Question

in an experiment i performed 3 EDTA complexometric tritrations to determine the amount of Al³⁺, Ca²⁺ and Mg²⁺ in 0.6499 g of antacid powder. i made my sample for titration following the method below.

A sample of antacid powder ("Hardy's indigestion powder") is supplied containing aluminium hydroxide, calcium carbonate and magnesium carbonate. Record this accurate mass. Add 5 mL of water and then 10 mL of 3 mol L-1 HCl Heat the mixture (microwave for 30 seconds) to boiling and shake until dissolved. Add about 100 mL of water and then quantitatively transfer the solution to a 250.0 mL volumetric flask and dilute to volume.

in 25 ml of the sample i determined that there are 0.000029 Moles of Al³⁺, in 10 ml of sample there are 0.00094 mol of Ca²⁺ and 0.000828 mol of Mg²⁺.

How do i determine their weight percentage of the sample?

Answer 1

I think you have missed a few zeros in your data; I will add them. Without the zeros, the calculations will yield absurd results.

You prepared a 250 mL sample of the antacid following your dissolution procedure. 25 mL of the prepared sample contains 0.000029 mol Al³⁺ while 10 mL of the prepared sample was found to contain 0.00094 mol Ca²⁺ and 0.000828 mol Mg²⁺.

Mol of Al³⁺ in 250 mL prepared sample = (0.000029 mol/25 mL)*(250 mL) = 0.00029 mol.

Mol of Ca²⁺ in 250 mL prepared sample = (0.000094 mol/10 mL)*(250 mL) = 0.00235 mol.

Mol of Mg²⁺ in 250 mL prepared sample = (0.0000828 mol/10 mL)*(250 mL) = 0.00207 mol.

The atomic masses of Al, Ca and Mg are 26.981 g/mol, 40.078 g/mol and 24.305 g/mol respectively.

Mass of Al³⁺ in the sample = (0.00029 mol)*(26.981 g/mol) = 0.007284 g.

Mass of Ca²⁺ in the sample = (0.00235 mol)*(40.078 g/mol) = 0.09418 g.

Mass of Mg²⁺ in the sample = (0.00207 mol)*(24.305 g/mol) = 0.05031 g.

Percentage of Al³⁺ in the sample = (0.007284 g)/(0.6499 g)*100 = 1.121% (ans).

Percentage of Ca²⁺ in the sample = (0.09418 g)/(0.6499 g)*100 = 14.491% (ans).

Percentage of Mg²⁺ in the sample = (0.05031 g)/(06499 g)*100 = 7.741% (ans).