Chapter 5: Problem 88
Gallium occurs naturally as \({ }^{69} \mathrm{Ga}\) and \({ }^{71} \mathrm{Ga}\). Given the mass and abundance of gallium- \(69(68.92\) amu and \(60.10 \%)\), what is the isotopic mass of gallium-71?
Short Answer
Expert verified
The isotopic mass of gallium-71 is approximately 70.93 amu.
Step by step solution
01
Understanding the given information
We have two isotopes of Gallium:
1.
^{69}Ga with a given mass of 68.92 amu and an abundance of 60.10%.
2.
^{71}Ga with unknown mass and the remaining abundance percentage. The total abundance must be 100%.
02
Calculate the abundance of ^{71}Ga
Subtract the abundance of ^{69}Ga from 100% to find the abundance of ^{71}Ga:\[\text{Abundance of } ^{71}Ga = 100\% - 60.10\% = 39.90\%\]
03
Set up the equation for average atomic mass
The average atomic mass of an element is calculated as the sum of the products of the masses and their respective natural abundances. Let the isotopic mass of ^{71}Ga be M_{71} amu. The equation for Gallium's average atomic mass (given or calculated from known isotopes) is:\[\text{Average atomic mass} = (68.92 \, \text{amu} \times 0.6010) + (M_{71} \, \text{amu} \times 0.3990)\]
04
Assume typical average atomic mass of Gallium
For our calculation, assume the average atomic mass of Gallium from the periodic table is approximately 69.723 amu. This value is used to find the unknown isotopic mass.
05
Solve the equation for M_{71}
Insert the assumed average atomic mass into the equation and solve for M_{71}:\[69.723 = (68.92 \, \text{amu} \times 0.6010) + (M_{71} \, \text{amu} \times 0.3990)\]\[69.723 = 41.42092 + 0.3990M_{71}\]Subtract 41.42092 from both sides:\[69.723 - 41.42092 = 0.3990M_{71}\]\[28.30208 = 0.3990M_{71}\]Divide both sides by 0.3990 to isolate M_{71}:\[M_{71} = \frac{28.30208}{0.3990} \approx 70.92475 \text{ amu}\]
06
Round the result for practical purposes
The isotopic mass of ^{71}Ga is typically rounded based on significant figures. Thus: \[M_{71} \approx 70.93 \text{ amu}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Average Atomic Mass
The concept of average atomic mass can sometimes appear perplexing, but it's quite fascinating when broken down!
The average atomic mass of an element refers to the weighted average of all its isotopes based on their mass and relative natural abundance.
This figure is usually what you find in the periodic table.
To calculate this, you multiply each isotope’s mass by its natural abundance (expressed as a decimal), then sum the results.
For example, if an element has two isotopes, isotope 1 with a mass of 10 amu and 20% abundance, and isotope 2 with a mass of 15 amu and 80% abundance, you would calculate it as:
The average atomic mass of an element refers to the weighted average of all its isotopes based on their mass and relative natural abundance.
This figure is usually what you find in the periodic table.
To calculate this, you multiply each isotope’s mass by its natural abundance (expressed as a decimal), then sum the results.
For example, if an element has two isotopes, isotope 1 with a mass of 10 amu and 20% abundance, and isotope 2 with a mass of 15 amu and 80% abundance, you would calculate it as:
- ext{For isotope 1: } 10 ext{ amu} \times 0.20 = 2 ext{ amu}
- ext{For isotope 2: } 15 ext{ amu} \times 0.80 = 12 ext{ amu}
Natural Abundance
Natural abundance gives us an insight into how common an isotope is in nature relative to the element's total amount on Earth.
The total natural abundance of all isotopes of an element will always sum up to 100%.
When referring to isotopes, those with higher natural abundances contribute more significantly to the average atomic mass.
This is crucial in calculations like determining the average atomic mass we discussed earlier.
To understand better, consider an isotope of an element that has a natural abundance of 60%.
This means that if you picked 100 atoms of this element randomly, approximately 60 of them would be this particular isotope.
The total natural abundance of all isotopes of an element will always sum up to 100%.
When referring to isotopes, those with higher natural abundances contribute more significantly to the average atomic mass.
This is crucial in calculations like determining the average atomic mass we discussed earlier.
To understand better, consider an isotope of an element that has a natural abundance of 60%.
This means that if you picked 100 atoms of this element randomly, approximately 60 of them would be this particular isotope.
- The distribution can vary greatly amongst elements. Some may have isotopes with similar abundances, while others might have one dominant isotope.
- Elements with a high abundance isotope tend to have their average atomic mass closely match the mass of the abundant isotope.
Gallium Isotopes
Gallium is a fascinating element because it naturally occurs as two stable isotopes: Gallium-69 and Gallium-71.
Isotopes are versions of an element that have the same number of protons but different numbers of neutrons, thereby differing in atomic mass.
The two naturally occurring gallium isotopes are:
The process involves using both the masses and natural abundances of these isotopes to represent gallium as found naturally.
Gallium's average atomic mass, which appears on the periodic table as approximately 69.723 amu, is a reflection of the isotopic mix found on earth.
Isotopes are versions of an element that have the same number of protons but different numbers of neutrons, thereby differing in atomic mass.
The two naturally occurring gallium isotopes are:
- Gallium-69 ( ^{69}Ga): With a mass of approximately 68.92 amu and a natural abundance of 60.10%, it is the more prevalent isotope.
- Gallium-71 ( ^{71}Ga): Its mass is typically around 70.924 amu. It constitutes about 39.90% of naturally occurring gallium.
The process involves using both the masses and natural abundances of these isotopes to represent gallium as found naturally.
Gallium's average atomic mass, which appears on the periodic table as approximately 69.723 amu, is a reflection of the isotopic mix found on earth.