Problem 20
Question
Strontium has four stable isotopes. Strontium-84 has a very low natural abundance, but \(^{86} \mathrm{Sr},^{87} \mathrm{Sr},\) and \(^{88} \mathrm{Sr}\) are all reasonably abundant. Knowing that the atomic weight of strontium is \(87.62,\) which of the more abundant isotopes predominates?
Step-by-Step Solution
Verified Answer
Strontium-88 (\(^{88}\mathrm{Sr}\)) is the predominant isotope.
1Step 1: Understand Atomic Weight
The atomic weight of an element is a weighted average of the masses of its isotopes based on their natural abundance. In this case, it is given as 87.62 for strontium.
2Step 2: Identify the Isotopes
Consider the isotopes of strontium mentioned: \(^{86}\mathrm{Sr}\), \(^{87}\mathrm{Sr}\), and \(^{88}\mathrm{Sr}\) and their respective atomic masses (86, 87, and 88).
3Step 3: Compare Isotope Masses to Atomic Weight
Compare each isotope's mass to the given atomic weight of 87.62 to determine proximity. The isotope with a mass closest to 87.62 is likely predominant since it will have the greatest influence on the weighted average.
4Step 4: Analyze Predominance
Since 87.62 is closer to 88 than to 87 or 86, \(^{88}\mathrm{Sr}\) is more likely to be the predominant isotope among the three.
Key Concepts
Atomic weightNatural abundanceStrontium isotopes
Atomic weight
Atomic weight, also known as atomic mass, is a crucial concept in understanding elements and their isotopes. It represents the average mass of all the isotopes of a specific element, weighted by their natural abundance. This value is not simply the sum or simple average of the isotopes' masses, but rather accounts for how common each isotope is in nature. The atomic weight of an element can be found on the periodic table, often displayed as a decimal to reflect these weighted averages.
To calculate atomic weight, consider an element like strontium which has isotopes with different atomic masses. You would multiply the mass of each isotope by its natural abundance (expressed as a decimal). Then, sum these values to get the atomic weight. For instance, if isotopes have abundances of 50% (0.50 in decimal form), 30% (0.30), and 20% (0.20), and respective masses of 86, 87, and 88, the weighted contributions would be:
To calculate atomic weight, consider an element like strontium which has isotopes with different atomic masses. You would multiply the mass of each isotope by its natural abundance (expressed as a decimal). Then, sum these values to get the atomic weight. For instance, if isotopes have abundances of 50% (0.50 in decimal form), 30% (0.30), and 20% (0.20), and respective masses of 86, 87, and 88, the weighted contributions would be:
- 86 x 0.50
- 87 x 0.30
- 88 x 0.20
Natural abundance
Natural abundance refers to the relative amount of each isotope of an element that can be found in nature. Each isotope of an element has a different proportion or percentage of the total natural composition of that element. This concept is integral in determining the element's atomic weight since it is based on the prevalence of each isotope and their respective masses.
Natural abundance is expressed as a percentage or fraction. For example, if an isotope has a natural abundance of 75%, it means that in any naturally occurring sample of the element, 75% of the atoms are of this particular isotope. Calculating the atomic weight takes these natural abundances into account. Consequently, the contribution of an isotope to the atomic weight is more pronounced if its natural abundance is high.
In the case of strontium, isotopes like \(^{86} ext{Sr}\), \(^{87} ext{Sr}\), and \(^{88} ext{Sr}\) each have their own distinct natural abundances which ensure that some isotopes exert more influence on the overall atomic weight than others. Understanding natural abundance helps in predicting the isotope that dominates in samples of the element and is vital in studies involving isotopic composition.
Natural abundance is expressed as a percentage or fraction. For example, if an isotope has a natural abundance of 75%, it means that in any naturally occurring sample of the element, 75% of the atoms are of this particular isotope. Calculating the atomic weight takes these natural abundances into account. Consequently, the contribution of an isotope to the atomic weight is more pronounced if its natural abundance is high.
In the case of strontium, isotopes like \(^{86} ext{Sr}\), \(^{87} ext{Sr}\), and \(^{88} ext{Sr}\) each have their own distinct natural abundances which ensure that some isotopes exert more influence on the overall atomic weight than others. Understanding natural abundance helps in predicting the isotope that dominates in samples of the element and is vital in studies involving isotopic composition.
Strontium isotopes
Strontium isotopes are atoms of the element strontium that have the same number of protons but different numbers of neutrons. This difference in neutron number is what defines an isotope. Strontium occurs naturally in four stable isotopic forms:
Strontium isotopes have significant applications beyond just altering atomic weights. They play important roles in fields like geology for dating rocks and in understanding various Earth processes through isotope ratio analysis. The differing natural abundances cause slightly different isotope ratios in different materials, which can provide insights into the origins and history of those materials. By knowing the specific isotopes of strontium, scientists can deduce such valuable geological information.
- \(^{84}\text{Sr}\)
- \(^{86}\text{Sr}\)
- \(^{87}\text{Sr}\)
- \(^{88}\text{Sr}\)
Strontium isotopes have significant applications beyond just altering atomic weights. They play important roles in fields like geology for dating rocks and in understanding various Earth processes through isotope ratio analysis. The differing natural abundances cause slightly different isotope ratios in different materials, which can provide insights into the origins and history of those materials. By knowing the specific isotopes of strontium, scientists can deduce such valuable geological information.
Other exercises in this chapter
Problem 18
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