms of an element, taking into account the abundance and mass of each isotope. It is a weighted average, and is generally expressed with a unit called unified atomic mass unit (u or Da). Step 3:
03

Identify Key Differences

1. Atomic mass is the mass of a single atom or isotope, while atomic weight is the weighted average mass of an element considering all its isotopes. 2. Atomic mass is expressed in atomic mass units (amu), whereas atomic weight is expressed in unified atomic mass units (u or Da). 3. Atomic mass deals with individual atoms, whereas atomic weight pertains to all the isotopes of an element collectively.

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Most popular questions from this chapter

Compute the percent ionic character of the interatomic bonds for the following compounds: \(\mathrm{TiO}_{2}, \mathrm{ZnTe}, \mathrm{CsCl}, \mathrm{InSb}\), and \(\mathrm{MgCl}_{2}\).

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What type(s) of bonding would be expected for each of the following materials: solid xenon, calcium fluoride \(\left(\mathrm{CaF}_{2}\right),\) bronze, cadmium tel luride (CdTe), rubber, and tungsten?

Calculate the force of attraction between a \(\mathrm{K}^{+}\) and an \(\mathrm{O}^{2-}\) ion whose centers are separated by a distance of \(1.5 \mathrm{~nm}\).

For \(\mathrm{a} \mathrm{K}^{+}-\mathrm{Cl}^{-}\)ion pair, attractive and repulsive energies \(E_{A}\) and \(E_{R}\), respectively, depend on the distance between the ions \(r\), according to $$ \begin{aligned} E_{A} &=-\frac{1.436}{r} \\ E_{R} &=\frac{5.86 \times 10^{-6}}{r^{9}} \end{aligned} $$ For these expressions, energies are expressed in electron volts per \(\mathrm{K}^{+}-\mathrm{Cl}^{-}\)pair, and \(r\) is the distance in nanometers. The net energy \(E_{N}\) is just the sum of the preceding two expressions. (a) Superimpose on a single plot \(E_{N}, E_{R}\), and \(E_{A}\) versus \(r\) up to \(1.0 \mathrm{~nm}\). (b) On the basis of this plot, determine (i) the equilibrium spacing \(r_{0}\) between the \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)ions, and (ii) the magnitude of the bonding energy \(E_{0}\) between the two ions. (c) Mathematically determine the \(r_{0}\) and \(E_{0}\) values using the solutions to Problem \(2.14\) and compare these with the graphical results from part (b).
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