by
Sandra7 » Mon Apr 11, 2011 4:16 pm
Hi Sandra
I was wondering how you were getting on with this.
How did you calculate the ECN for decanol ?
We are doing another lab experiment but the report for the GC one is due today.
I used dodecane as the internal standard to normalize peak areas. Then, for the standard, to determine whether FID responds to carbon consistently or not, I first take the ratio of normalized peak area to ppm of carbon in the six alkanes, and then the result of these ratios were checked to be within 15% of each other. It's 5.8% between the highest and the lowest ratio (of normalized peak area/ppm of C in alkane) result.
The question of statistical significance turns on the variability in the results - and not only multiple injections of the same solution (presuming here that you had all your standards and the decanol dissolved together and injected together), but also replicates of the the various weighings and dilutings that went on to prepare the solutions.
Peter
Yes, three replicate analyses were done for the standards as well as for the sample containing decanol. With the latter analyses, I were to determine the effective carbon number for 1-decanol (X1). So I set up like this:
X1/Y1 = X2/Y2 where
X1 = ECN for 1-decanol; X2 = ECN for decane
Y1 = Ratio of normalized peak area/ ppm of carbon in 1-decanol
Y2 = Ratio of normalized peak area/ ppm of carbon in decane
I got X1 to be 9.88. 9.57, and 9.98. Average is 9.81. Now I need to make a conclusion statement and I am confused about whether the answer is yes or no.
Since there are ten carbon in 1-decanol, according to the definition given by the professor in the handout (as stated in my initialpost), ECN for 1-decanol would be 10, wouldn't it? In fact, I do not understand the way EC was calculated for 1-decanol in the response by DSP007 (Decanol is C10H22O EC= 12*12/(12*12+22*1+16*1)= 0.77 versus
Decane is C10H22 EC= 12*10/(12*10+22*1)=0.83).
Why 12*12 for the carbon in decanol? Please help.