these results it was determined that the overall exper eed and a success. uestio
ID: 1044035 • Letter: T
Question
these results it was determined that the overall exper eed and a success. uestions- 6-56. Gas chromatographic analysis of a mixture of organic compounds gave the following pesk areas (m) heane-27 heptane 1.6, hexanol 1.8, and toluene 05. (a) Calculate the mole percent composition of the mixture. Assume that the response of the detector (area per mole) is the same for each component. (b) Calculate the weight percent composition of the mixture, using the same assumptions as in lal. analysis: Ch ?Br CH,CH e-Pentyl ethyl etheExplanation / Answer
Q6.56) Area under the peak of a chromatogram is proportional to the amount of the component of the peak.
So mole fraction of a component=peak area of the component/total peak area
% mol composition=100*mole fraction
Now, total peak area=2.7+1.6+1.8+0.5=6.6
mole fraction of hexane=2.7/6.6=0.409
mole fraction of heptane=1.6/6.6=0.242
mole fraction of hexanol=1.8/6.6=0.273
mole fraction of toluene=0.5/6.6=0.0757
mol % hexane=0.409*100=40.9%
mol % heptane=0.242*100=24.2%
mol % hexanol=0.273*100=27.3%
mol % toluene=0.0757*100=7.57%
b) weight composition=mol fraction *molar mass
weight composition of hexane=mol fraction * (86.18g/mol)=0.409*(86.18g/mol)=35.247 g/mol
weight composition of heptane=mol fraction * (100.21g/mol)=0.242*(100.21g/mol)=24.251g/mol
weight composition of hexanol=mol fraction * (102.16g/mol)=0.273*(102.16 g/mol)=27.890 g/mol
weight composition of toluene=mol fraction * (92.14g/mol)=0.0757*(92.14g/mol)=6.975 g/mol
total weight=6.975 +27.890+24.251+35.247=94.363
weight %=100(weight/total weight)
weight % composition of hexane =100*(35.247/94.363)=37.3%
weight % composition of heptane =100*(24.251/94.363)=25.7%
weight % composition of hexanol =100*(27.890/94.363)=29.5%
weight % composition of toluene =100*(6.975/94.363)=7.4%
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