September 20, 2020, 03:46:51 PM
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Topic: How much tin is in a wheat penny (lab question)  (Read 299 times)

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Offline Krynvel

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How much tin is in a wheat penny (lab question)
« on: September 14, 2020, 11:24:29 PM »
Hey guys I need some help with a lab question.

A density of a penny is 8.8g/cm3
the composition of the penny is 95% copper, 5% tin and zinc
density of copper is 8.96g/cm3
density of tin is 7.31g/cm3
density of zinc is 7.13g/cm3

The question is how much tin is in the penny?

Here's how I approached this problem,
d(coin)=(d[Cu]*95%)+(d[Zn]*5%-[Sn]%)+(d[Sn]*[Sn]%)
8.8=(8.96*0.95)+(7.13*[Zn]%)+(7.31*[Sn]%)
(0.288-(7.31*[Sn]%))/7.13=[Zn]% <- Plug this back into the original equation

I ended up here and I got stuck.
0.288=(0.288-7.31*[Sn]%)+(7.31*[Sn]%)

I tried using limits on the original equation: if [Zn]% approached 0, what % composition would Tin approach and it gave me an answer that I couldn't accept which was less than 1%.

A guiding hand to point me to the fault of my logic would be greatly appreciated.



Online chenbeier

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Re: How much tin is in a wheat penny (lab question)
« Reply #1 on: September 15, 2020, 01:37:59 AM »
Not Impossible,  You dont know tin eiher zinc. Your calculation is wrong. 5% = Sn + Zn

Offline Borek

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Re: How much tin is in a wheat penny (lab question)
« Reply #2 on: September 15, 2020, 03:04:54 AM »
d(coin)=(d[Cu]*95%)+(d[Zn]*5%-[Sn]%)+(d[Sn]*[Sn]%)

Right direction, but wrong execution. Assume x to be the fraction of tin and y to be the fraction of zinc and rewrite the equation in terms of these unknowns.

What do you know about sum of x and y?
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Offline Krynvel

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Re: How much tin is in a wheat penny (lab question)
« Reply #3 on: September 15, 2020, 06:20:28 PM »
I know the sum of x and y to add to 5% of the total density which is 0.44 g/cm3
or 5%-%[Sn]=%[Zn]
Plug that into the originial equation...

So what I got was:
(0.05)(8.8g/cm3)=(7.13g/cm3*(0.05-%[Sn]))+7.31%[Sn]
Simplify:
0.44g/cm3=0.3565g/cm3+0.18%[Sn]
Simplify:
0.0835=0.18%[Sn]
Simplify:
0.46[of 5% of total density]=%[Sn]
Reworded:
%[Sn]=2.3%

Now my question...
Why couldn't I use dcoin=(dcopper*%copper)+(dzinc*%zinc)+(dtin*%tin)

Whenever I used that equation I got
0.288g/cm3for the 5% of total density by subbing in copper density and %copper and subtracting that value from the density of the coin.

Offline Borek

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Re: How much tin is in a wheat penny (lab question)
« Reply #4 on: September 16, 2020, 03:19:05 AM »
I know the sum of x and y to add to 5% of the total density which is 0.44 g/cm3
or 5%-%[Sn]=%[Zn]
Plug that into the originial equation...

So what I got was:
(0.05)(8.8g/cm3)=(7.13g/cm3*(0.05-%[Sn]))+7.31%[Sn]

No, this is not correct, I feel like you are trying to make a shortcut and it fires away. Please write the whole equation, including information you have on all three metals.
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Offline Enthalpy

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Re: How much tin is in a wheat penny (lab question)
« Reply #5 on: September 16, 2020, 05:01:32 AM »
The method is basically wrong. Whoever asked the question is mistaken.

In alloys, volumes don't add. Bell bronze and Invar have a bigger volume than their constituents. NiTi changes its volume completely between austenitic and martensitic forms. TiAl has less volume than its constituents.

But there is an interesting correlation between the volume change when alloying and Young's modulus (the stiffness if you wish).

The same happens with liquids, for instance a mixture of ethanol and water occupies less volume than the sum of the constituents.

Offline Borek

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Re: How much tin is in a wheat penny (lab question)
« Reply #6 on: September 16, 2020, 06:23:37 AM »
The method is basically wrong. Whoever asked the question is mistaken.

Yes, but let's tackle the problem in only possible way first, then we will discuss why it is only an approximation.

As of now we have to correct math behind the simplified model.
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Offline AWK

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Re: How much tin is in a wheat penny (lab question)
« Reply #7 on: Today at 10:28:31 AM »
The discussion on this topic has died down.
After a week, it's time to explain what the error is. Calculating this problem is possible assuming the additivity of the volume. As Enthalpy rightly pointed out, there may be surprises, but for the total content of Zn + Sn = 5%, surprises are unlikely to be expected.
For both 5% of binary alloys of Cu with Sn or Zn, manufacturers give a density of 8.85 (assuming volume additivity, the calculated values ​​are 8.860 and 8.846 respectively - i.e. the error is less than 2%).
There are no data for three-component alloys with such a low content of Zn and Sn, but with a high probability it can be expected that their density will be within these limits (8.846 to 8.860). And only for the density within such limits the problem is solvable. For Sn content 4 to 1% and Zn up to 5% I got densities 8.857, 8.855, 8.852 and 8.849, respectively.
AWK

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