When using an internal standard analysis, it’s important to demonstrate that the

Question

When using an internal standard analysis, it’s important to demonstrate that the response factor is constant across a range of analyte concentrations. The chemical contaminant naphthalene is analyzed by a chromatography method using the compound biphenyl as its internal standard. (Hence, the nature of the analytical signal is in peak areas; either way, it’s just a signal!) Calculate the response factor, F, for each of the analyses shown below. Then use the average response factor to analyze a 0.0234 g sample containing naphthalene spiked with 10 ug of internal standard. The analysis yielded the following peak areas: (naphthalene = 582; biphenyl = 1865).

a) How many ug of naphthalene are in the sample?

b) What is the mixing ratio of naphthalene in the sample (in ppm)?

naphthalene biphenyl naphthalene biphenyl (Hg) (Hg) eak area eak area 1.00 10 00 303 2992 5.00 10.00 1591 3052 3023 2819 10.00 10.00

Explanation / Answer

Response Factor is calculated by using following formula.

(Peak area of naphthalene/ concentration of naphthalene) = F x (Peak area of biphenyl / concentration of biphenyl)

The response factor, F, for analyses 1-

303/1.00 = F x 2992/10

F = 1.013

The response factor, F, for analyses 2

1591/5 = F x 3052/10

F = 3182/3052 = 1.043

The response factor, F, for analyses 3

3023/10 = F x 2819/10

F = 302.3/281.9 = 1.072

The average response factor = (1.013 + 1.043 + 1.072)/3 = 1.04

a)

10 µg of internal standard

The analysis yielded the following peak areas: (naphthalene = 582; biphenyl = 1865).

  µg of naphthalene is in the sample -

(Peak area of naphthalene/ µg of naphthalene) = F x (Peak area of biphenyl / µg of biphenyl)

582/µg of naphthalene = 1.04 x 1865/10 µg

582/µg of naphthalene = 193.96

µg of naphthalene = 582/193.96

      = 3 µg of naphthalene

b) The mixing ratio of naphthalene in the sample (in ppm)

0.0234 g sample contains 3 µg of naphthalene

There are 1,000,000 micrograms in 1 gram

So, 0.0234 g sample = 0.0234 x 1,000,000 micrograms

                                     = 23400 µg

23400 µg sample contain 3 µg of naphthalene

1,000,000 micrograms sample contain 3 µg x 1,000,000/23400 of naphthalene

= 128µg

128 microgram/gram = 128 part per million

The mixing ratio of naphthalene = 128 ppm

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