1.350g of insoluble carbonate, MCO3 was dissolved in 250cm^3 of a 0.203 mol/dm^3 HCL solution. The resulting solution was boiled off to remove all carbon dioxide produced. 25.0 cm^3 of the solution required 24.10cm^3 of 0.100 mol/dm^3 sodium hydroxide solution. Determine the relative atomic mass of M
This problem is confusing. Is that second titration a 25 mL aliquot of the original or has the solution been boiled so that the new volume is only 25 mL. If the latter, then
MCO3 + 2HCl ==> MCl2 + CO2 + H2O
mols HCl initially = 0.250 L x 0.203 M = 50.75/1000 = ?
mols NaOH added = mols HCl in excess = 0.0241 x 0.1M = ?
HCl used in the reaction = initial-excess
mols MCO3 used in the rxn = mols HCl/2 (from the coefficients in the balanced equation).
Then molar mass MCO3= grams/mols
Atomic mass M = molar mass MCO3-atomic mass C - 3*atomic mass O.
To determine the relative atomic mass of M, we need to start by calculating the number of moles of MCO3 reacted with HCL.
Step 1: Calculate the number of moles of HCL used.
To find the moles, we can use the equation:
moles = concentration × volume
moles of HCL = 0.203 mol/dm^3 × 250 cm^3 / 1000 cm^3/dm^3
Step 2: Calculate the number of moles of MCO3 reacted.
Since the reaction is 1:1 between HCL and MCO3, the number of moles of MCO3 reacted will be the same as HCL:
moles of MCO3 = moles of HCL
Step 3: Calculate the relative atomic mass of M.
Since we know the mass of MCO3 that was dissolved (1.350g) and the number of moles of MCO3 reacted, we can calculate the molar mass of MCO3 and derive the relative atomic mass of M:
moles of M = mass of MCO3 / molar mass of MCO3
Finally, we have the formula of MCO3, which is MCO3. Therefore, the molar mass of MCO3 is given by the sum of the molar masses of M and CO3.
molar mass of MCO3 = molar mass of M + molar mass of CO3
Once we have the relative atomic mass of M, we can compare it with the atomic masses of known elements to determine which element it corresponds to.