Page 277 - Modern Analytical Chemistry

1400-CH08 9/9/99 2:18 PM Page 260

260 Modern Analytical Chemistry

Volatilization gravimetry is also used to determine biomass in water and waste-
water. Biomass is a water quality index, providing an indication of the total mass of
organisms contained within a sample of water. A known volume of the sample is
passed through a preweighed 0.45-mm membrane filter or a glass-fiber filter and
dried at 105 °C for 24 h. The residue’s mass provides a direct measure of biomass. If
samples are known to contain a substantial amount of dissolved inorganic solids,
the residue can be ignited at 500 °C for 1 h, thereby removing all organic materials.
The resulting residue is wetted with distilled water to rehydrate any clay minerals
and dried to a constant weight at 105 °C. The difference in weight before and after
ignition provides an indirect measure of biomass.

Quantitative Calculations When needed, the relationship between the analyte and
the analytical signal is given by the stoichiometry of any relevant reactions. Calcula-
tions are simplified, however, by applying the principle of conservation of mass.
The most frequently encountered example of a direct volatilization gravimetric
analysis is the determination of a compound’s elemental composition.

EXAMPLE 8.5

A 101.3-mg sample of an organic compound known to contain Cl is burned in
pure O 2 and the combustion gases collected in absorbent tubes. The tube used
to trap CO 2 increases in mass by 167.6 mg, and the tube for trapping H 2 O
shows a 13.7-mg increase. A second sample of 121.8 mg is treated with
+
concentrated HNO 3 producing Cl 2 , which subsequently reacts with Ag ,
forming 262.7 mg of AgCl. Determine the compound’s composition, as well as
its empirical formula.
SOLUTION

Applying the principle of conservation of mass to carbon, we write
Moles C = moles CO 2
Converting from moles to grams, and rearranging to solve for milligrams of
carbon gives
g CO ´ AW C ´ 1000 mg/g 0 1676 g ´ 12.011 g/mol ´ 1000 mg/g
.
2
=
FW CO 44.011 g/mol
2
.
= 45 74 mg C
Thus, the %w/w C in the sample is
.
mg C 45 74 mg
´100 = ´100 =45 15.% w/w C
mg sample 101.3 mg
The calculation is then repeated for hydrogen
Moles H = 2 ´moles H 2 O
.
2 ´ g H O ´ AW H ´ 1000 mg/g 2 ´0 0137 g ´ 1.008 g/mol ´ 1000 mg/g
2
=
FW H O 18.015 g/mol
2
.
=1 533 mg H
mg H 1 533 mg
.
1
´100 = ´100 = 51.% w/w H
mg sample 101.3 mg