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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

Sodium chloride \((\mathrm{NaCl})\) exhibits predominantly ionic bonding. The \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions have electron structures that are identical to which two inert gases?

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

To what group in the periodic table would an element with atomic number 114 belong?

The net potential energy between two adjacent ions, \(E_{N}\), may be represented by the sum of Equations \(2.8\) and \(2.9\); that is, $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B\), and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\), and then set the resulting expression equal to zero, because the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0 \text { - }}\) 2\. Solve for \(r\) in terms of \(A, B\), and \(n\), which yields \(r_{0}\), the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substituting \(r_{0}\) into Equation \(2.11\).

Chromium has four naturally occurring isotopes: \(4.34 \%\) of \({ }^{50} \mathrm{Cr}\), with an atomic weight of \(49.9460\) amu; \(83.79 \%\) of \({ }^{52} \mathrm{Cr}\), with an atomic weight of \(51.9405 \mathrm{amu} ; 9.50 \%\) of \({ }^{53} \mathrm{Cr}\), with an atomic weight of \(52.9407 \mathrm{amu} ;\) and \(2.37 \%\) of \({ }^{54} \mathrm{Cr}\), with an atomic weight of \(53.9389\) amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Cr}\) is \(51.9963 \mathrm{amu}\)
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