Titration and moles question
6) A student carries out an experiment to identify an unknown carbonate.
• The student weighs a sample of the solid carbonate in a weighing bottle.
• The student tips the carbonate into a beaker and weighs the empty weighing bottle.
• The student prepares a 250.0 cm3 solution of the carbonate.
• The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.
(a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and thesolution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.
The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acidto a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.
The equation below represents the reaction between the carbonate and hydrochloric acid.
M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)
(i) Calculate the amount, in mol, of M2CO3 used in the titration.
ii) The student’s mass readings are recorded below.
Mass of weighing bottle + carbonate / g 14.92
Mass of weighing bottle / g 13.34
Use the student’s results to identify the carbonate, M2CO3.
The answer for (i) was 0.00111. How? I can do (ii) obviously if I knew the answer to (i) because that's just rearranging n=M/mr.
So can someone please explain how to get the answer? Thanks.
http://www.ocr.org.uk/Images/171752-unit-h032-02-depth-in-chemistry-sample-assessment-materials.pdf
That's the paper and the mark scheme is on the bottom - question 6.
• The student weighs a sample of the solid carbonate in a weighing bottle.
• The student tips the carbonate into a beaker and weighs the empty weighing bottle.
• The student prepares a 250.0 cm3 solution of the carbonate.
• The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.
(a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and thesolution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.
The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acidto a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.
The equation below represents the reaction between the carbonate and hydrochloric acid.
M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)
(i) Calculate the amount, in mol, of M2CO3 used in the titration.
ii) The student’s mass readings are recorded below.
Mass of weighing bottle + carbonate / g 14.92
Mass of weighing bottle / g 13.34
Use the student’s results to identify the carbonate, M2CO3.
The answer for (i) was 0.00111. How? I can do (ii) obviously if I knew the answer to (i) because that's just rearranging n=M/mr.
So can someone please explain how to get the answer? Thanks.
http://www.ocr.org.uk/Images/171752-unit-h032-02-depth-in-chemistry-sample-assessment-materials.pdf
That's the paper and the mark scheme is on the bottom - question 6.
Reply 1
Original post by _Lizabeth_
6) A student carries out an experiment to identify an unknown carbonate.
• The student weighs a sample of the solid carbonate in a weighing bottle.
• The student tips the carbonate into a beaker and weighs the empty weighing bottle.
• The student prepares a 250.0 cm3 solution of the carbonate.
• The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.
(a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and the solution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.
The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acid to a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.
The equation below represents the reaction between the carbonate and hydrochloric acid.
M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)
(i) Calculate the amount, in mol, of M2CO3 used in the titration.
ii) The student’s mass readings are recorded below.
Mass of weighing bottle + carbonate / g 14.92
Mass of weighing bottle / g 13.34
Use the student’s results to identify the carbonate, M2CO3.
The answer for (i) was 0.00111. How?
• The student weighs a sample of the solid carbonate in a weighing bottle.
• The student tips the carbonate into a beaker and weighs the empty weighing bottle.
• The student prepares a 250.0 cm3 solution of the carbonate.
• The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.
(a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and the solution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.
The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acid to a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.
The equation below represents the reaction between the carbonate and hydrochloric acid.
M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)
(i) Calculate the amount, in mol, of M2CO3 used in the titration.
ii) The student’s mass readings are recorded below.
Mass of weighing bottle + carbonate / g 14.92
Mass of weighing bottle / g 13.34
Use the student’s results to identify the carbonate, M2CO3.
The answer for (i) was 0.00111. How?
moles = molarity x volume (for the HCl)
Then use the equation and you see that the moles of carbonate = moles of HCl/2
I can do (ii) obviously if I knew the answer to (i) because that's just rearranging n=M/mr.
So can someone please explain how to get the answer? Thanks.
http://www.ocr.org.uk/Images/171752-unit-h032-02-depth-in-chemistry-sample-assessment-materials.pdf
That's the paper and the mark scheme is on the bottom - question 6.
n(HCL) is 0.1 x 22.20/1000 which gives you 0.00222 (to get the values for the titre and conc of HCL you need to use the answer from the question before this). Using the equation, for every one mole of M2CO3 you need 2 moles of HCL. Therefore using the stoichiometry of the equation we divide 0.00222 by 2 to give us the moles of M2CO3 which is 0.00111.
Original post by sera14
n(HCL) is 0.1 x 22.20/1000 which gives you 0.00222 (to get the values for the titre and conc of HCL you need to use the answer from the question before this). Using the equation, for every one mole of M2CO3 you need 2 moles of HCL. Therefore using the stoichiometry of the equation we divide 0.00222 by 2 to give us the moles of M2CO3 which is 0.00111.
I'm so confused - this post is 7 years old and I just answered that exact same question from you in the Chemistry forum just now?
Original post by dbhc2411
I'm so confused - this post is 7 years old and I just answered that exact same question from you in the Chemistry forum just now?
I saw this question and answered it while I was waiting for a reply on my own forum since people may still come and try find an answer to this part (i) of the question but it was the part (ii) after this that you answered for me that I was confused on. Thank you again!
(edited 1 year ago)
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