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Topic: Average Atomic Mass  (Read 10308 times)

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Forumz

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Average Atomic Mass
« on: March 05, 2006, 04:42:26 PM »
Hello,
There were two problems on a sheet of 12 which no one in my class could figure out.  I'm sure i'm missing something simple, but We're not too informed about isotopes yet, etc.  

How would you determine the average atomic mass of neon, if it's isotopes are present in the following proportions:

90.92%    Ne - 20     mass = 20.0 amu
  0.26%    Ne - 21     mass = 21.0 amu
  8.82%    Ne - 22     mass = 22.0 amu

And a bonus question on the same seet was to calculate the percentage of each isotope in a mixture of ag - 107 and ag - 109 having an average atomic mass of 107.9 amu.  

Thanks in advance for any *delete me* :D
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Offline Bakegaku

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Re:Average Atomic Mass
« Reply #1 on: March 05, 2006, 04:48:35 PM »
to find the average atomic mass of neon, you'd add the products of each isotope's mass in AMU's and their percentage.  I.E. the percentage of Ne-20 is 90.92%, which in decimal form would be .9092.  You'd multiply that by 20 AMU.  Then do the same for Ne-21 and Ne-22, and add the numbers.

This may be a bit ill-explained, but if you understand this then you should be able to figure out the next problem.
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Offline Albert

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Re:Average Atomic Mass
« Reply #2 on: March 05, 2006, 04:49:39 PM »
There is a formula you can use with all the isotopes:

average atomic mass = (AMa * %a)+(AMb * %b)+(AMc * %c)+...+(AMn * %n)
                                                            100
« Last Edit: March 05, 2006, 04:50:02 PM by Albert »
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Offline Borek

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Re:Average Atomic Mass
« Reply #3 on: March 05, 2006, 04:52:26 PM »
It is called weighted average.
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Forumz

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Re:Average Atomic Mass
« Reply #4 on: March 05, 2006, 04:56:32 PM »
Thanks for the quick replys guys.

What I did was:

(20 x 0.9092) + (21 x 0.0026) + 22 x 0.0882) = 20.179amu.  On my periodic table, it says 20.1797.  Both formulas gave me the same answer, thanks guys!

For the second question, Wouldn't you have to be given at least one percentage or isotope to find the other?  I'm sure i'm missing something.  'cause couldn't there be many posibilities that add up to get an avegage amu of 107.9?  Thanks again for any help in advance!  
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Offline Albert

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Re:Average Atomic Mass
« Reply #5 on: March 05, 2006, 05:02:11 PM »
You can use a system of two equations. The former is the one aforementioned and the latter is: 100 = %Ag1 + %Ag2
« Last Edit: March 05, 2006, 05:06:26 PM by Albert »
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Forumz

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Re:Average Atomic Mass
« Reply #6 on: March 05, 2006, 05:13:06 PM »
Let % Ag-107 = Y
Let % Ag-109 = 100-Y

(107 x Y ) + (109 x (100 - Y) ) = 107.9
           107Y + 10 900 - 109 Y = 107.9
                                      -2Y = 107.9 - 10 900
                                      - 2Y = -10 972.1
                                          Y = 5 396.05
That's what I came up with, Obviously Y can not be 5 396.05  :-\ .  Any help on where I went wrong would be great!  Thanks!

EDIT:  Unless Y = 5 396.05/100 = 53,9605%?  
 ???
« Last Edit: March 05, 2006, 05:18:42 PM by Forumz »
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Offline Albert

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Re:Average Atomic Mass
« Reply #7 on: March 05, 2006, 05:21:12 PM »
Check this out.
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Forumz

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Re:Average Atomic Mass
« Reply #8 on: March 05, 2006, 05:33:42 PM »
OH!  Thanks Albert, So...

Let % Ag-107 = Y
Let % Ag-109 = 100-Y

(107 x Y ) + (109 x (100 - Y) ) = 107.9
                   100
          107Y + 10 900 - 109 Y = 107.9 x 100
                                      -2Y = 10 790 - 10 900
                                      - 2Y = -110
                                          Y = 55%
To Check...

(107 x .55) + (109 x (100-55) )  = 107.9
(107 x .55) + (109 x (45) )        = 107.9
(58.85)      +    (49.05)             = 107.9
             107.9                         = 107.9

Is that right?
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Offline jdurg

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Re:Average Atomic Mass
« Reply #9 on: March 08, 2006, 09:22:21 AM »
Quote from: Forumz on March 05, 2006, 05:33:42 PM
OH!  Thanks Albert, So...

Let % Ag-107 = Y
Let % Ag-109 = 100-Y

(107 x Y ) + (109 x (100 - Y) ) = 107.9
                   100
          107Y + 10 900 - 109 Y = 107.9 x 100
                                      -2Y = 10 790 - 10 900
                                      - 2Y = -110
                                          Y = 55%
To Check...

(107 x .55) + (109 x (100-55) )  = 107.9
(107 x .55) + (109 x (45) )        = 107.9
(58.85)      +    (49.05)             = 107.9
             107.9                         = 107.9

Is that right?

Exactly right!   ;D  Now you see why they forced you to take algebra?  ;)  :D
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