In a chromatography experiment, a solution was prepared by mixing 12.98 mg of isopropanol plus 10.00 mL of an unknown containing some acetone, and diluting to 25.00 mL. Signal of 5.97 and 6.38 were observed for the acetone and isopropanol, respectively. In an earlier experiment, the Response Factor for acetone (analyte) to isopropanol (internal standard) was found to be 2.11. What is the concentration of acetone in the unknown solution?
Answer is around .576
I'm trying to use this equation: [X] = Ix/Is [S]/F but I how do I solve for [S] first?
To solve for [S], you need to use the equation given: [X] = (Ix/Is) * [S]/F.
In this equation:
[X] represents the concentration of acetone in the unknown solution.
Ix is the signal observed for acetone (5.97).
Is is the signal observed for isopropanol (6.38).
[S] is the concentration of isopropanol (internal standard) in the unknown solution.
F is the response factor (2.11).
To solve for [S], let's rearrange the equation:
[X] = (Ix/Is) * [S]/F
Multiply both sides by F:
[X] * F = (Ix/Is) * [S]
Divide both sides by (Ix/Is):
[S] = ([X] * F) / (Ix/Is)
Now you have the equation to solve for [S]. Plug in the known values:
[S] = ([X] * F) / (Ix/Is)
[S] = (0.576 * 2.11) / (5.97/6.38)
To evaluate this expression, first divide 5.97 by 6.38, then multiply the result by 0.576 and 2.11.