Question

A 0.4 g sample of CMA was dissolved in HCL and diluted to 100 mL. A 10 mL aliquot of the sample was titrated with 25.5 mL of 0.01 M EDTA. The calcium was removed from a second 10 mL aliquot by precipitation with ammonium oxalate, and the remaining filtrate was titrated with 17.25 mL of 0.01 M EDTA.

a) Multiply the concentration (molarity) of EDTA by the volume, in liters, of EDTA added in the first titration to determine the combined amount (moles) of calcium and magnesium ions in a 10 mL CME aliquot.

b) Calculate the amount (moles) of magnesium and calcium in a 10 mL CMA aliquot.

c) determine the mass of magnesium acetate and the mass of calcium acetate in a 10.0 mL CMA aliquot. (Molar mass: Calcium acetate = 158.2 g/mole; magnesium acetate = 142.3 g/mole)

d) divide the original mass of the CMA sample by the initial solution volume, before removing aliquots, and multiply by the aliquot volume to determine the mass of CMA and a 10 ML aliquot.

e) calculate the percent composition of calcium acetate and the percent composition of magnesium acetate in CMA.

Answer 1

SOLUTION:

(a) Amount of Ca and Mg (combined) in moles = mols of EDTA consumed

Moles of EDTA consumed = Molarity X Volume in liters = 0.01 X 0.0255L = 2.55 X 10^-4 Moles.

Because 25.5 mL = 25.5 /1000 = 0.0255L

Hence , Amount of Ca and Mg (combined) in moles = 2.55 X 10^-4 Moles.

(b) The magnesium requires 17.25 mL of 0.01M EDTA.

Hence moles of Magnesium = 0.01 M X 17.25mL = 0.01 M X 0.01725L = 1.725 X 10^-4 moles

Therefore moles of Calcium = Total moles - Hence moles of Magnesium

= 2.55 X 10^-4 Moles - 1.725 X 10^-4 moles = 8.25 X 0-5 Moles

(c) Number of moles = Given Mass / Molar mass

Given mass = Number of moles X Molar mass

Formulla of Magnisium and Calcium acetate Mg(CH3COO)2 and Ca(CH3COO)2 . Each mole of Ca and /or magnesium acetate contain one mole of Ca and/or Mg respectivelly.

Therefore moles of Ca or magnesium acetate = moles of Ca or Mg

Mass of Calcium acetate = 8.25 X 10^-5 Moles X 158.2g / mol = 0.013g

Similarly mass of Magnesium acetate = 2.55 X 10^-4 Moles X 142.3g / mol = 0.036g

(d) Mass of CMA in 10mL aliquot = (Mass of CMA / Initial Volume) X 10mL

= (0.4 / 100mL ) X 10mL = 0.04g.

(e) % composition of Calcium acetate = (Mass of Calcium aceta / Total mass of CMA ) X 100

= (0.013 / 0.4 ) X 100 = 3.25 %

Similaly % comosition of Magnisium acetate = 0.036 / 0.4 ) X 100 = 9 %