Question

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

Answer 1

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

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

Step-by-step explanation:

With a basis of 100 mL:

Mass of n-hexane:

85 mL ×
`(0,655g)/(1mL)` = 55,7 g

Mass of dichloromethane

15 mL ×
`(1,33g)/(1mL)` = 20,0 g

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

Mass fraction of n-hexane:

`(55,7g)/(75,7g)` =73,6%

Mass fraction of dichloromethane:

`(20,0g)/(75,7g)` = 26,4%

Moles of n-hexane:

55,7 g ×
`(1mol)/(86,18 g)` = 0,65 moles

Mass of dichloromethane

20,0g ×
`(1mol)/(84,93 g)` = 0,24 moles

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

Molar fraction of n-hexane:

`(0,65moles)/(0,89moles)` =73,0%

Molar fraction of dichloromethane:

`(0,24moles)/(0,89moles)` = 27,0%

I hope it helps!