Finding the percent weight an element in an Alloy

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Discussion Overview

The discussion revolves around determining the weight percent of lithium required to create an Al-Li alloy with a specified density of 2.55 g/cm³. Participants explore various approaches to solving this problem, which is framed as a homework question involving concepts from chemistry and materials science.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how to start the problem and requests hints.
  • Another suggests using the individual densities of aluminum and lithium and proposes creating a chart or table for calculations.
  • Some participants discuss the assumption that the total mass and volume of the mixture can be calculated by summing the masses and volumes of the components, while noting that volume additivity is an approximation.
  • A participant questions whether they can assume the volumes of aluminum and lithium can be added together directly.
  • There is a discussion about setting up equations based on volume and density relationships, with some participants expressing difficulty in following the logic presented.
  • One participant provides a calculation for the mass of aluminum and lithium based on assumed volumes, but another critiques the approach as chaotic and unclear.
  • Participants suggest writing equations for the sum of masses and volumes, with some proposing specific equations to relate the variables involved.
  • One participant arrives at a calculated weight percent of lithium, stating it as approximately 1.45%, while another confirms this result as plausible but emphasizes the need for clarity in the steps taken to reach it.
  • There is acknowledgment of the confusion experienced during the problem-solving process, with a participant reflecting on the importance of clear communication in mathematical reasoning.

Areas of Agreement / Disagreement

Participants generally agree on the approach to use equations relating mass and volume but exhibit disagreement on the clarity and correctness of the steps taken to arrive at the solution. The discussion remains somewhat unresolved as participants navigate through the complexities of the problem.

Contextual Notes

Some assumptions regarding volume additivity and the relationships between mass, volume, and density are discussed but not fully resolved. The discussion reflects a range of understanding and approaches to the problem, highlighting the challenges faced in applying theoretical concepts to practical calculations.

Who May Find This Useful

This discussion may be useful for students studying materials science, chemistry, or engineering who are grappling with similar problems involving alloy composition and density calculations.

KMjuniormint5
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Homework Statement


Determine the wt% of lithium required to obtain an Al-Li alloy with a density of 2.55 g/cm^3.


Homework Equations





The Attempt at a Solution


I am so lost in this problem, any hints of how to start this problem with GREATLY appreciated!
 
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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.
 
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").
 
so in this problem I am allowed to assume I can add the two volumes together?

so say:
V(Al)+V(Li)=V(total)
 
Unless you were just discussing volume contraction tables or excess volumes - which can be part of some advanced physical chemistry course - yes.
 
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 . .
 
Show your system of equations.
 
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)
 
Volume is not enough. You need another equation, that will combine volumes and densities.

Reread my first post in this thread.
 
  • #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?
 
  • #11
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.
 
  • #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**
 
  • #13
KMjuniormint5 said:
Volume(Li)+Volume(Al)=2 mL

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

OK

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

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.
 
  • #14
[0.534(V(Li))/0.534]+[2.7*(2-V(Li))/2.7]=[5.1/2.55]

but in this case doesn't everything cancel?
 
  • #15
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.
 
  • #16
ok you that's 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
 
  • #17
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:
  • #18
ok so my original thought of 1.45% was within the ballpark of an answer?
 
  • #19
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.
 
  • #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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