Page 270 - Modern Analytical Chemistry

1400-CH08 9/9/99 2:18 PM Page 253

Chapter 8 Gravimetric Methods of Analysis 253

Moles Mg = moles Mg(C 9 H 6 NO) 2 Moles Al = moles Al(C 9 H 6 NO) 3
g Mg g Mg(C H NO) 2 g Al g Al(C H NO) 3
6
9
6
9
= =
AW Mg FW Mg(C H NO) 2 AW Al FW Al(C H NO) 3
9
6
9
6
g Mg(C H NO) 2 ´ AW Mg g Al(C H NO) 3 ´ AW Al
9
6
g Mg = g Al = 9 6
FW Mg(C H NO) 2 FW Al(C H NO) 3
9
6
9
6
.
7 5125 g ´ 24.305 g/mol 0 .3029 g ´ 26.982 g /mol
= =
312.61 g/mol 459.45 g /mol
.
= 0 5841 g = 0 0178 g
.
g Mg g Al
%Mg = ´100 %Al = ´100
g sample g sample
0 5841 g 0 0178 g
.
.
´100 =95 55 w/w Mg ´100 =2 91 w/w Al
.%
.%
0.6113 g 0.6113 g
In an indirect analysis the precipitate does not contain the analyte, but is the
product of a reaction involving the analyte. Despite the additional complexity, a
stoichiometric relationship between the analyte and the precipitate can be written
by applying the conservation principles discussed in Section 2C.
EXAMPLE 8. 3
An impure sample of Na 3 PO 3 weighing 0.1392 g was dissolved in 25 mL of
water. A solution containing 50 mL of 3% w/v mercury(II) chloride, 20 mL of
10% w/v sodium acetate and 5 mL of glacial acetic acid was then prepared. The
solution containing the phosphite was added dropwise to the second solution,
3– 3–
oxidizing PO 3 to PO 4 and precipitating Hg 2 Cl 2 . After digesting, filtering,
and rinsing, the precipitated Hg 2 Cl 2 was found to weigh 0.4320 g. Report the
purity of the original sample as %w/w Na 3 PO 3 .
SOLUTION
This is an example of an indirect analysis since the isolated precipitate, Hg 2 Cl 2 ,
does not contain the analyte, Na 3 PO 3 . The stoichiometry of the redox reaction
in which Na 3 PO 3 is a reactant and Hg 2 Cl 2 is a product was given earlier in the
chapter. It also can be determined by using the conservation of electrons.
3– 3–
Phosphorus has an oxidation state of +3 in PO 3 and +5 in PO 4 ; thus,
3– 3–
oxidizing PO 3 to PO 4 requires two electrons. The formation of Hg 2 Cl 2 by
reduction of HgCl 2 requires 2 electrons as the oxidation state of each mercury
3–
changes from +2 to +1. Since the oxidation of PO 3 and the formation of
Hg 2 Cl 2 both require two electrons, we have
Moles Na 3 PO 3 = moles Hg 2 Cl 2
Converting from moles to grams leaves us with

2
3
g Na PO 3 g Hg Cl 2
=
2
FW Na PO 3 FW Hg Cl 2
3