3 pt Question 17 The response factor between an alcohol and an alkane is 2.54. A

Question

3 pt Question 17 The response factor between an alcohol and an alkane is 2.54. An internal standard with a concentration of 25.9 nM has a peak area of 92832. The peak area for the alcohol is 23422. What is the concentration (nM) of alcohol in the sample? Hint: Realize that the response factor is multiplied by the ratio of area to concentration of the internal standard. Plug the values into the equation and solve for [X). Question 18 3 pts To quantify the amount of ethanol in an organic sample, a technician prepares a standard solution containing 0.1473 grams ethanol and 0.1847 grams decane in 1.203 grams dichlormethane. GC analysis returns peak areas of 9,930 for ethanol and 7,893 for decane. Next, 0.9831 grams unknown and 0.2932 grams decane are added to 1.2119 grams dichloromethane. GC analysis returns peak areas of 6,382 for ethanol and 11,283 for decane. What is the percent by mass ethanol in the sample? Hint: Use the mass ethanol and decane in the known sample as well as the peak areas to calculate the response factor F. Next, use the peak area for ethanol and decane as well as the mass decane added n. Divide the mass ethanol by the mass unknown and multiply by 100 to get the percent by mass ethanol. enoVO

Explanation / Answer

I am solving the first question (Question 17) as per Chegg guidelines, post multiple question to get the remaining answers

Q17)

Response Factor = Peak Area/Concentration

In case of internal standard

Concentration A = Peak area A / Peak area B x   1 / RRF x   Concentration B

(where B is an internal standard)

Concentration A = 23422/92832 * 1/2.54 * 25.9 = 2.57 nM

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