A tablet of weight $\pu{200 mg}$ was analysed for its morphine content using a HPLC system in conjunction with a single wavelength UV detector operating at $\pu{260 nm}$.

The tablet was prepared for analysis by crushing and extraction using a solvent system similar to the HPLC mobile phase i.e. $70\%$ acetonitrile, $30\%$ water. $\pu{50 mg}$ of codeine was added as the internal standard and the mixture made up to $\pu{50 mL}$. The resulting solution was analysed by HPLC using the instrument conditions given below. A standard solution was obtained by taking $\pu{50 mg}$ of each component, dissolving in mobile phase and making up to $\pu{50 mL}$.

Using the results obtained below, calculate the morphine content of the unknown tablet (both as %w/w and mg/tablet).

Standard Solution Analysis

\begin{array}{c|cc}\hline &\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline \textbf{Morphine} & 2.00 & 153000 \\ \textbf{Codeine} & 3.25 & 212000\\ \hline \end{array}

Analysis of tablet solution

\begin{array}{c|cc}\hline &\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline \textbf{Morphine} & 2.02 & 17600 \\ \textbf{Codeine} & 3.29 & 212500\\ \hline \end{array}

I understand the use of the internal standard methodology, however I am unsure which values I should use in the equation as although I have been given the internal standard solution I also have been given the second standard solution of both codeine and morphine.

The question is asking to find the content of morphine in the sample tablet (200 mg) via the use of an internal standard solution of codeine (50 mg) which was made up to 50 mL ([codeine] = $\pu{1gL^-1}$). This was given as well as another standard solution of codeine (50 mg) and morphine (50 mg) made up to 50 mL ([codeine] = $\pu{1gL^-1}$).

The equation I have been given is $$D_\textrm{rfx} = \frac{A_x}{ A_\textrm{is}} \times \frac{C_\textrm{is}}{ C_x}$$

where subscripts and symbols have the following meaning: $x$ = unknown, $\text{is}$ = internal standard, $A$ = peak area, $C$ = concentration.

First we need to calculate the detector response factor ($D_\text{rf}$) of the standard solution - but which standard solution?

Assuming that the standard solution conc. of both morphine and codeine are $\pu{1gL^-1}$, this gives a $D_\text{rf}$ for the standard solution of 0.722.

Using the calculated $D_\text{rf}$ value, the $D_\text{rf}$ equation can be rearranged to find the concentration of codeine in the sample tablet.

However my question is which values need to be used for the $A_x$ and $A_\textrm{is}$ - do they need to be both morphine or just the two values from the sample peak areas?

enter image description here


1 Answer 1


However my question is which values need to be used for the $A_\mathrm x$ and $A_\mathrm{IS}$ - do they need to be both morphine or just the two values from the sample peak areas?

To answer just this part of the question, \begin{align} A_\mathrm{x} &= A_\mathrm{unknown} \mathrm{(Morphine)} \\ A_\mathrm{IS} &= A_\mathrm{standard} \mathrm{(Codeine)} \\ C_\mathrm{x} &= C_\mathrm{unknown} \mathrm{(Morphine)} \\ C_\mathrm{IS} &= C_\mathrm{standard} \mathrm{(Codeine)} \\ \end{align}

Looking at your picture, it looks like you successfully calculated your $\mathrm {D_{rf}({morphine})}$ (though it's next to a misleading picture, as the peak area of codeine is larger than that of morphine when their concentrations are equal).

Moving down, you've successfully solved your $\mathrm{D_{rf}}$ equation for unknown concentration, but the substitution is a little off. When you're dealing with the dissolved tablet solution, $A_\mathrm{unknown}$ is indeed $17.6$ but $A_\mathrm{standard}$ would be the codeine area of $212.5$, not the earlier morphine value.

If I had to guess, the confusion is between the terms "internal standard," a known spike of (usually) a single standard (here codeine) and "standard solution," which is an reference standard of known morphine and internal standard concentrations. Two common uses of reference standards are as initial calibrants and a good way to check for interference between analytes.

Getting back to the solution, once you have your unknown concentration, in units of $\pu{g/L} = \pu{mg/mL}$, you will simply multiply by the $\pu{50 mL} = \pu{0.05 L}$ volume to have the morphine concentration of the solution and thus of the whole $\pu{200mg}$ tablet. %w/w is also easy at that point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.