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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Da4gotten Da4gotten
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8 years ago Edited: 8 years ago, Da4gotten
The percent by mass octane is quantified for an organic sample via GC using decane as the
internal standard. A GC vial containing 0.5023 grams organic sample and 0.2384 grams
decane is prepared and analyzed, returning peak areas of 119,993 for octane and 83755 for
decane. If a response factor of 2.50 from previous work analyzing octane with a decane
internal standard is used, what is the percent by mass octane in the organic sample?

Please, I need help.
Post Merge: 8 years ago

using formula: area of analyte signal/mass of analyte = F(area of internal standard signal/mass of internal standard signal)

(11,993/X)=(2.50)(83755/0.2384g) = 0.1366 g octane /0.5023g organic sample x 100 = 27%

Is this correct?
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