Question

You separated a mixture containing 20 wt% CH3OH, 35 wt% C2H5OH, and the balance water into two fractions. You then drew and analyzed samples of both product streams and reported that one stream contained 42.2% CH3OH and 29.3% C2H5OH and the other stream contained 16.5% CH3OH and 45.2% C2H5OH. After examining the report your professor indicated that the results must be wrong and asked you to repeat the stream analyses. Show why the professor is right

Answer 1

Let moles of total sample= 1

This divided into two streams. Let the two stream be S1 and S2.

S1+S2= 1

Given moles of methanol in the sample= 0.20, ethanol= 0.35 and rest water= 1-0.20-0.35=0.45

Stream-1 : methanol= S1*0.422, ethanol = S1*0.293, water = S1*(1-0.422-0.293)=S1*0.285

Stream 2 : Methanol =S2*0.165, ethanol = S2*0.452, water =S2*(1-0.165-0.452)= S2*0.383

Writing methanol balance, 0.20= S1*0.422 +S2*0.165 or 0.20/0.422= S1+S2*0.165/0.422

0.47= S1+0.39*S2.................................................................(1)

Writing ethanol balance 0.35= S1*0.293 +S2*0.452,

0.35/0.293= S1+S2*0.452/0.293

1.194= S1+S2*1.542.......................................................... (2)

Eq.2-Eq.1 gives:

S2*(1.542-0.39)= 1.194-.47 ,

S2= 0.63, S1= 1-0.63=0.37

Accordingly water in stream-1 = 0.37*0.285 = 0.105,

water in S2= 0.63*0.383= 0.241

Total water= 0.105+0.241= 0.346 which is less than the water entering the system. Hence the analysis is incorrect.