Back Titration Of Metal Carbonate

[Lancelot59]

I'm having an issue solving a back titration problem. We need to identify the metal in a metal carbonate. It can either be a group 1A or 2A

1.8500 grams of the carbonate are diluted to .2500L and then 0.02000L is taken and mixed with 0.01500L of 0.1789M HCl in a flask. After shaking the mixture around 0.01756 L of 0.1013M NaOH was used to titrate the remaining HCl.

I started this by first determining how many moles of HCl were added to the flask, and how much titrated with the NaOH:

0.01500L*0.1789M = 0.0026835 mol HCl total in flask
0.01756 L NaOH * 0.1013M x 1mol HCl/1mol NaOH = 0.001778828 mol HCl titrated.

So 0.000904672 mols of the HCl reacted with the metal carbonate. From here I would go to determine how many mols of the carbonate there are, but I'm not sure how to do that. Writing the equation could turn out two different ways becuase I don't know how the metals are grouped. Just picking one and running with it gave me a molar mass of about 3400...not good.

This is what I think the equation should be:
M₂CO₃ + 2HCL -> H₂CO₃ + 2MCL₂

How can I solve this?

 

[Borek]

Ignore metal - it is just a spectator. Calculate number of moles of carbonate anion.

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[Lancelot59]

What do I use for a mole ratio though? If I just ignore the metal I came up with:

CO₃^-2 + 2HCL --> H₂CO₃ + 2Cl^-

So using that setup:

0.000452336 mol HCL x 1mol CO₃^-2/2mol HCL = 0.000226168 mol CO₃^-2

Then since g=mols*molar mass --> molar mass = g/mols. The molar mass turns out to be 8179.760178. A bit much...

 

[Borek]

Perhaps there is something more, but seems like you forgot about dilution.

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[Lancelot59]

Right...oops. /facepalm

So there's 0.000226168 mol in 0.02000L, so 0.113084M x 0.250L = 0.028271 mols of the anion in the main sample bucket. So using that my molar mass is 65.4ish... well it's getting better at least. I'll do some more math and edit it in.

No, still not working. I dropped a 0. The molar mass I got was 327ish. I have no clue what's going wrong.