1) In a GC-separation of a mixture of three compounds consisting of 1 mol A, 1 m

Question

1) In a GC-separation of a mixture of three compounds consisting of 1 mol A, 1 mol B, and 1 mol C, the peak area (integral) produced by the response of the detector is: Compound A: 50 Compound B: 20 Compound C: 10 The separation of a second mix of the same compounds with an unknown composition leads to the following peak areas: Compound A: 20 Compound B: 400 Compound C: 10 Please calculate the response ratio of your compounds if you choose compound C as your standard. What is the molar composition of the second mixture (show your calculations)?

Explanation / Answer

Ratio of compound versus peak area is 1 molA : 1 molB : 1 molC :: 50 : 20 : 10

As C is taken as standard compound, moles of C in both mixtures is same as peak area is same. Therefore

x molA : y mol B : 1 mol C :: 20 : 400 : 10

Comparing the ration of concentration versus peak area in both mixtures, moles of A, x = 1*20/50 = 0.4 mol

moles of B, y = 1*400/20 = 20 mol

So, the molar composition of second mixture is 0.4 mol A, 20 mol of B and 1 mol C

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.