Relative Molecular Mass of a Gr2 Carbonate

Hello, I'm new and I found this site through Google. 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 he only ends up ridiculing you for not knowing how, and in the end he doesn't tell you how to do it anyway, so I'm fed up of going to him.

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 weigh 1.34grams of the solid A in a container.
We then use a pipette to measure 50cm^3 of 1 mol dm^-3 aqeous 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 fromthe burette itno the conical flask untilt he end point if reached.

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

RESULTS
Weight of Solid A 1.34g
Container 4.51g
Container + Solid A 5.85g

Mean Titre = 23.6cm^3 of .1 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 om moles) of hydrocholic acid reacting with NaOH calculated in (a). Calculat 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 wtih 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

Reply 1

Ben.S.

11

ramroff

Hello, I'm new and I found this site through Google. 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 he only ends up ridiculing you for not knowing how, and in the end he doesn't tell you how to do it anyway, so I'm fed up of going to him.

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 weigh 1.34grams of the solid A in a container.
We then use a pipette to measure 50cm^3 of 1 mol dm^-3 aqeous 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 fromthe burette itno the conical flask untilt he end point if reached.

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

RESULTS
Weight of Solid A 1.34g
Container 4.51g
Container + Solid A 5.85g

Mean Titre = 23.6cm^3 of .1 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 om moles) of hydrocholic acid reacting with NaOH calculated in (a). Calculat 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 wtih 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

a) To find the number of moles of NaOH required: moles = molarity x volume (best to always use SI units of dm^3 or mol/dm^3 if you're unsure of yourself).

Moles = 0.1 x (23.6/1000) = 0.00236 moles

b) Reaction stoichiometry is 1:1, therefore 0.00236 moles of HCl reacted. That means there were 0.00236 x 10 = 0.0236 moles of HCl in the 250cm^3 flask.

c) Moles = 1 x (50/1000) = 0.05 moles HCl. Therefore 0.05-0.0236 = 0.0264 moles of HCl reacted with the carbonate.

d) 0.0264/2 = 0.0132 moles of carbonate in 1.34g of MCO3 (reaction stoichiometry is 1:2 this time). Mr = mass/moles, so A has a molecular mass of 101.52 The metal has an atomic mass, therefore of 101.52 - 12 - (16 x 3) = 41.52 and is likely to be calcium (what a surprise).

Ben

Reply 2

BCHL85

12

Remind you a bit with your first equation.
MCO3 + 2HCl -> MCl2 + H2O + CO2 (you missed MCl2)
Anyway. there's no change in Ben's solution. :smile:

It's perfectly fit with Ca ... Mr = 40 and it's in group II as well. Some errors in the experiment make Mr be slightly greater.

Relative Molecular Mass of a Gr2 Carbonate

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