[hanita304]

so i got this problem

Hi,

. It involves the combustion of three different hydrated salts: Na2CO3.10H2O, MgSO4.7H2O and CuSO4.5H2O:

You are given a mass of three hydrated salts: (see above). The total mass of the mixture is 12.123 grams. When the mixture is gently heated, the following two reactions occur:

Na2CO3.10H2O (s) + _______________> Na2CO3.7H2O (s) +3H2O (g)

MgSO4.7H2O (s) ______________> MgSO4.H2O (s) +6H2O (g)

After these reactions are complete, the mass of the mixture has decreased to 9.049 grams. This mixture is then heated more strongly and the following additional reactions occur:

Na2CO3.7H2O (s) ___________________> Na2CO3 (s) + 7H2O

MgSO4.H2O (s) _____________________> MgSO4 (s) + H2O (g)

CuSO4.5H2O (s) _____________________> CuSO4 (s) + 5H2O (g)

After this final heating, the mass of the mixture has decreased to 6.412 grams. From this information, calculate the masses of each of the three compounds in the original mixture.

ANSWERS GIVEN: 1.374 grams of Na2CO3,.10H2O, 6.418 grams of MgSO4.7H2O and 4.331 grams of CuSO4.5H2O. 

so a friend helped me and explained that you make a system of three equations and in front of each of the variable you put the molar mass of the changed compound over the molar mass of the initial compound. for example one of my equations was
(106/286)x + (120.4/246.6)y +(159.7/249.7)Z=6.412.
My question was, why do you do this. i dont understand the reasoning behind this. if you could explain this that would be great. feel free to ask me to clarify if some things arent clear in my question

thank you, hannah

[magician4]

My question was, why do you do this.

do what, exactly? the whole process?

situation is: you have a problem (what is the composition of...) and you want to know the answer.
to get this answer, somebody in the long ago past figured out, that making clever use of what we know about the behaviour of different hydrates at different temperatures might hold the key to the solution of our problem.

he also must have had a certain idea of "molar mass belonging to" as as of the law of constant composition, and he also must have had a at least basic knowledge of math, i.e. must have known that and how a system of x equations with x variables a,b,c, .. can be calculated for.

after all this, he transformed his chemical problem to a mathematical problem, and constructed those three equations, thereby making use of above named knowledge.

last not least, he applied a well known algorithm to solve this system of equations, and recalculated those mathematical results to chemical composition of his sample

now, what exactly from all this don't you understand?


regards

Ingo

[blaisem]

EDIT: I for some reason was dividing by molecular mass when I of course needed to be multiplying by molecular mass to get units of grams.  I also solved the matrix wrong.  I'll upload the correct things.  I did it a different way than your friend did, Hannah, so I'm sorry to confuse you.  My post is already here, though, so I will put it up.  If I could explain how your friend did it, I would.  Hopefully that doesn't confuse you.  This was good practice for me because I was doing the matrices wrong.

Edit 2: Ok, you guys all responded, thank you for input!  Curiouscat, I tried your method first actually, but for some reason I botched it on paper, so it appears that I abandoned the easiest route right at the outset.

As for Hannah, I have no idea how the formula your friend used worked.  But you can use either Cat's or my way to do it.  Cat's is simple because it deals with just the molecular weight of water, which makes it very easy and keeps the numbers small.  If you like to do things complicated, I'll explain my way.

Given that we know the molecular ratio between each salt in each reaction was 1:1 (both have molecular coefficients of one), we know that they are equal in moles.  After that, we are given 3 different masses, so we can set up an equation for each mixture mass.  We know that Moles * Molecular Weight = grams.  So our next step is to set up our equations, where we add all the masses of each individual salt and set it equal to the corresponding mixture mass.

It's a property of multivariate equations (equations where we have more than one variable) that you need as many equations as you have variables.  Since there are 8 different salts, we potentially have 8 different variables and would need 8 different equations.  However, as stated above, due to the 1:1 molar ratios, we can say that 5 of the salts are equal in molar amount as the original three salts.  This reduces us down to 3 variables.  Good, because we only have 3 equations.  Now we can solve them.

