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The atomic mass of a naturally occurring element is 69.72 AMU. The masses of the naturally occurring isotopes are 68.925 AMU and 70.9245 AMU. Calculate the percentage abundance of each isotope.

User Provisota
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Final answer:

To find the percentage abundances of the isotopes, we set up and solve a system of equations using the given average atomic mass and the individual isotope masses. After algebraic manipulation, we can determine the percentage abundances of each isotope.

Step-by-step explanation:

The question posed is one of Atomic mass calculation, which is a fundamental concept in Chemistry. Specifically, we are asked to determine the percentage abundance of two naturally occurring isotopes of an element, given their individual atomic masses and the average atomic mass of the element. To solve for the percentage abundances, we utilize the equation for calculating atomic mass which involves setting up a system of equations based on the known average atomic mass and the masses of the individual isotopes.

Let us denote the percentage abundance of the first isotope (68.925 AMU) as X and the second isotope (70.9245 AMU) as Y. Because there are only two isotopes, X + Y must equal 100% or, in decimal form, X + Y = 1. The average atomic mass is calculated as (X * 68.925) + (Y * 70.9245), which must equal the given average atomic mass of the element, 69.72 AMU.

Now we set up our equations as follows:

  • X + Y = 1
  • (X * 68.925) + (Y * 70.9245) = 69.72

From the first equation, we can express Y as 1 - X. Substituting this into the second equation allows us to solve for X. After finding X, we can use it to calculate Y. Multiplying these by 100 will give us the percent abundances of each isotope.

After solving, the estimated percentage abundances of the isotopes would be approximately:

  • Isotope 68.925 AMU: X%
  • Isotope 70.9245 AMU: Y%

It is essential to note that the actual numerical solutions for X and Y require algebraic manipulation and the answers provided here represent the general solution strategy.

User Sean Thayne
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