Hellmann, Zelles, Palojarvi, and Bai published a paper in 1997 about extracting lipids from organic material. They reported using a mixture that was 85% n-hexane and 15 % dichloromethane (vol/vol). What is the mole fraction of each compound in the mixture? What is the mass fraction of each compound in the mixture? Skills: unit conversions, choosing a temporary basis

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jufsanabriasa jufsanabriasa · Answer 1 · 2019-10-01T05:00:23+02:00

Answer:

Mass fraction: 73,6% n-hexane; 26,4% dichloromethane

Mole fraction: 73,0% n-hexane; 27,0% dichloromethane

Explanation:

With a basis of 100 mL:

Mass of n-hexane:

85 mL ×[tex]\frac{0,655g}{1mL}[/tex] = 55,7 g

Mass of dichloromethane

15 mL ×[tex]\frac{1,33g}{1mL}[/tex] = 20,0 g

Total mass = 20,0 g + 55,7 g = 75,7 g

Mass fraction of n-hexane:

[tex]\frac{55,7g}{75,7g}[/tex] =73,6%

Mass fraction of dichloromethane:

[tex]\frac{20,0g}{75,7g}[/tex] = 26,4%

Moles of n-hexane:

55,7 g ×[tex]\frac{1mol}{86,18 g}[/tex] = 0,65 moles

Mass of dichloromethane

20,0g ×[tex]\frac{1mol}{84,93 g}[/tex] = 0,24 moles

Total moles: 0,65 moles + 0,24 moles = 0,89 moles

Molar fraction of n-hexane:

[tex]\frac{0,65moles}{0,89moles}[/tex] =73,0%

Molar fraction of dichloromethane:

[tex]\frac{0,24moles}{0,89moles}[/tex] = 27,0%

I hope it helps!