While dealing with gas chromatography, how do you calculate the response factor with only the area known?

I am given the equation

$${k_1\over k_2} = {f_1 \times O_1 \over f_2 \times O_2}$$

where $O_1 = 21.28$, $O_2 = 178.35$, and $f_1 = 1.00$. $k_1$ and $k_2$ are the weights (in moles) that we added into the mixture. I calculated $0.05\,\mathrm{mol}$ for each component.

Using the formula (above) and the given quantities, I get

$$1 = {1.0 \times 21.28 \over f_2 \times 178.35}$$

Solving for $f_2$, I get 0.119, and I do not believe that is right. Is my approach valid?


1 Answer 1


This is a bit convoluted.

A simply dimensional analysis shows that your approach doesn't work. The units of $O$ are unit area. The conversion factor $f$ has units moles per unit area. The variable $k$ is in moles. If you divide the two equations you get a relative sensitivity of $f_2$ as compared to $f_1$, but the value is dimensionless. So each compound has its own response factor. So for two unknowns you have:

$f_1 = \dfrac{O_1}{k_1}$
$f_2 = \dfrac{O_2}{k_2}$

The twist here is the GC typically uses a standard either on a run before the unknown or spiking the unknown sample. So the standard is known to be $x$ ppm and you'll get some area from the run. The hope is that if the gain of the detector drifts them the area of the standard will change too by the same amount. So a 10% gain in area means that the detector gain has drifted up by 10%. Thus measuring area relative to the standard compensates for the drift.

The sensitivity changes a bit from day to day because the ionization in the detector changes due to aging, the gas flow rate isn't 100.0000% the same each time, the column ages and so on.

So the ratio of two unknowns is meaningless. You ratio each of the unknowns to a standard.


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