Titration and Moles question

A student carries out an experiment to identify an unknown carbonate.
The student weighs a sample of the solid carbonate in a weighing plate --> 14.92g
the student weighs the empty weighing bottle --> 13.34g
student prepares 250.0cm3 solution of the carbonate
the student carries out a titration using 25.0cm3 of the solution with 0.100mol dm-3 hydrochloric acid in the burette.
EQUATION
M2CO3(aq) + 2HCl (aq) --> 2MCl(aq) + CO2(g) +H2O(l)
TITRATION
Trial - 22.5 (titre), 1 - 22.20 , 2 - 22.20.
MOLES of HCl used: 2.22 x 10-3mol
MOLES of M2CO3 in 250.0cm3: 0.0111mol
QUESTION
Identify the carbonate M2CO3

The answer is Potassium but how??
The mark scheme and my answer was 41.171 which would be Calcium on the periodic table, so why would it not be Ca2CO3?

Reply 1

dbhc2411

I haven't done the calculation just yet, but straight away I know it cannot be Calcium, because of the equation.

M2CO3 + 2HCl -> 2MCl + CO2 + H2O

The carbonate ion has a -2 charge, so for the conservation of charge, M must be a group 1 metal.
Titrations aren't perfectly accurate, so the closest mass of a group 1 metal from the answer must be potassium.

With your suggestion of Ca2CO3, this would not be correct as the charge of this compound would be 2+
(Ca = 2+) (CO3 = 2-) 2 x Ca = +4, CO3 = 2- = +2 overall.

Hope this helps!

(edited 3 months ago)

Reply 2

sera14

That makes sense now, thank you!

Reply 3

dbhc2411

Original post by sera14

That makes sense now, thank you!

No problem! I just double checked the calculation, looks good - I got the same answer.

This is just one of those situations where it's highlighting that experimentally, equipment is sadly not perfect. The slightly high mass calculated will be down to the equipment errors of the mass, burette, volumetric flask (tiny), measuring cylinders etc all combined 🙂