# Question bc73b

Jan 20, 2014

You divide the area of each peak by the total area of all the peaks and multiply by 100 %.

#### Explanation:

Each component in the mixture causes the same response in the detector.

So, we can use the areas under the curves to calculate the mole percent of each component.

The formula for the area of a peak is:

"Area" = "height" × "width at half-height" = h×w_(1/2)

"Mole % A" = "area of Peak A"/"total area" × 100%

Here is a typical chromatogram of a mixture of two components, $1$ and $2$.

$\text{Peak 1}$ has a height of $\text{34 mm}$ and a width at half-height of $\text{15 mm}$.

$\text{Peak 2}$ has a height of $\text{41.5 mm}$ and a width at half-height of $\text{16 mm}$.

${\text{Peak 1 area = 34 mm × 15 mm = 510 mm}}^{2}$
${\text{Peak 2 area = 41.5 mm × 16 mm = 664 mm}}^{2}$
${\text{Total area = (510 + 664) mm^2 = 1174 mm}}^{2}$

"Mole % 1" = (510 color(red)(cancel(color(black)("mm"^2))))/(1174 color(red)(cancel(color(black)("mm"^2)))) × 100 % = 43.4 %

"Mole % 2" = (664 color(red)(cancel(color(black)("mm"^2))))/(1174 color(red)(cancel(color(black)("mm"^2)))) × 100 % = 56.6 %

Note: If you have a solvent peak and impurity peaks as well as $\text{Peaks 1}$ and $2$, you can ignore them and still use just $\text{Peaks 1}$ and $2$ in your calculation.

Many modern gas chromatographs calculate and report the areas of the peaks in arbitrary units.

If your instrument had reported $\text{Peak 1 = 117.4856}$ and $\text{Peak 2 = 153.1145}$, you would have calculated:

$\text{Total area = 117.4856 + 153.1146 = 270.6002}$

"Mole % 1" = 117.4856/270.6002 × 100 % = 43.4 %

"Mole % 2" = 153.1146/270.6002 × 100 % = 56.6 %#