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Sample Calculations 1. Mole Percentage of Ethanol from Weight Percentage Mol % EtOh= Wt % EtOH×Mass÷MM etOH ( Wt % EtOH×Mass÷MM etOH )−( Wt % Water× Mass÷ MM Water ) Basis of 100 g in a mixture: Mol % EtOh = 0.2468 × 100 ÷ 46 ( 0.2468 × 100 ÷ 46)−(( 1− 0.2468) × 100 ÷ 18) Mol % EtOh=8.85 % 2. Weight Fraction 2.1 From Refractive Index, n Equation obtained from calibration graph: n=−2 × 10 −10 ( w etOH ) 4 −2 × 10 −8 ( w etOH ) 3 + 2 × 10 −2 +0.0004 10 −8 ( w etOH ) +1.331 1.3373=−2 × 10 −10 ( w etOH ) 4 −2 × 10 −8 ( w etOH ) 3 + 2 × 10 −2 +0.000410 −8 ( w etOH ) + 1.331 w etOH =14.8379 % 2.2 From Density Equation obtained from calibration curve: y = -457.65x + 462.28

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Sample Calculations1. Mole Percentage of Ethanol from Weight Percentage

Basis of 100 g in a mixture:

2. Weight Fraction2.1 From Refractive Index, nEquation obtained from calibration graph:

2.2 From DensityEquation obtained from calibration curve:y = -457.65x + 462.28

3. Average Distillate Concentration, xDave Rayleigh Equation is used:

Plotting vs. xB, an equation of the line can be obtained.

This is used in the calculation of xBf. The xBi used is the first obtained bottoms concentration.A graphical method was used. Simpsons 1/3 Rule was applied.

Performing a component mass balance:

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