[Rutherford]

A sample of mineral dolomite (m=0.7g) is completely dissolved in HCl, the solution is neutralized and diluted in a normal court of 100cm³ to the line. 25cm³ of that solution is measured and (NH₄)₂C₂O₄ is added in excess. The precipitate was then flushed, strained and dissolved in dilute H₂SO₄. The obtained solution is titrated with KMnO₄ solution (c=0.02mol/dm³). 18cm³ of KMnO4 solution was spent. Calculate the % of Ca in the dolomite sample.

Dolomite is MgCO₃*CaCO₃.
After dissoluting in HCl I got: MgCl₂ and CaCl₂.
Here I stopped. What makes the precipitate with C₂O₄^2-? Mg,Ca or both?

[Borek]

Both.

[Rutherford]

Then I have both MgC₂O₄ and CaC₂O₄. After dissolving in H₂SO₄ I get MgSO₄ and CaSO₄. I don't know now how MgSO₄ or CaSO₄ react with KMnO₄.
Is it this way:
3CaSO₄+2KMnO₄+10H⁺-->3Ca(OH)₂+K₂SO₄+2MnSO₄+2H₂O and the same for Mg.

[Borek]

It is not sulfate that reacts with permanganate, SO₄^2- is just a spectator.

Note that your reaction equation:

3CaSO₄+2KMnO₄+10H⁺-->3Ca(OH)₂+K₂SO₄+2MnSO₄+2H₂O

is not balanced - balanced means not only the same amount of atoms, but also the same charge on both sides. Charge is conserved just like mass is conserved.

[Rutherford]

Oh, right. Then MgC₂O₄ and CaC₂O₄ are reacting with KMnO4 in the presence of H⁺ ions. I will try now to write the correct equation.

[Rutherford]

I wrote this equation:
5CaC₂O₄+5MgC₂O₄+4MnO₄^-+32H⁺-->5Ca²⁺+5Mg²⁺+4Mn²⁺20CO₂+16H₂O and after some calculations I got the right result (20.57%). Thanks for the help.

[Borek]

I wrote this equation:
5CaC₂O₄+5MgC₂O₄+4MnO₄^-+32H⁺-->5Ca²⁺+5Mg²⁺+4Mn²⁺20CO₂+16H₂O and after some calculations I got the right result (20.57%). Thanks for the help.

That's surprising, as the equation is wrong. You have two separate reactions occurring, not a one. You can't write them as a single equation, as it means assuming already that they are present in 1:1 molar ratio.

As a result of your approach reaction equation is not unique. Why not

5CaC₂O₄+10MgC₂O₄+6MnO₄^-+48H⁺-->5Ca²⁺+10Mg²⁺+6Mn²⁺+30CO₂+24H₂O ?

It is possible to write as many different reaction equations as you wish.

Would they yield different answers?

[Rutherford]

I have balanced the Ca and Mg equation separately and I thought to write them as a single equation because in dolomite the molar ratio of Mg and Ca must be 1:1 I think.

[AWK]

If name dolomite is use one can expect exact molar ratio CaCO₃ to MgCO₃ as 1:1.
Usually balancing separate reactions followed by addition them needs less effort.

[Borek]

Now that I think about it, 1:1 assumption is necessary to solve the question. Trick is, real dolomite has a rather variable composition (±several percent), so the assumption doesn't have to be true.

[Rutherford]

Came back to this problem and couldn't solve it again. This would be the reaction:
5CaC₂O₄+5MgC₂O₄+4MnO₄^-+32H⁺-->5Ca²⁺+5Mg²⁺+4Mn²⁺+20CO2+16H2O
n of KMnO4 is 3.6*10^-4, n of the oxalates would be 9*10^-4
If the molar ration of Mg and Ca is 1:1, then n(Ca)=9*10^-4/2=4.5*10^-4.
n of Ca in the 100cm³ is 4.5*10^-4/0.025*0.1=1.8*10^-3
The mass is m=1.8*10^-3*40=0.072g
And the mass share 10.286%. Two times lower than the correct one. Can't find a mistake.

[Borek]

As far as I can tell 10.3% is a correct answer. If 18 mL of 0.02M permanganate was needed to titrate the sample containing equimolar amounts of Mg/Ca oxalates, 9 mL of the solution was spent for CaC₂O₄ - that means 0.18 mmole, and whole sample would need four times as much titrant - so 0.72 mmole or 7.2x10^-4 mole. That means there was 0.07214 g of calcium in the sample.

Sorry if the image is too wide.

[Rutherford]

I got the book answer once, but accidentally. Now, the other one (10.3%) seems correct to me. Thanks for the confirmation.

[Borek]

Make a habit of solving problems in a notebook and not throwing solutions away, you will be able to check them later. Also, that's a good way of learning how to properly keep a lab notebook.