Organic Chemistry Lab, Any help will be appreciated! I have mole ratios for cycl

Question

Organic Chemistry Lab, Any help will be appreciated!

I have mole ratios for cyclohexane : toluene for early and late fractions from both simple and fractional distillations determined by GC analysis.
1st fraction simple = 2.98
2nd fraction simple = 0.62
1st fraction fractional = 8.63
2nd fraction fractional = 0.12

So how do I use these ratios to calculate volume and mole % composition of each? Show Calculations

Compare these moles % composition results to explain whether the experimental data supports the theory?

Explanation / Answer

mole% (compound) = (mole ratio (compound)/overall mole) *100

when mole ratio is 2.98:1, overall ratio is 2.98+1 = 3.98

(1st fraction simple) mole% (Cyclohexane): (2.98/3.98)*100 = 75%

(1st fraction simple) mole% (toluene): (1/3.98)*100 = 25%

vol = (M*mole%)/(density*100)

M(Cyclohexane) = 84 g/mol, density = 0.78 g/mL

vol (1st fraction simple, cyclohxane) = (84*75)/(0.78*100) = 80.8 mL

M(toluene) = 92 g/mol, density = 0.87 g/mL

vol(1st fraction simple, toluene) = (92*25)/(0.87*100) = 26.4 mL

vol%= (vol/overall vol)*100

vol% (cyclohexane) (80.4mL/107.2mL)*100 = 75%

because M/density of both solvents are very simmilar mole% and vol% are the same in this case.

2ndfraction simple: cyclohexane: 38%, toluene: 62%

1st fraction fractional: cyclohexane: 90%, toluene: 10%

2nd fraction fractional: cyclohexane: 11%, toluene: 89%

In theory a fractional distillation of two liquids should gave a better seperation of the compounds and as you can see in the experiment it worked. :-)

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