What is the empirical formula of a hydrocarbon if complete combustion or 2.900 mg of the hydrocarbon produced 9.803 mg of CO2 an
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Answer:
a. 0,903
b. 0,985
c. 0,996
Explanation:
Partition coefficient between ether and water is:
a. As k = 3,10, a single extraction with 300mL ether will give:
9,30 - 9,30q = q
9,30 = 10,30q
<em>q = 0,903</em>
b. Three extractions of 100 mL will extract:
First extraction:
3,10 - 3,10q = q
3,10 = 4,10q
<em>q = 0,756</em>
The second extraction will 0,756 times of the solute that is in water. This solute is 1-0,756=0,244
Second extraction will extract 0,244*0,756 = <em>0,184</em>
In the same way, third extraction will extraxt: (0,244-0,184)*0,756 = <em>0,045</em>
Total solute extracted:
<em>q = 0,756 + 0,184 + 0,045 = 0,985</em>
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c. Six extractions with 50mL of ether extract:
First extraction:
3,10 - 3,10q = 2q
3,10 = 5,10q
<em>0,608 = q</em>
Second extraction = (1-0,608)*0,608 = 0,234
Third extraction =(1-0,608-0,234)*0,608 = 0,096
Fourth extraction =(1-0,608-0,234-0,096)*0,608 = 0,038
Fifth extraction =(1-0,608-0,234-0,096-0,038)*0,608 = 0,015
Sixth extraction =(1-0,608-0,234-0,096-0,038-0,015)*0,608 = 0,005
q = 0,608+0,234+0,096+0,038+0,015+0,005 = <em>0,996</em>
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I hope it helps!