It is sometimes advantageous to analyze a hydrocarbon mixture quantitatively by
ID: 1019670 • Letter: I
Question
It is sometimes advantageous to analyze a hydrocarbon mixture quantitatively by oxidizing the column effluent prior to detection. The water vapor formed, can be removed by a trap, and the CO2 admitted to the TCD detector. In a particular experiment, seven components gave peaks with integrated areas as follows:
Peak # Compound Relative area
1 n-pentane 2.00
2 n-hexane 5.72
3 3 -methylhexane 2.21
4 n-heptane 8.15
5 2,2, 4-trimethylpentane 1.92
6 toluene 3.16
7 n-octane 5.05
a) List advantages and disadvantages of this process.
b) Should the oxidation step precede or follow passage through the column?
Compute the composition of the sample, giving rise to the sample cited in terms of the mole % of total hydrocarbons.
Explanation / Answer
a) Advantages= The amount of the hydrocarbon is exactly proportional to the amount of carbon dioxide + water produced by it on oxidation.Hence ,it gives an appropriate approximation of the mol % of hydrocarbon by comparing the parameter of area under the peak for each hydrocarbon.
Disadvantages=While oxidation ,the hydrocarbons needs to be burnt to give out CO2 + H2O ,so some amount of carbon in the hydrocarbon may be lost due to conversion to CO (monoxide) .
Also, combustion/oxidation cannot be 100% efficient, leading to inappropriate CO2 amount in detection.
b) Area of peak is proportional to the mole % of hydrocarbon
% composition of a hydrocarbon= Area of peak /total area*100%
n-pentane=2.00/(2.00+5.72+2.21+8.15+1.92+3.16+5.05)*100%=2.00/28.21*100=7.09 mol %
n-hexane=5.72/28.21*100%=20.8%
3 -methylhexane 2.21/28.21*100=7.83%
4 n-heptane 8.15/28.21*100=28.9 %
5 2,2, 4-trimethylpentane 1.92/28.21*100=6.8 mol%
6 toluene 3.16/28.21*100=11.2%
7 n-octane 5.05/28.21*100=17.9%
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.