A 3.00 microliter standard sample of 50% 1-propanol and 50% n-butanol is injected and the results are below. 1-propanol is the internal standard. The peak areas are:
1-propanol: 6.550 cm^2
n-butanol: 5.321 cm^2
A 3.00 microliter unknown sample of 50% 1-propanol and an unknown mixture concentration of n-butanol is analyzed with the same gas chromatograph. The peak areas are:
1-propanol: 6.181 cm^2
n-butanol: 3.216 cm^2
Calculate the percent composition of the n-butanol in the unknown mixture. (hint, the total will not be 100%, there are other components in the sample).
2007-12-17 16:34:32 · 1 answers · asked by Lobster 4 in Science & Mathematics ➔ Chemistry
This is just a scaling problem, using your standard. Since sample sizes are the same, you can assume, for the propanol,
percentage = 6.181/6.550 x 50%
and so on.
2007-12-18 06:47:53 · answer #1 · answered by Facts Matter 7 · 0⤊ 0⤋