Question

During this distillation experiment, a student collects a 10.00 mL fraction consisting of 70% ethyl acetate and 30% n-butyl acetate (by volume). What is the mole fraction of n-butyl acetate in this sample?

Answer 1

In 10.0ml sample, we have 70% ethyl acetate by volume i.e 0.7*10ml= 7.0ml and 30% n-butyl acetate by volume i.e. 0.3*10.0ml= 3.0ml

Now, density of ethyl acetate= 0.90 g/ml

and Molecular weight of ethyl acetate C4H8O2= 4*molar mass of carbon+8*molar mass of hydrogen+2*molar mass of Oxygen

=> 4*12+8*1+2*16=> 48+8+32=> 88 g/mol

So, number of moles of ethyl acetate=> (7.0 ml*0.90g/ml)/88g/mol

=> 6.3g/88g/mol=> 0.07 moles

Density of n-butyl acetate=> 0.88 g/ml

Molecular weight of n-butyl acetate C6H12O2=> 6*molar mass of carbon+12*molar mass of hydrogen+2*molar mass of Oxygen=> 6*12+12*1+2*16=> 72+12+32=> 116 g/mol

then, number of moles of n-butyl acetate=> (3.0 ml*0.88 g/ml)/116g/mol=> 2.64g/116g/mol=> 0.02 moles

So mole fraction of n-butyl acetate=> moles of n-butyl acetate/total moles=> 0.02moles/(0.02+0.07) moles=> 0.02moles/0.09moles=> 0.22 or 22 % is mole fraction of n-butyl acetate.