analyze a mixture

Question

A mixture of CrCl₂ and inert material is analyzed to determine the Cr content. First the mixture is dissolved in water. Then all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate.

CrCl₂(aq) + 2AgNO₃(aq) 2AgCl(s) + Cr(NO₃)₂(aq)

In one experiment, a 0.6067 g sample of the mixture resulted in 1.0921 g of AgCl. Determine the percent (by mass) of Cr in the mixture.

Answer 1

Approach:

Find moles of AgCl formed. Convert moles AgCl to moles CrCl₂ using mole ratio of the balanced equation. Convert moles CrCl₂ to grams Cr and finally use this value to determine mass % of Cr.

moles AgCl formed = 1.0921 g AgCl x 1 mol AgCl / 143.3 g = 0.007621 moles AgCl

moles CrCl₂ present = 0.007621 moles AgCl x 1 mol CrCl₂ / 2 mols AgCl = 0.003811 moles CrCl₂

grams Cr = 0.003811 moles CrCl₂ x 1 mol Cr / mol CrCl₂ x 52.00 g Cr / mol Cr = 0.1982 g Cr

mass % Cr = mass Cr / mass of sample (x100%) = 0.1982 g / 0.6067 g (x100%) = 32.66% Cr