# Plotting a calibration curve for Gas-Liquid Chromatography

Textbook solution:

Why aren't the points like this:

$(20,5) (40,10) (60,15) (a,7)$

then to plot the curve we use $y=mx+c$ and then find the value of '$a$'

• I get (60,16) for the third peak. – LDC3 Oct 5 '14 at 4:59
• It's (40,11) in graph. – Freddy Oct 5 '14 at 5:00
• @hey, no the numbers are not on the correct lines. – LDC3 Oct 5 '14 at 5:02
• @LDC3 Oh ok i see! – Freddy Oct 5 '14 at 5:05

If I remember correctly, the amount of your compound isn't proportional to the peak height but to the peak area.

Back in the days, when plotters were just plotters (=no numerical integration of the peaks), people used to cut out the curves and weighted the paper pieces on a balance to determine the peak area.

• I still do that! – user8016 Oct 5 '14 at 4:46
• If you have sharp, narrow peaks, the height is a good substitute for the area. But the area is more accurate. – LDC3 Oct 5 '14 at 5:01

You are actually correct, the formatting of the vertical axis in the solution graph has resulted in the vertical (Peak Height) values written below the actually gridline it is referring to - you can see the same for peak height values of 5, 10 and 15.

Using the linear equation formula $y=mx+c$ for the 3 known values yields a relationship of:

$$P = 0.25C$$

(Where $P$ is Peak height and $C$ is concentration)

Substituting the known peak height for the 4th value:

$$7=0.25C$$

$$C=28$$

Which is exactly where it is shown on the graph in the answer.