Page 207 - Modern Analytical Chemistry

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190 Modern Analytical Chemistry

When the target population is segregated, or stratified, equation 7.5 provides a poor
estimate of the amount of sample needed to achieve a desired relative standard de-
viation for sampling. A more appropriate relationship, which can be applied to both
segregated and nonsegregated samples, has been proposed. 7
A B
2
s s = + 7.6
mn s n s
where n s is the number of samples to be analyzed, m is the mass of each sample, A is
a homogeneity constant accounting for the random distribution of analyte in the
target population, and B is a segregation constant accounting for the nonrandom
distribution of analyte in the target population. Equation 7.6 shows that sampling
variance due to the random distribution of analyte can be minimized by increasing
either the mass of each sample or the total number of samples. Sampling errors due
to the nonrandom distribution of the analyte, however, can only be minimized by
increasing the total number of samples. Values for the homogeneity constant and
heterogeneity constant are determined using two sets of samples that differ signifi-
cantly in mass.

EXAMPLE .7
7
To develop a sampling plan for the determination of PCBs in lake sediments,
the following two experiments are conducted. First, 15 samples, each with a
mass of 1.00 g, are analyzed, giving a sampling variance of 0.0183. In a second
experiment, ten samples, each with a mass of 10.0 g, are analyzed, giving a
sampling variance of 0.0069. If samples weighing 5.00 g are to be collected, how
many are needed to give a sampling variance of 0.0100? If five samples are to be
collected, how much should each sample weigh?
SOLUTION
Substituting known values for the two experiments into equation 7.6 gives the
following pair of simultaneous equations
A B
0 0183 = +
.
)(
( 100 15) 15
.
A B
.
0 0069 = +
.
)(
( 10 0 10) 10
Solving for A and B gives values of 0.228 and 0.0462, respectively. The number
of 5.00-g samples is determined by solving
.
0 228 0 0462
.
0 0100 = +
.
(. n
500)n
for n, giving n = 9.2 » 9 samples. When using five samples, the mass of each is
given by the equation
.
.
0 228 0 0462
0 0100 = +
.
m 5 () 5
for which m is 60.0 g.