Question

Figure out the RI mixture mole fraction of ethyl acetate and butyl acetate with these numbers: A- Pure ethyl acetate: 1.3715 B-Pure butyl acetate: 1.3945 Fraction 1: 1.3715 Fraction 2: 1.384 and fraction 3: 1.395 all at a temperature of 20 degrees.

the formula is: RI mixture= Xa (RI pure A) + Xb (RI pure B). 

Answer 1

RI mixture= Xa (RI pure A) + Xb (RI pure B)

Or

RI mixture= a (RI pure A) + b (RI pure B)/(a+b)

Where a and b are ratios of ethyl acetate and butyl acetate respectively.

Fraction 1: 1.3715

It is pure A

a= 1 , b= 0

Xa = 1/ (0+1) = 1

Xb = 0/ (0+1) = 0

Fraction 2

RI mixture = 1.384

1.384 = 1.3715a + 1.3945 b /(a+b)

Cross multiply above equation

1.384(a+b) = 1.3715a + 1.3945 b

1.384 a +1.384 b = 1.3715a + 1.3945 b

0.0125 a = 0.0114 b

a : b = 125 : 114

Mole fraction of a = a/ (a+b) = 125/239 = 0.523

Mole fraction of b = b/ (a+b) = 114/239 = 0.467

Fraction 2

RI mixture = 1.394

1.394 = 1.3715a + 1.3945 b /(a+b)

Cross multiply above equation

1.394(a+b) = 1.3715a + 1.3945 b

1.394 a +1.394 b = 1.3715a + 1.3945 b

0.0225 a = 0.0005 b

a : b = 2250 : 5

Mole fraction of a = a/ (a+b) = 2250/2255 = 0.998

Mole fraction of b = b/ (a+b) = 2250/2255 = 0.002