Question

An certain element Y has two naturally occurring isotopes, 6Y (isotopic mass = 6.015123 amu) and 7Y (isotopic mass = 7.016005 amu). If the atomic mass of the element is 6.9412 amu. What is the percent abundance of 6Y?
Do not include the symbol "%" in your answer.

Report the correct answer to three significant figures

Answer 1

The percentage abundance of  6Y is 0.102.

An Isotope refers to two or more atoms that has the same atomic number but different mass  numbers.

We have been informed from the question that the atomic mass of the element is 6.9412 amu.

Let the abundance of isotope 6Y be x

Let the abundance of isotope 7Y  be 1-x

Hence;

6.9412 = 6.015123x + 7.016005(1 - x)

6.9412 = 6.015123x + 7.016005 - 7.016005x

6.9412 - 7.016005 = 6.015123x - 7.016005x

-0.102 = -1x

x = -0.102/-1

x = 0.102

Hence the percentage abundance of isotope 6Y is 0.102.

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Answer 2

The percentage abundance of 6Y is 7.47%

Isotopy is the phenomenon whereby two or element have the same atomic number but different mass number due to the difference in their neutron number.

We'll begin the solving as follow:

Let A represent isotope 6Y

Let B represent isotope 7Y

From the question given above, the following data were obtained:

For isotope A (6Y):

Mass of A = 6.015123 amu

Abundance of A (A₀) =…?

For isotope B (7Y):

Mass of B = 7.016005 amu

Abundance of B (B₀) = (100 – A₀%

Relative Atomic mass of Y = 6.9412 amu

The abundance of 6Y can be obtained as follow:

$RAM = \frac{MassA * A_{0} }{100} + \frac{MassB * B_{0} }{100}\\\\6.9412 = \frac{6.015123 * A_{0}}{100} + \frac{7.016005 (100 - A_{0})}{100}\\\\6.9412 = 0.06015123 A_{0} + \frac{701.6005 - 7.016005A_{0}}{100}\\\\6.9412 = 0.06015123 A_{0} + 7.016005 - 0.07016005A_{0}$

Collect like terms

$6.9412 - 7.016005 = 0.06015123A_{0} - 0.07016005A_{0}\\-0.074805 = -0.01000882A_{0}$

Divide both side by –0.0100882

$A_{0} = \frac{-0.074805}{-0.01000882}$

A% = 7.47%

Therefore, the abundance of 6Y is 7.47%

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