The amount of camphor in an analgesic ointment can be determined by GC using the

Question

The amount of camphor in an analgesic ointment can be determined by GC using the method of internal standards A standard sample was prepared by placing 45.21 mg of camphor and 2.000 mL of a 6.000 mg/L internal standard solution of terpene hydrate in a 25-mL volumetric flask and diluting it to volume with CCl_4 When an approximately 2 microliter sample of the standard was injected, the FID signals for the two components were measured as 67 CA for camphor and 19 CB for terpene hydrate A 53.61 mg sample of an analgesic ointment was prepared for analysis by placing it in a 50 mL Erienmeyer flask along with 10 mL of CCl_4. After heating to 50 degree C in a water bath, the sample was cooled to below room temperature and filtered The residue was washed with two 5-mL portions of CCl_4, and the combined filtrates were collected in a 25-mL volumetric flask After adding 2.000 mL of the internal standard solution, the contents of the flask were diluted to volume with CCl_4. Analysis of an approximately 2 microliter sample gave FID signals of 13. AB for the terpene hydrate and 24.0 for the camphor. Report the %w/w camphor in the analgesic ointment.

Explanation / Answer

45.21 mg camphor and 0.012 mg internal standard (in 2 ml) diluted to 25 ml with CCl4

concentration of camphor = 45.21 mg/152.23 g/mol = 0.30 mmol/25 = 0.012 M

concentration of internal standard = 0.012/190.28 = 6.31 x 10^-5 mmol/25 = 2.52 x 10^-6 M

2 ml of solution gave value, camphor = 67 for camphor and 19 for standard

Unknown gave a value for the same 25 ml solution, 24 for camphor and 13 for standard

0.012 M gave a value of 67, so 24 would be for = 0.0043 M

In 25 ml we have = 0.0043 x 25 = 0.11 mmol of camphor

mass of camphor = 0.11 x 152.23 = 16.36 mg

w/w% of camphor in the unknown = (16.36/53.61) x 100 = 30.50%

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