Question

(a) The atomic absorption signal shown here was obtained with$\mathbf{0}\mathbf{.}\mathbf{0485}\mathbf{\mu gFe}\mathbf{/}\mathbf{mL}$in a graphite furnace. The root-mean-square noise in the baseline, measured by the instrument's computer, is s=0.30 vertical units, where each horizontal line on the chart is 1 vertical

unit. Estimate where the baseline is beneath the tall signal and measure the height of the signal. Estimate the detection limit for Fe, defined as the concentration of Fe that gives a signal height of 3 s.

(b) Seven replicate measurements of a standard containing$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ngHg}\mathbf{/}\mathbf{L}$gave readings of 0.88,1.48,0.94,1.12,1.03,1.40, andin cold vapor atomic absorption (Box 21-1). From Equations 5-5 and 5 -6, estimate the detection and quantitation limits. (Note that in Equations 5-5 and 5-6, the quotient s / m is the standard deviation in concentration.)

Short answer

1. Fe has a detection limit of $0.0021\mathrm{\mu g}/\mathrm{mL}$.
2. The quantitation limit can be computed as$Q_c=\frac{\left|N{O}^4\right|H_{2}O{|}^6}{\left|N{H}_3{|}^4\right|O_2{|}^5}$

Step-by-step solution

Determine the formula for active power and apparent power

• The detection limit is (informally) the lowest analyte concentration that can be consistently identified, and it reflects the precision of the instrumental response obtained by the method when the analyte concentration is zero.
• The smallest amount or lowest concentration of a substance that can be determined using a certain analytical process with the established accuracy, precision, and uncertainty is referred to as the limit of quantification, or LOQ.

Determine the concentration of Fe that gives a signal height of 3 s.

(a)

It is necessary to calculate the Fe detection limit.

Fe has a detection limit of$0.0021\mathrm{\mu g}/\mathrm{mL}$.

In the baseline, the root-mean-square noise is 0.30 vertical units.

The height of the Fe signal is estimated to be 21.2 vertical units.

As,

s = 0.303

s = 0.90

The Fe concentration required to provide a signal height of 0.90 can be computed as follows:

$$\begin{gathered} \left(\frac{0.90}{21.2}\right)\left(0.0485\mu g/mL\right)=\left(0.04245\right)\left(0.0485\mu g/mL\right) \\ =0.00205\mu g/mL \\ =0.0021\mu g/mL \end{gathered}$$

The Fe concentration was estimated as follows $0.0021\mu g/mL$

Determine the detection and quantitation limits

(b)

It is necessary to estimate the detection and quantitation limits.

The benchmark measurement was replicated seven times 1.00ngHg/L gave 0.88,1.48,0.94,1.12,1.03,1.40, and 1.14ng/L

The standard deviation for the seven values above is determined as follows: 0.22ng/L

This is the same as s / m.

As demonstrated below, the detection limit can be computed.

Limit of detection = 3 s / m

=3(0.22ng/L)

=0.66ng/L

The quantitation limit can be computed using the formula below.

Quantitation limit = 10 s / m

$$\begin{gathered} =10(0.22ng/L) \\ =2.2ng/L{Q}_c \\ {Q}_c=\frac{\left|N{O}^4\right|H_{2}O{|}^6}{\left|N{H}_3{|}^4\right|O_2{|}^5} \end{gathered}$$

The quantitation limit can be computed as$Q_c=\frac{\left|N{O}^4\right|H_{2}O{|}^6}{\left|N{H}_3{|}^4\right|O_2{|}^5}$