Naturally occurring strontium consists of four isotopes, Sr-84, Sr-86, Sr-87 and Sr-88.

Below the data concerning strontium:
Sr-84 83.913 amu ,0.56%
Sr-86 85.909amu ,9.86%
Sr-87 86.909amu ,7.00%
Sr-88 ? ?

Based on the above information and the average atomic mass of Strontium, what is the mass a percent abundance of Strontium-88?

Respuesta :

Answer:

7+9.86+.56=17.42, 100-17.42= 82.58%

Explanation:

Add them all together then minus by 100%

From all the naturally occurring strontium isotopes (Sr-84, Sr-86, Sr-87, and Sr-88), the atomic mass and percent abundance of Sr-88 is 78.071amu and 82.58%, respectively.      

The average atomic mass of Strontium is given by:

[tex] A_{Sr}*100\% = A_{Sr-84}*\%_{Sr-84} + A_{Sr-86}*\%_{Sr-86} + A_{Sr-87}*\%_{Sr-87} + A_{Sr-88}*\%_{Sr-88} [/tex]    (1)

Where:

%: is the percent abundance of each isotope

A: is the atomic mass of each isotope

[tex] 87.62*100\% = 83.913*0.56\% + 85.909*9.86\% + 86.909*7.00\% + A_{Sr-88}*\%_{Sr-88} [/tex]

To find the percent abundance of Sr-88 we need to rest all the values of percent abundance of the isotopes of Sr to 100.

[tex] 100\% = 0.56\% + 9.86\% + 7.00\% + \%_{Sr-88} [/tex]

[tex] \%_{Sr-88} = 100\% - (0.56\% + 9.86\% + 7.00\%) = 82.58\% [/tex]

Now, after changing the percent values of equation (1) to decimal ones, we have:

[tex] 87.62*1 = 83.913*0.0056 + 85.909*0.0986 + 86.909*0.007 + A_{Sr-88}*0.8258 [/tex]                

Solving for [tex]A_{Sr-88}[/tex]:

[tex] A_{Sr-88} = \frac{87.62 - (83.913*0.0056 + 85.909*0.0986 + 86.909*0.007)}{0.8258} = 94.54 amu [/tex]  

 

Therefore, the mass and percent abundance of Sr-88 is 78.071amu and 82.58%, respectively.                

You can find another example of average atomic mass here: https://brainly.com/question/21536220?referrer=searchResults

I hope it helps you!

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