Question

Atomic absorption spectroscopy was used to determine the amount of aluminum in a sample of sunscreen. A blank solution was measured seven times to determine the detection and quantitation limits of the spectrometer. The blank readings were as follows: 0.0086, 0.0107, 0.0095, 0.0103, 0.0113, 0.0104, and 0.0094 AU. Determine the signal at the detection limit and the signal at the limit of quantitation for this method.

Answer 1

Given balnk Reading are as 0.0086, 0.0107, 0.0095, 0.0103, 0.0113, 0.0104, and 0.0094 AU.

A) Mean = (0.0086+ 0.0107 + 0.0095 + 0.0103+ 0.0113+ 0.0104 + 0.0094 ) / 7 = 0.00892

B) calculate standard deviation for blank

STDEV =√ [ ∑(x-mean)^2 / N ] here N= 7

STDEV = √[(0.0086-0.00892)² + (0.0107-0.00892)² + (0.0095-0.00892)² + (0.0103-0.00892)² +(0.0113-0.00892)² +(0.0104-0.00892)² + (0.0094-0.00892)² ] /7

STDEV ( σ )= 0.0036

C) calculate Limit of detection & Limit of quantitation

Limit of detection = Mean + 3 σ =0.00892 +3 x 0.0036 =0.01977

Limit of detection = 0.020

Limit of quantitation = Mean + 10 σ = =0.00892 + 10 x 0.0036 =0.04508

Limit of quantitation = 0.045