Question

Internal standard graph -- Data are shown below for chromatographic analysis of naphthalene $\mathbf{\left(}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{8}}\mathbf{\right)}$, using deuterated naphthalene $\mathbf{(}{\mathbf{C}}_{\mathbf{10}}{\mathbf{D}}_{\mathbf{8}}$, in which D is the isotope ${}^{\mathbf{2}}\mathbf{H}$) as an internal standard. The two compounds emerge from the column at almost identical times and are measured by a mass spectrometer.

(a) Using a spreadsheet such asFigure 4-15, prepare a graph of Equation 5-12 showing peak area ratio$\left(C_{10}{H}_8/C_{10}{D}_8\right)$versus concentration ratio role="math" localid="1663559632352" $\left(\left[C_{10}{H}_8\right]/\left[C_{10}{D}_8\right]\right)$ . Find the least-squares slope and intercept and their standard uncertainties. What is the theoretical value of the intercept? Is the observed value of the intercept within experimental uncertainty of the theoretical value?

(b) Find the quotientrole="math" localid="1663559638520" $\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{8}}\mathbf{\right]}\mathbf{/}\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{D}}_{\mathbf{8}}\mathbf{\right]}$for an unknown whose peak area ratio $\left(C_{10}{H}_8/C_{10}{D}_8\right)$ is 0.652. Find the standard uncertainty for the peak area ratio.

(c) Here is why we try not to use 3-point calibration curves. For n = 3 data points, there is n - 2 = 1 degree of freedom, because 2 degrees of freedom are lost in computing the slope and intercept. Find the value of Student's for confidence and 1 degree of freedom. From the standard uncertainty in (b), compute the 95 % confidence interval for the quotient$\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{8}}\mathbf{\right]}\mathbf{/}\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{D}}_{\mathbf{8}}\mathbf{\right]}$ . What is the percent relative uncertainty in the quotient$\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{8}}\mathbf{\right]}\mathbf{/}\mathbf{\left[}{\mathbf{C}}_{\mathbf{10}}{\mathbf{D}}_{\mathbf{8}}\mathbf{\right]}$? Why do we avoid 3-point calibration curves?

Short answer

a) The least squares slope and intercept and their standard deviations and the theoretical value was calculated.

b) The standard uncertainty ${\mathrm{u}}_{\mathrm{x}}$for the peak ratio is ${\mathrm{u}}_{\mathrm{x}}=0.0355$

c) The uncertainty from a 3-point calibration is large that is 75% because students t is 12.706 where there is only 1 degree of freedom.

If we had just one more data point, which gives 2 degree of freedom,, 95 % confidence interval would decrease by a factor of 3.

Step-by-step solution

Concept used

LINEST:

The LINEST is a Microsoft excel function which is uses the least squares method to measure the statistics for a straight line and array describing that line it is built-in function that can be categorized.

LINEARITY:

It is measure of how well a calibration curve follows a straight line and displaying that the response is proportional to the quantity of the analyte.

${\mathrm{R}}^2=\frac{{\left[{\displaystyle \sum _{}}\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)\left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)\right]}^2}{\sum _{}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2{\left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)}^2}$

Where R means the square of the correlation coefficient.

Standard additions to one solution:

The graph can be drawn using the equation

plot $\underset{\mathrm{Function}\mathrm{to}\mathrm{plot}\mathrm{on}\mathrm{y}-\mathrm{axis}}{\underbrace{{\mathrm{I}}_{\mathrm{s}+\mathrm{x}}\left(\frac{\mathrm{V}}{{\mathrm{V}}_0}\right)}}$versus $\mathrm{IX}+\frac{{\mathrm{I}}_{\mathrm{x}}}{{\left[\mathrm{X}\right]}_{\mathrm{i}}}\underset{\mathrm{function}\mathrm{to}\mathrm{plot}\mathrm{on}\mathrm{X}-\mathrm{axis}}{\underbrace{{\left[\mathrm{S}\right]}_{\mathrm{i}}\left(\frac{{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{V}}_0}\right)}}$

Confidence Intervals:

The confidence interval is given by the equation:

Confidence interval $\overline{\mathrm{x}}\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}}$

$=\overline{\mathrm{x}}\pm {\mathrm{tu}}_{\mathrm{x}}$( since standard uncertaintylocalid="1663560297056" $\left({\mathrm{u}}_{\mathrm{x}}\right)=s/\sqrt{n}$)

Where,

$\overline{\mathrm{x}}$ is mean

s is standard deviation

t is Student's

n is number of measurements

${\mathrm{u}}_{\mathrm{x}}$is standard uncertainty

Calculate the least squares slope and intercept, and their standard deviations and the theoretical value

a)

A spreadsheet with x-axis and y-axis are

Every solution is made to constant volume.

Then [X] / [S] vs ${\mathrm{A}}_{\mathrm{x}}/{\mathrm{A}}_{\mathrm{s}}$ is plotted.

The intercept of the graph is 0.0084 and ${\mathrm{u}}_{\mathrm{m}}=0.0517$, ${\mathrm{u}}_{\mathrm{b}}=0.0335$; the theoretical value of the intercept is 0.

The observed value is less than one standard uncertainty away from 0, which lies with the experimental error of 0.

Calculate the standard uncertainty for the peak ratio

b)

A spreadsheet with x-axis and y-axis is

The x -intercept is calculated in B19 and B20 and its uncertainty $\left[{\mathrm{C}}_{10}{\mathrm{H}}_8\right]/\left[{\mathrm{C}}_{10}{\mathrm{D}}_8\right]$is 0.598.

The standard uncertainty is$u_x=0.0355$

Calculate the confidence interval and degree freedom

c)

A spreadsheet with x-axis and y-axis is

The -intercept is calculated in B19 and B 20 and its uncertainty $\left[{\mathrm{C}}_{10}{\mathrm{H}}_8\right]/\left[{\mathrm{C}}_{10}{\mathrm{D}}_8\right]$is 0.598.

The standard uncertainty is $u_x=0.0355$

The students t for 95% confidence and 1 Degree freedom is 12.706. The 95% confidence for $\left[{\mathrm{C}}_{10}{\mathrm{H}}_8\right]/\left[{\mathrm{C}}_{10}{\mathrm{D}}_8\right]$is

$$\begin{gathered} =\overline{\mathrm{x}}\pm {\mathrm{tu}}_{\mathrm{x}} \\ =0.598\pm \left(12.706\right)\left(0.035\right) \\ =0.598\pm 0.451 \end{gathered}$$

The relative uncertainty $=\frac{0.451}{0.598}=75\%$

The uncertainty from a 3-point calibration is large that is 75% because students t is 12.706, where there is only 1 degree of freedom.

If we had just one more data point which gives 2 degree of freedom, the 95% confidence interval would decrease by a factor of 3.