Chromatography Forum

I tried to assay of cinnamaldehyde in some solid powder(it may contain some amount of cinnamaldehyde).
I apply standard addition method to this powder.
Add each 1g of power to each five 20 mL volumetic flask.
And I make 100 ug/mL of cinnamaldehyde standard solution with methanol.
And add 0, 1, 2, 3, 4 mL of 100 ug/mL of cinnamaldehyde standard solution to above five 20 mL volumetic flask.
and apply to HPLC.
I take result below.

Standard solution Peak
added amount area
0 mL 88992
1mL 121659
2mL 158103
3mL 183828
4mL 211855

I want to get result of concentration of cinnamaldehyde in powder(ug/g).
How to calculate this problem? and how to get unit of ug/g result?

Dilution factor for your sample is 20x
x ug/ml * 20 = ug/g

1) you disolve 1g of powder in 20 ml volumetric flask so its 1g/20ml = 0,05g/ml

2) 100 ug/ml of standard solution you add to flasks:

0 ml of 100ug/ml (CA) to 20 ml - 0ug (CA)/ml = 0ug (CA)/ml
1 ml of 100ug/ml (CA) to 20 ml - 100ug (CA)/20ml = 5ug (CA)/ml
2 ml of 100ug/ml (CA) to 20 ml - 200ug (CA)/20ml = 10ug (CA)/ml
3 ml of 100ug/ml (CA) to 20 ml - 300ug (CA)/20ml = 15ug (CA)/ml
4 ml of 100ug/ml (CA) to 20 ml - 400ug (CA)/20ml = 20ug (CA)/ml

So those are our concentrations ug/ml (CA) in our 5 flasks from our standard addition

The peak areas are :

0ml - 88992 (area for your CA in powder)
1ml - 121659 (area for your CA in powder + your CA from standard addition)
2ml - 158103 - // -
3ml - 183828 - // -
4ml - 211855 - // -

Now we calculate what are is from standard addition :

1ml -> 121659 - 88992 = 32667 (so you can say that this are is given by concentration of 5ug/ml)
2ml -> 158103 - 88992 = 69111 (this are is given by concentration of 10ug/ml)
3ml -> 183828 - 88992 = 94846 (this are is given by concentration of 15ug/ml)
4ml -> 211855 - 88992 = 122863 (this are is given by concentration od 20ug/ml)

So now we will calcualte the concentration of CA in your powder (0 ml)
I will use only one point for the clarity

88992 -> x (CA)
32667 -> 5 (CA) (ug/ml)

88995*5=32667*x
444975=32667*x
x= 13.62 ug/ml

We get te result in ug/ml
Now our dilution factor

13.62ug/ml * 20 ml/g = 272ug/g

Why in this way ?
13.62ug in 1ml so in 20 ml we will get 272ug
272ug/20ml now you remember that you put 1g of powder to this 20 ml
so 272ug/1g powder will be true also

p.s.
sry for my english low level...