Page 278 - Modern Analytical Chemistry

1400-CH08 9/9/99 2:18 PM Page 261

Chapter 8 Gravimetric Methods of Analysis 261

and for chlorine
Moles Cl = moles AgCl
.
g AgCl ´ AW Cl ´ 1000 mg/g 0 2627 g ´35 453 g/mol ´ 1000 mg/g
.
=
FW AgCl 143.32 g/mol
.
= 64 98 mg Cl
mg Cl 64 98 mg
.
´100 = ´100 =53 35.% w/w Cl
mg sample 121.8 mg
Adding together the weight percents for C, H, and Cl gives a total of 100.01%.
The compound, therefore, contains only these three elements. To determine the
compound’s empirical formula, we assume that we have 1 g of the compound,
giving 0.4515 g of C, 0.0151 g of H, and 0.5335 g of Cl. Expressing each element
in moles gives 0.0376 mol C, 0.0150 mol H, and 0.0150 mol Cl. Hydrogen and
chlorine are present in a 1:1 molar ratio. The molar ratio of C to moles of H or
Cl is

.
Moles C moles C 0 0376
.
= = =251 »2 5
.
.
Moles H moles Cl 0 0150
Thus, the simplest, or empirical, formula for the compound is C 5 H 2 Cl 2 .
In an indirect volatilization gravimetric analysis, the change in the sample’s
weight is proportional to the amount of analyte. Note that in the following example it
is not necessary to apply the conservation of mass to relate the analytical signal to the
analyte.

EXAMPLE 8.6
A sample of slag from a blast furnace is analyzed for SiO 2 by decomposing a 0.5003-g
sample with HCl, leaving a residue with a mass of 0.1414 g. After treating with HF
and H 2 SO 4 and evaporating the volatile SiF 4 , a residue with a mass of 0.0183 g
remains. Determine the %w/w SiO 2 in the sample.
SOLUTION
In this procedure the difference in the residue’s mass before and after volatilizing
SiF 4 gives the mass of SiO 2 in the sample. Thus the sample contained

0.1414 g – 0.0183 g = 0.1231 g SiO 2
The %w/w SiO 2 , therefore, is

.
g SiO 2 0 1231 g
´100 = ´100 =24 61.% w/w SiO 2
.
g sample 0 5003 g

Finally, in some quantitative applications it is necessary to compare the result for a
sample with a similar result obtained using a standard.