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Finding the percent weight an element in an Alloy

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the wt% of lithium required to obtain an Al-Li alloy with a density of 2.55 g/cm^3.


    2. Relevant equations



    3. The attempt at a solution
    I am so lost in this problem, any hints of how to start this problem with GREATLY appreciated!
     
  2. jcsd
  3. Jan 17, 2009 #2

    symbolipoint

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    Could you use the individual densities of Aluminum and of Lithium? They should be listed in a handbook. Maybe try setting up a chart or table for a given number of grams or cubic centimeters of all materials involved.
     
  4. Jan 17, 2009 #3

    Borek

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    As Symbolipoint suggests, you are most likely expected to assume that mass of the mixture is sum of the masses of components (always true) and sume of the volumes is the sum of the volumes. The latter assumption is only an approximation, as volumes are in general not additive (google "ethanol water volume contraction").
     
  5. Jan 18, 2009 #4
    so in this problem I am allowed to assume I can add the two volumes together?

    so say:
    V(Al)+V(Li)=V(total)
     
  6. Jan 19, 2009 #5

    Borek

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    Unless you were just discussing volume contraction tables or excess volumes - which can be part of some advanced physical chemistry course - yes.
     
  7. Jan 19, 2009 #6
    ok i have almost got it . . .for some odd reason I am just unable to solve for the mass of each element. No idea why . . .but I just realized that someone else has posted the same problem . . .sorry, still new to this forum. I saw he got

    mass Al = 2.7VAl
    mass Li = .543VLi

    I have no clue where he got those two. Its been a long time since I have taken chemistry . .
     
  8. Jan 19, 2009 #7

    Borek

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    Show your system of equations.
     
  9. Jan 19, 2009 #8
    V(Al)+V(Li)=V(total)
    V(Al)+V(Li)=5 mL (where 5 is just an arbitrary number)

    solve for each:
    V(Al)=5-V(Li)
    V(Li)=5-V(Al)
     
  10. Jan 19, 2009 #9

    Borek

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    Volume is not enough. You need another equation, that will combine volumes and densities.

    Reread my first post in this thread.
     
  11. Jan 19, 2009 #10
    density = mass/volume

    saying that 1mL of each

    2mL total

    mass of Al = 2.7*1= 2.7 grams
    mass of Li =0.534*1= 0.534 grams
    mass of the entire system = (2.55)*2 = 5.1 grams

    putting that into the equation:

    Mass(Li)*Volume(Li)+Mass(Al)*Volume(Al)=5.1

    solving gets to about 1.45 percent weight of Li?
     
  12. Jan 19, 2009 #11

    Borek

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    This is chaotic. You first assume you have to mix 1mL of each, then you list their masses, suggesting that 2.7+0.534=5.1 (no, it doesn't equal). Then you write some unrelated equation...

    Hard to follow, if possible. Could be you think OK, but you have to show it. Slow down.

    You have to mix VLi and VAl, these are your unknowns (alternatively your unknowns can be mLi and mAl). Decide which unknowns do you use, then write equations that you want to use. Write them here.
     
  13. Jan 19, 2009 #12
    mass Al = 2.7V(Al)
    mass Li = .543V(Li)

    V(Al)=2-V(Li)
    V(Li)=2-V(Al)

    putting the 4 equations together if the form of:

    Volume(Li)+Volume(Al)=2 mL

    Mass(Li)/Density(Li)+Mass(Al)/Density(Al)=Mass/Density

    0.534(2-V(Al))/0.534+2.7*(2-V(Li))/2.7=5.1/2.55

    **This is where I think go wrong**
     
  14. Jan 19, 2009 #13

    Borek

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    OK

    It is not wrong, it just doesn't make sense. You have substituted both volumes by (2-other volume) - do it with one volume only, you will get one equation in one unknown.
     
  15. Jan 19, 2009 #14
    [0.534(V(Li))/0.534]+[2.7*(2-V(Li))/2.7]=[5.1/2.55]

    but in this case doesnt everything cancel?
     
  16. Jan 19, 2009 #15

    Borek

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    Everything cancels out. I got lost in the chaos.

    This is again equation for sum of volumes. You have already used it.

    Write equation for the sum of masses. That's what I suggested in my first post, that's what I asked you to do 3 hours ago.
     
  17. Jan 19, 2009 #16
    ok ya thats what started to do but for some odd reason. . .im stupid:

    Mass(Li)+Mass(Al)=Mass

    density(Li)*volume(Li)+density(Al)*volume(Al)=(density)*(volume)

    0.534*V(Li)+2.7*(2-V(Li))=2.55*2

    solving for Volume one gets .13876 mL
    and to get mass (.534*.13876)=0.0741 grams
     
  18. Jan 19, 2009 #17

    symbolipoint

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    How he expresses what he is thinking is difficult to follow, but he seems to be thinking correctly - at least mostly correctly. The stated result of 1.45% Lithium is correct.

    Try to follow this: Assume 100 grams of alloy, use m1 for mass of lithium and m2 for mass of aluminum. For this 100 grams, continuing with d for density, and v for volume, we can say that the volume of this 100 gram of alloy is
    (m1/d1 + m2/d2)

    We know the given density of the alloy of 2.55 grams/cubic centimeter, so we can say
    (m1 + m2)/(m1/d1 + m2/d2) = 2.55

    The other equation which we can use is for the assumed arbitrary 100 g. sample,
    100 = m1 + m2

    Those are the two equations in two unknowns. The rest is simple properties of equality for Real numbers; a few tedious but easy algebra steps. We could find two equivalent expressions for m2 from each equation, equate those expressions, and then find the resulting formula for m1. We will find the value which KMjuniormint5 did, 1.45%.
     
    Last edited: Jan 19, 2009
  19. Jan 19, 2009 #18
    ok so my original thought of 1.45% was within the ballpark of an answer?
     
  20. Jan 20, 2009 #19

    Borek

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    Looks like. I told you - you could be right, it was just hidden :wink:

    Whenever you are trying to explain what you are doing try to write explicitely each step down. That may be tedious at first, but if you will get back later to what you wrote and you will try to understand it you will see where and why you are hard to follow - and you should learn then what can be left out, what has to be included to be sure others can understand.
     
  21. Jan 20, 2009 #20
    ok i understand where the confusion comes into play as i was helping someone with physics on here. . . .thanks again for the help and getting me totally frustrated but in the end. . .i understand now where I need to go with problems like this. . .thanks so much kudos to ya
     
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