Question

The level of ethyl acetate (from which colchicines is precipitated from) is determined in colchicine by headspace gas chromatography. A 10 mL aqueous solution containing 0.1361g of the sample of colchicine was prepared. The USP allowed limit of ethyl acetate in colchicines is 8% by weight. An aqueous standard containing 0.1078 mg/10mL of ethyl acetate was also prepared. Both solutions also had n-propyl alcohol added in equal proportions representing the internal standard. Each chromatogram had two peaks, one from ethyl acetate and one from n-propyl alcohol. The colchicines sample had a peak area for the ethyl acetate at 13875 and for the n-propyl alcohol at 1975. The ethyl acetate standard had a peak area for the ethyl acetate at 14638 and for the n-propyl alcohol at 1971. Calculate the ethyl acetate content in ppm and %w/w. Was the limit test exceeded? Yes or No. Thank you in advance, and please explain it detail!

Answer 1

Answer: According to given informations we have :

10 ml of solution contains 0.1361 g of Colchicine [ molar mass = 399.43 ]

number of moles of colchine = 0.1361 / 399.43 = 0.0003407 mol

Now Molarity of colchine = 0.0003407 * 1000 / 10 = 0.03407355 M

Now , we have 8% by weight ethyl acetate means 8 g in 100 g solution

we required 10 ml for = 0.1078 * 10^-3 g

Hecne for 8 g the required volume is = 10 * 8 / 0.1078 * 10^-3  = 742.11 * 10³ ml

Now using M1V1 = M2V2 we get

0.03407355 * 10 = 742.11 * 10³ * M2

Hence M2 = 0.459 ppm

hence the required concentraton of ethyl acetate is 0.459 ppm .

Thank you :)