2. A GC separation was conducted on a sample containing a pesticide analyte, yme

Question

2. A GC separation was conducted on a sample containing a pesticide analyte, ymetozine. This sample was treated with an internal standard of chlorobenzene, giving a concentration of 5.00 ppm AT 0 ??!nection of this solution onto the GC gave an ID detector response of 1012 (AU (i.c. Arbitrary Unit) for chlorobenzene and 3411) for Pymetozine. When a 1.0 ??standard solution containing 30.0 ppm of chlorobenzene and 15.0 ppm of Pymetozine was injected onto the same GC, the response of the same detéctor was 899 and 2791, respéctively. What is the concentration of Pymctozine in the sample? (Clint: You may use the following quantity celled response factor (RF) of the detector: Ajs IS Aw GC detector signal (or peak area measured by the GC detector) of an analyte, Ca: concentration of an analyte, As: GC detector signal (or peak area measured by the GC detector) of an intermal standard C concentration of an internal standard) 8t9 15

Explanation / Answer

First we need to find out the RF value by using given infomation -

AAN = 2791, CAN = 15.0ppm

AIS = 899, CIS = 30

So Use the given formula for RF and plug these value to get the RF value.

RF = (2791 / 15) divided by (899/30)

RF = 186.07/29.97

RF = 6.21

Now let's move to the next step to find out the value of concentration of Pymetozine in the sample.

We have RF = 6.21

AAN =3411, CAN = X

AIS =1012, CIS = 15ppm

Plug the values in RF equation,

6.21 = (3411/CAN) divided by (1012/15.00)

6.21 = (3411/CAN) divided by 67.47

After solving this equation we will get,

3411/CAN = 6.21 * 67.47

3411/CAN = 418.99

CAN = 3411/418.99

CAN = 8.14ppm

So concentration of Pymetozine in the sample is 8.14ppm.

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