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In some cases, the most abundant isotope of an element can be established by rounding off the atomic mass to the nearest whole number, as in \(^{39} \mathrm{K},^{85} \mathrm{Rb}\), and \(^{88} \mathrm{Sr}\). But in other cases, the isotope corresponding to the rounded-off atomic mass does not even occur naturally, as in \(^{64} \mathrm{Cu}\). Explain the basis of this observation.

Short Answer

Expert verified
Not all atomic masses, when rounded to the nearest whole number, represent a naturally occurring isotope of that element. This can occur when there are multiple isotopes of an element with varying natural abundances and mass numbers that lead to an average value in between the naturally occurring isotopes. A good example is Copper - \(^{64} \mathrm{Cu}\) is not a naturally occurring isotope, even though 64 is the rounded atomic mass, because its natural isotopes are \(^{63} \mathrm{Cu}\) and \(^{65} \mathrm{Cu}\) with similar abundances.

Step by step solution

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Step 1. Understand atomic mass and isotopes

The atomic mass of an element is the weighted average of the masses of all the naturally occurring isotopes. Each isotope contributes in proportion to its natural abundance, and the isotopes of an element do not necessarily have the same atomic masses. Note that an isotope is identified by the sum of the number of protons and neutrons in its nucleus, called the mass number, which is a whole number.
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Step 2. Understand the concept of abundance and weighting

In calculating the atomic mass, more abundant isotopes weigh more heavily. Hence, in some cases where there is a single most abundant isotope, the atomic mass of the element can seem to be close to the mass number of this isotope, when rounded off to the nearest whole number.
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Step 3. Understand why round-off does not always work

In other cases, the atomic mass may not seem close to the mass number of any single isotope, if there are several stable isotopes of significantly different mass numbers. This is because the atomic mass reflects the average of all these mass numbers, weighted by the respective natural abundances. Therefore, simply rounding off the atomic mass to the nearest whole number does not necessarily yield a mass number that corresponds to an existing isotope.
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Step 4. Apply the concepts to the given examples

The cases of \(^{39} \mathrm{K},^{85} \mathrm{Rb},\) and \(^{88} \mathrm{Sr}\) are instances where the atomic mass, rounded off to the nearest whole number, coincides with the mass number of the most abundant isotope. But in the case of \(^{64} \mathrm{Cu},\) there are two stable isotopes, \(^{63} \mathrm{Cu}\) and \(^{65} \mathrm{Cu},\) with almost equal abundance, leading to a weighted average that does not coincide with the mass number of either isotope when rounded off to the nearest whole number.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotope Abundance
In chemistry and physics, understanding isotope abundance is crucial for grasping how atomic masses are determined. Different isotopes of an element have varying numbers of neutrons, which leads to slight variations in their mass. However, these isotopes don’t occur in equal amounts in nature.

For example, chlorine has two main isotopes, chlorine-35 and chlorine-37. If chlorine-35 is more abundant, it will play a larger role in determining the atomic mass of chlorine found on the periodic table. The natural abundance of an isotope is the percentage of that isotope found when sampling a large amount of the element. It represents how much of the total composition of an element that isotope accounts for.

The concept of isotope abundance is essential when calculating atomic mass. As not all isotopes are present in equal amounts, scientists must consider each isotope’s frequency of occurrence. For elements with a single dominant isotope, the atomic mass is closer to the mass number of that isotope. Conversely, for elements with multiple isotopes of comparable abundance, the result is a weighted average that may not match the mass numbers exactly.
Mass Number
The mass number is a fundamental concept when discussing atoms and their isotopes. Whereas the atomic number defines the number of protons in an atom and hence its identity as an element, the mass number focuses on the total count of protons and neutrons in an atom’s nucleus.

For instance, in an isotope named carbon-14 (or \(^{14}C\)), 14 is the mass number. It implies that the sum of protons and neutrons in the nucleus of carbon-14 is 14. Since the number of protons in carbon is always 6 (its atomic number), this means carbon-14 has 8 neutrons.

Every isotope of an element has a different mass number due to the variation in the neutron count. This difference in mass number is crucial because it's the primary way to distinguish between isotopes of the same element. The mass number is always a whole number; however, when discussing atomic mass - the average mass of all naturally occurring isotopes - the number is typically not whole due to the varying isotope abundances.
Weighted Average of Isotopes
When working with atomic mass specifics, one encounters the weighted average of isotopes. This average takes into account not just the individual masses of isotopes but also their relative abundances. By combining these factors, the atomic mass reflects the overall 'weight' of all isotopes as they naturally occur.

Take copper as an example. With its two primary isotopes, \(^{63}Cu\) and \(^{65}Cu\), their respective natural abundances and mass numbers must be considered. If \(^{63}Cu\) comprises roughly 70% of natural copper and \(^{65}Cu\) the remaining 30%, their masses would contribute proportionally to the overall atomic mass of copper. This is calculated by multiplying the mass of each isotope by its abundance and adding the results together. The atomic mass found on the periodic table is the weighted average, representing a careful balancing of these numerous atomic contributions.

It is important to note that the weighted average may not align perfectly with the mass numbers of the isotopes. This discrepancy explains why, for certain elements, rounding the atomic mass to the nearest whole number does not always result in a number that corresponds to one of its natural isotopes.

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Most popular questions from this chapter

The packing fraction of a nuclide is related to the fraction of the total mass of a nuclide that is converted to nuclear binding energy. It is defined as the fraction \((M-A) / A,\) where \(M\) is the actual nuclidic mass and \(A\) is the mass number. Use data from a handbook (such as the Handbook of Chemistry and Physics, published by the CRC Press) to determine the packing fractions of some representative nuclides. Plot a graph of packing fraction versus mass number, and compare it with Figure \(25-6 .\) Explain the relationship between the two.

Iodine-129 is a product of nuclear fission, whether from an atomic bomb or a nuclear power plant. It is a \(\beta^{-}\) emitter with a half-life of \(1.7 \times 10^{7}\) years. How many disintegrations per second would occur in a sample containing \(1.00 \mathrm{mg}^{129} \mathrm{I} ?\)

A nuclide has a decay constant of \(4.28 \times 10^{-4} \mathrm{h}^{-1}\). If the activity of a sample is \(3.14 \times 10^{5} \mathrm{s}^{-1},\) how many atoms of the nuclide are present in the sample? (a) \(2.64 \times 10^{12} ;\) (b) \(7.34 \times 10^{8}\) (c) \(2.04 \times 10^{5}\) (d) \(4.40 \times 10^{10} ;\) (e) none of these.

Hydrogen gas is spiked with tritium to the extent of \(5.00 \%\) by mass. What is the activity in curies of a \(4.65 \mathrm{L}\) sample of this gas at \(25.0^{\circ} \mathrm{C}\) and 1.05 atm pressure? [Hint: Use 3.02 u as the atomic mass of tritium and data from elsewhere in the text, as necessary.]

One of the following elements has eight naturally occurring stable isotopes. We should expect that one to be (a) \(\mathrm{Ra} ;\) (b) \(\mathrm{Au} ;\) (c) \(\mathrm{Cd} ;\) (d) Br.

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