I put them in matrix form, which is just a slightly condensed form of a system of equations.  Google augmented matrix to see an example.  The gist of it is that it's the same as writing out the equations as I did in my set up picture below, but when I write it as a matrix, I leave out the x, y and z, and just write the coefficients in.  Then I solve the system of equations using the method of elimination, which you should have seen in Algebra.

I show in the picture below how I set this up (I counted wrong and at some point said 6 variables instead of 8--don't ask how I managed that!).  Let us know if you have more questions!

[magician4]

my impression is, that your problem might result from the way you set up you matrix

in my opinion, in the first row of "My matrix setup" it should have been for example "x*a + y*b + z * c = 12.123  "  (i.e. multiplying insted of dividing)*⁾, and you repeated that approach from there on [and your friend seems to have fallen into a subtype the same trap]

you might wish to check whether, with this corrected, you'd gain the propper results

regards

Ingo



*⁾
m_total = m_a + m_b + m_c = n(a)*M(a) + n(b)*M(b) + n(c)*M(c)

[curiouscat]

I get 1.44, 6.4 and 4.3 gm as answers.

Close enough, I think. I used approximate MWs

The way I found easiest was to use a differential water balance.

My equations are:

3x+6y=0.171,

7x+y+5z=0.147,

286 x + 246 y + 250 z = 12.123

x,y,z are initial moles of each hydrated salt.

[Big-Daddy]

The actual answers are not exactly as given but rather 4.33261 g of the copper hydrate, 6.41808 g of the magnesium sulphate hydrate and 1.3723 g of the copper carbonate hydrate. ("Actual" answers with molecular mass to 2 decimal places, 3 for H, for each element; fractions used otherwise.) http://www.wolframalpha.com/input/?i=c%2Bm%2Bn%3D12.123%2C+c%2B%28232.102%2F286.15%29*n%2B%28138.376%2F246.472%29*m%3D9.049%2C+%28159.61%2F249.69%29*c%2B%28105.99%2F286.15%29*n%2B%28120.36%2F246.472%29*m%3D6.412

[magician4]

@ Big-Daddy:

though from a mathematical point of view your calculation is correct, you might wish to contemplate for a moment a situation where initial masses are given with three decimals (i.e. the hypothetical fourth decimal being unknown), whether it would be wise to give results with like five decimals (for, as a scientist, you're "responsible" for any number you're writing down):

would it be wise for example, if you knew that the total mass of a mixture of three components was 1,000 g, and if you additionally knew that (let's say for stoichiometric reasons) each component is 1/3 of the total, that the commitment of each component was 0.33333 grams (give or take whatever might follow in further decimals) ?


... let alone that, case in point , i.e. the solving of 3 linear equations with 3 unknowns, would demand to be more cautious, still, due to the laws of error propagation (that should be applied here)
 
regards

Ingo


btw.:

and 1.3723 g of the copper carbonate hydrate.

you mean copper sulfate pentahydrate, i take it

[curiouscat]

@ Big-Daddy:

though from a mathematical point of view your calculation is correct, you might wish to contemplate for a moment a situation where initial masses are given with three decimals (i.e. the hypothetical fourth decimal being unknown), whether it would be wise to give results with like five decimals (for, as a scientist, you're "responsible" for any number you're writing down):

+1

I agree with @magician

[Big-Daddy]

though from a mathematical point of view your calculation is correct, you might wish to contemplate for a moment a situation where initial masses are given with three decimals (i.e. the hypothetical fourth decimal being unknown), whether it would be wise to give results with like five decimals (for, as a scientist, you're "responsible" for any number you're writing down):

I agree, but that was not the main point of my post. Surely you would agree that that given answers of m₀(CuSO4·5H2O)=4.331 g and m₀(Na2CO3·10H2O)=1.374 g are not correct in light of the more exact method (my only approximation was the Mr values used, and at some level Mr is always approximate AFAIK) which gives m₀(CuSO4·5H2O)=4.333 g and m₀(Na2CO3·10H2O)=1.372 g. It's got nothing to do with reported accuracy. Only a small quibble but I wanted to save people going back over their M_r calculations to check if they plugged something in wrong (which is what I did initially).

Anyway I see your point with the 5dp.