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Author Topic: Problem of the week - 06/05/2013  (Read 9375 times)

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Borek

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Problem of the week - 06/05/2013
« on: May 06, 2013, 05:34:10 AM »

What is formula of a basic calcium phosphate present in the tooth enamel, if its molar mass is 502 g/mol?
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delta609

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Re: Problem of the week - 06/05/2013
« Reply #1 on: May 06, 2013, 07:22:46 AM »

Ca5(PO4)3OH
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Big-Daddy

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Re: Problem of the week - 06/05/2013
« Reply #2 on: May 06, 2013, 11:00:26 AM »

I think it is Ca5(PO4)3OH. Write equations:

-v[OH]+2v[Ca]-3v[PO4]=0 (total charge is assumed to be 0)
v[OH]·Mr[OH]+v[PO4]·Mr[PO4]+v[Ca]·Ar[Ca]=502

And since we don't have a third equation, we can't solve this system rigorously (i.e. by direct substitution or elimination), so we'll have to guess one of the variables (which is ok because Mr=502 only, hopefully it will not be too high). Start with v[OH]=1 (plan to move onto v[OH]=2, v[OH]=3 etc. if you don't get integer values for the other coefficients when v[OH]=1) and you calculate v[PO4]=3, v[Ca]=5 immediately.
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stewie griffin

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Re: Problem of the week - 06/05/2013
« Reply #3 on: May 06, 2013, 01:51:19 PM »

I agree with delta609. It's hydroxyapatite.
I'm confused on how to solve this problem though without just "knowing" what hydroxyapatite is.
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delta609

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Re: Problem of the week - 06/05/2013
« Reply #4 on: May 06, 2013, 01:59:27 PM »

trial and error and the fact that the question states it is a basic phosphate helps to narrow it down a little
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Borek

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Re: Problem of the week - 06/05/2013
« Reply #5 on: May 06, 2013, 09:50:20 PM »

I'm confused on how to solve this problem though without just "knowing" what hydroxyapatite is.

Big-Daddy nailed it (I mean the method, he was not the first to post the answer).

You are told it is a basic (OH-) calcium (Ca2+) phosphate (PO43-). Basic means no chances of HPO4- and H2PO4-. Molecule is Cak(PO4)l(OH)m. Then it is just a matter of writing correct charge balance (molecule must be neutral) and molar mass equations:

2k - 3l - m = 0
40k + 95l + 17m = 502

solving them in terms of a parameter and poking the solution space with small integers (which is just a fancy way of calling trial and error ;)).

For example you can solve these equations for k:

[tex]k = \frac{502-44l}{74}[/tex]

and plug l=1, 2, 3... into the equation. When l is 3, you get a nice integer value of 5 for k, so you just plug l=3 and k=5 into original set of equations and you get m=1 - so Ca5(PO4)3OH is the correct answer.

In general this kind of a problem doesn't have to have a unique solution.
« Last Edit: May 06, 2013, 11:36:36 PM by Borek »
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Big-Daddy

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Re: Problem of the week - 06/05/2013
« Reply #6 on: May 09, 2013, 10:03:53 AM »

I plotted a graph varying v[OH] and find v[PO4] and v[Ca] at each integer value of v[OH] from 1 to around 29 (no point in plotting v[OH]>29 because then v[OH]·Mr[OH] on its own is higher than Mr[enamel] - in reality v[PO4] becomes negative if v[OH]>13). Couple of interesting points:

v[Ca] increases as v[OH] increases, and v[PO4] decreases as v[OH] increases - linearly. Didn't completely expect this but given that the charge equation is linear I suppose it makes sense. The gradient of both lines appears to be independent of the Mr[enamel]. Definitely not sure what to make of that, given that Mr[enamel] features in my function.

Lastly, there is no solution at which both v[PO4] and v[Ca] are positive integer values, except the one deduced before. So this problem is totally consistent mathematically.
« Last Edit: May 09, 2013, 10:30:03 AM by Big-Daddy »
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delta609

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Re: Problem of the week - 06/05/2013
« Reply #7 on: May 09, 2013, 01:22:12 PM »

I agree with delta609. It's hydroxyapatite.
I'm confused on how to solve this problem though without just "knowing" what hydroxyapatite is.

Or you could use some simple algebra

given the two equations:

2k - 3l - m = 0
40k + 95l + 17m = 502

solve each for m:

(1) m = 2k -3l
(2) m = (502 - 40k - 95l)/17

since m = m you can solve for l (or k, whichever you want):

l = (502 - 74k)/44

substitute l into (1) or (2).  I used (1).

m = 2k - 3((502-74k)/44)

find (k,m) in which both are whole numbers on a graphing calculator's table feature which turns out to be (5,1)

now you have your k and m values.  solve for l

2(5) - 3l - 1 = 0
9 - 3l =0
-3l = -9
l=3

Ca5(PO4)3OH

Big Daddy and Borek's methods might be better, but I'm just showing you another way of solving it without trial and error.


 
« Last Edit: May 09, 2013, 01:34:11 PM by delta609 »
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Borek

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Re: Problem of the week - 06/05/2013
« Reply #8 on: May 09, 2013, 09:04:48 PM »

find (k,m) in which both are whole numbers on a graphing calculator's table feature which turns out to be (5,1)


Technically what you did is you solved the diophantine equation. You could do it earlier, when you got to the equation

l = (502 - 74k)/44

Converting it into equation in (m,k) doesn't change anything - you still have a diophantine equation in two integer unknowns. Equation above is equivalent to

22l + 37k = 251

This is guaranteed to have an infinite number of solutions: http://en.wikipedia.org/wiki/Bézout's_identity, but the first set of coefficients has to be find by trial and error. And from what I understand that's what you did, you just used calculator to do so.
« Last Edit: May 09, 2013, 09:18:20 PM by Borek »
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Big-Daddy

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Re: Problem of the week - 06/05/2013
« Reply #9 on: May 10, 2013, 05:54:46 AM »

I agree with delta609. It's hydroxyapatite.
I'm confused on how to solve this problem though without just "knowing" what hydroxyapatite is.

Or you could use some simple algebra

given the two equations:

2k - 3l - m = 0
40k + 95l + 17m = 502

solve each for m:

(1) m = 2k -3l
(2) m = (502 - 40k - 95l)/17

since m = m you can solve for l (or k, whichever you want):

l = (502 - 74k)/44

substitute l into (1) or (2).  I used (1).

m = 2k - 3((502-74k)/44)

find (k,m) in which both are whole numbers on a graphing calculator's table feature which turns out to be (5,1)

now you have your k and m values.  solve for l

2(5) - 3l - 1 = 0
9 - 3l =0
-3l = -9
l=3

Ca5(PO4)3OH

Big Daddy and Borek's methods might be better, but I'm just showing you another way of solving it without trial and error.

The component of trial and error is exactly the same. Although taking it from a table is the best way of checking for all possible integer solutions.

In this case, there are no positive integer solutions besides (1,3,5). I checked by calculating the results for the whole table, until v[PO4] ceases to be positive and thus we cannot get results as we need.
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delta609

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Re: Problem of the week - 06/05/2013
« Reply #10 on: May 11, 2013, 08:48:45 AM »

You're right.  It's still trial and error, but it is the simplest method for trial and error.  Which is the way I should have phrased my answer in the first place.  Sorry about that folks. 
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