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Mon 21 Feb, 2005 09:26 am
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
Algebra Algebra Algebra
The reaction with the Acid (HCL) is
mco3+2HCL->mcl2+H2CO3
the effervescence is CO2 coming off
H2CO3->H2O+CO2^(bubbles of gas)
The Neutralization with NaOH is with the excess HCl
HCl+NaOH->H2O+NaCl (1)
now lets look at the reaction
You have 1.34g of MCO3 with an unknown molecular weight so what you have to do is determine the number of moles reacted. You've added 0.05 moles HCl (50ml/1000ml*1 mol HCl) into a solution and reacted all of the MCO3. From the reaction above you've got excesss HCl in the solution---set this up as an unknown--excess HCl in solution is then 0.05-2X moles. The X represents the moles of MCO3 and the 2 comes from the reaction quantities on equation 1 (2 moles of HCl to react with 1 mole MCO3).
You make this solution up to 250 ml and take 25mls for neutralization with NaOH (.1 molar soln). You use 23.6 ml to neutralize. This is 0.00235 ml NaOH (23.6ml/1000ml*.1 mol NaOH).
So I set up the algebraic equation as
(0.05-2X)*25ml/250ml=0.00236mol and solve for X
X in this case is 0.0132 mole which is equivalent to
1.34 g MCO3
and 1.34g MCO3/0.0132mol=101.5 g/mol
now CO3 has a molecular weight of about 60 (12+3*16), so the metal has a molecular weight of about 41.4 (101.5-60). So i'd look for a group 1 metal with a molecular weight of about 20 (40/2) or a group 2 metal with a molecular weight of about 40.
Looking at my handy dandy periodic table I see that sodium (Group 1 Na) has a mol weight of 23 and Calcium (Group 2 Ca) has a mol wt of 40--so calcium is a good fit (plus CaCO3 is a very common chemical salt and a consequently a cheap unknown).
If I wanted to refine my two possibilities, I'd look at physical descriptions of these two possible inorganic salts and see which matched the properties (color density solubility) of the unknown. Nevertheless it sounds like it would be CaCO3.
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