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III- Gravimetric Calculations
Reaction:
aA + rR
AaRr
ppt
where: a is # of moles of analyte A
r is # of moles of reagent R
AaRr is a pure, insoluble precipitate which we can dry
and weigh or ignite to convert to something we can
weigh
--------------------------------------------------ppt=precipitate
1
Example:
Ag+ + Cl– = AgCl
–
[Cl ] = ?
Ag+ + Cl– = AgCl
1 mol 1 mol 1 mol
x
x mol
Excess AgNO3
AgCl
0.4368 g
143.321g
0.4368g
x = 3.048 ×10–3mol
[Cl– ] = 3.048 ×10–3 mol / 10.00ml
–
= 0.3048 M
10.00 ml Cl
T.W. Richards : Nobel Prize-winning research
2
Percentage of analyte ( % A)
%A = weight of analyte x 100
weight of sample
wt analyte = wt ppt x gravimetric factor (G.F.)
% A = weight of ppt x gravimetric factor (G.F.) x 100
weight of sample
G.F. =
a (F.wt.analyte)
b ( F.wt.precipitate)
Where a, b are stoichiometric coefficients
G.F. = # gms of analyte per 1 gm ppt
3
Example 1:
Calculate the gravimetric factor for each of the following:
P Ag3PO4, K2HPO4 AgPO4,
Bi2S3 BaSO4
As2O3Ag3AsO4, K2OKB(C6H5)4
Answer
Analyte
ppt
G.F.
P
Ag3PO4
At.wt P/ M.wt Ag3PO4
K2HPO4
Ag3PO4
M.wt K2HPO4 / M.wtAg3PO4
Bi2S3
BaSO4
M.wt Bi2S3/ 3M.wt BaSO4
As2O3
Ag3AsO4
M.wt As2O3/2M.wt Ag3AsO4
K2O
KB(C6H5)4
M.wt K2O/ 2M.wt KB(C6H5)4
4
Example 2:
Calculate the gravimetric factor for each of the following:
Analyte
ppt
CaO
CaCO3
FeS
BaSO4
UO2(NO3)2.6H2O
U 3O 8
Cr2O3
Ag2CrO4
G.F.
5
Answer:
Analyte
ppt
G.F.
CaO
CaCO3
M.wt CaO/ M.wt CaCO3
FeS
BaSO4
M.wt FeS/ M.wt BaSO4
UO2(NO3)2
U3O8
3 M.wt UO2(NO3)2/ M.wt U3O8
Cr2O3
Ag2CrO4
M.wt Cr2O3/2 M.wt Ag2CrO4
6
Example 3:
Orthophosphate (PO43-) is determined by
weighing as ammonium phosphomolybdate,
(NH4)PO4.12MoO3. Calculate the percent P in the
sample and the percent P2O5 if 1.1682g precipitate
were obtained from a 0.2711 g sample.
Remember: Wt of analyte = Wt of ppt x Gravimetric Factor
Answer
Wt of P = 1.1682 x
At. wt P

Mwt (NH ) PO .12MoO
4 3
4
3
30.97
 1.1682 x
 0.0193 g
1876.5
% P = (0.0193/0.2711) x 100 = 7.11%
7
In general,
% ppt 
wt ppt x gravimetri c factor
wt of sample
x100
M wt P2O5
1.1682 x
2 M wt (NH4 )3 PO4 .12M oO3
 %P2O5 
x 100
0.2711
141.95
1.1682 x
2 x1876.5 x100  16.30%

0.2711
8
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