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Relative Molecular Mass of a Gr2 Carbonate

I hope you don't mind me asking a chemistry question, I am trying to find the relative molecular mass of a Group 2 carbonate in a back titration

M is unknown

MCO3 (s) + 2HCl (aq) -> H2O (l) + CO2 (g)
HCl (aq) + NaOH (aq) -> NaCl (aq) + H2O (l)

I would ask my chemistry teacher for help, but she only ends up ridiculing you for not knowing how, and in the end, she doesn't tell you how to do it anyway, so I'm fed up of going to her.

All help appreciated, thanks in advance

We are given a solid A, which is a Group 2 carbonate, MCO3 (M is unknown) and we have to carry out a 'back titration'.

We weighed 1.35grams of the solid A in a container.
We then use a pipette to measure 50cm^3 of 1 mol dm^-3 aqueous Hydrochloric acid into a 250cm^3 volumetric flask and then place a funnel in the top of the flask.
Carefully transfer A into the dry funnel and reweight the container. Wash A into the flask with distilled water, when the effervescence has stopped, make the solution in the vol. flask up to the mark with distilled water then shake it.

Fill a burette held in a clamp with NaOH solution provided (0.1mole concentration).

After rising it out with the solution made up in 3 use a pipette to transfer 25cm^3 of the solution fo a conical flask, then add four drops of indicator provided.

Then titrate the alkali from the burette into the conical flask until the endpoint is reached.

Then repeat until you obtain two titres within 0.20cm^3 of each other.



RESULTS
Weight of Solid A 1.35g
Container 16.70g
Container + Solid A 18.05g

Mean Titre = 27.10cm^3 of 0.0966 mol dm^-3 sodium hydroxide



Sorry for being so long-winded but I thought I should post the procedure so you could understand the back titration.

CALCULATIONS

a) calculate the amount (# of moles) of NaOH needed to react with 25cm^3 of the acidic solution made up in the volumetric flask.

b) Sate the amount (number of moles) of hydrochloric acid reacting with NaOH calculated in (a). Calculate the # of moles of HCl in excess in the volumetric flask.

c) Calculate the amount (# of moles) of HCl added to the volumetric flask. Hence, calculate the # of moles of HCl that reacted with the solid A.

d Calculate the relative molecular mass (molar mass) of A and suggest the identity of the carbonate.


Sorry for the long post, but could someone help me with the calculations? What steps do I take, and which formulas do I use?

Thanks
(edited 6 years ago)
Original post by champion1221221
I hope you don't mind me asking a chemistry question, I am trying to find the relative molecular mass of a Group 2 carbonate in a back titration

M is unknown

MCO3 (s) + 2HCl (aq) -> H2O (l) + CO2 (g)
HCl (aq) + NaOH (aq) -> NaCl (aq) + H2O (l)

I would ask my chemistry teacher for help, but she only ends up ridiculing you for not knowing how, and in the end, she doesn't tell you how to do it anyway, so I'm fed up of going to her.

All help appreciated, thanks in advance

We are given a solid A, which is a Group 2 carbonate, MCO3 (M is unknown) and we have to carry out a 'back titration'.

We weighed 1.35grams of the solid A in a container.
We then use a pipette to measure 50cm^3 of 1 mol dm^-3 aqueous Hydrochloric acid into a 250cm^3 volumetric flask and then place a funnel in the top of the flask.
Carefully transfer A into the dry funnel and reweight the container. Wash A into the flask with distilled water, when the effervescence has stopped, make the solution in the vol. flask up to the mark with distilled water then shake it.

Fill a burette held in a clamp with NaOH solution provided (0.1mole concentration).

After rising it out with the solution made up in 3 use a pipette to transfer 25cm^3 of the solution fo a conical flask, then add four drops of indicator provided.

Then titrate the alkali from the burette into the conical flask until the endpoint is reached.

Then repeat until you obtain two titres within 0.20cm^3 of each other.



RESULTS
Weight of Solid A 1.35g
Container 16.70g
Container + Solid A 18.05g

Mean Titre = 27.10cm^3 of 0.0966 mol dm^-3 sodium hydroxide



Sorry for being so long-winded but I thought I should post the procedure so you could understand the back titration.

CALCULATIONS

a) calculate the amount (# of moles) of NaOH needed to react with 25cm^3 of the acidic solution made up in the volumetric flask.

b) Sate the amount (number of moles) of hydrochloric acid reacting with NaOH calculated in (a). Calculate the # of moles of HCl in excess in the volumetric flask.

c) Calculate the amount (# of moles) of HCl added to the volumetric flask. Hence, calculate the # of moles of HCl that reacted with the solid A.

d Calculate the relative molecular mass (molar mass) of A and suggest the identity of the carbonate.


Sorry for the long post, but could someone help me with the calculations? What steps do I take, and which formulas do I use?

Thanks


Do you understand how a back-titration works?

Do you know how to work out moles of solute from concentration and volume?
First I understand back titration. Also, I did it and got 53.45g and suggested that it was Calcium but I feel it is wrong since calcium is around 40
Original post by champion1221221
First I understand back titration. Also, I did it and got 53.45g and suggested that it was Calcium but I feel it is wrong since calcium is around 40


Have you done an inaccuracy analysis to see if your answer is in the ball-park?

I agree with your answer btw.
(edited 6 years ago)
I did not
Original post by champion1221221
I did not


Then you should - your answer's good

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