Question

A student, Legna, massed out 10.12 g of chromium(iii)chloride hexahydrate to synthesize a transition metal coordination complex. She obtained 7.20 g of a complex containing Chromium (Cr), Ammonia (NH3), and Chlorine (Cl). Samples of this compound were used to perform all analyses for the remainder of the advanced study assignment. The following data were obtained.

analysis 1: no heat applied
mass of compound used .4988g
mass of crucible plus AgCl 19.2859g
mass of crucible 18.4628g
MM AgCl 143.32 g/mol
MM Cl- 35.45 g/mol

Analysis 2: heat applied
mass of compound used .4632g
mass of crucible plus AgCl 21.0926g
mass of crucible 20.3286g

a. determine the % Cl in each sample of compound
b. calculate the ration of % Cl(heated analysis)/%Cl(unheated analysis)
c. Which Cl-analysis gives the total chloride in the complex? Heated or unheated? Briefly explain.

Answer 1

(a)
analysis 1: no heat applied
mass of compound used .4988g
mass of crucible plus AgCl 19.2859g
mass of crucible 18.4628g

mass of AgCl = mass of crucible plus AgCl - mass of crucible $= 19.2859g-18.4628g=0.8231g$

MM AgCl 143.32 g/mol
MM Cl- 35.45 g/mol

0.8231g AgCl $= 0.8231g \times \dfrac {35.45}{143.32} = 0.2036$ g of chloride.

Percent chlorine in the sample $= 100 \times \dfrac {0.2036}{0.4988} =40.8$ %

Analysis 2: heat applied
mass of compound used .4632g
mass of crucible plus AgCl 21.0926g
mass of crucible 20.3286g

mass of AgCl = mass of crucible plus AgCl - mass of crucible $=21.0926g - 20.3286g=0.764g$

MM AgCl 143.32 g/mol
MM Cl- 35.45 g/mol

0.764g AgCl $= 0.8231g \times \dfrac {35.45}{143.32} = 0.2036$ g of chloride.$=0.764g \times \dfrac {35.45}{143.32} = 0.189$

Percent chlorine in the sample $= 100 \times \dfrac {0.189}{0.4632} =40.8$ %

(b) the ratio of % Cl(heated analysis)/%Cl(unheated analysis) $= \dfrac {40.8}{40.8}=1:1$

(c) Both Cl-analysis (Heated or unheated) gives the total chloride in the complex. This shows that chloride is present in the ionisation sphere (outer sphere) and can easily be removed without heating.