To a sample of a protein hydroysate, an analyst added 1.00 mg of tryptophan, which was labeled with 14C and exhibited a counting rate of 584 cpm above background. The labeled compound was thoroughly mixed with the sample and the mixture was passed through an ion-exchange column. The fraction effluent contaning only tryptophan was collected, and from it an 18.0 mg sample of pure tryptophan was isolated. The isolated sample had a counting rate of 204 cpm in the same counter. What was the weight of tryptophan in the original sample? A. 6.28 mg B. 20.8 mg C. 50.5 mg D. 58.4 mg Help me! Why is C correct. Please explain! Thank you.

Appl Of Ms Excel In Analytical Chemistry
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Chapter12: Spectrochemical Methods
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To a sample of a protein hydroysate, an analyst added 1.00 mg of tryptophan, which was labeled with 14C and exhibited
a counting rate of 584 cpm above background. The labeled compound was thoroughly mixed with the sample and the
mixture was passed through an ion-exchange column. The fraction effluent contaning only tryptophan was collected,
and from it an 18.0 mg sample of pure tryptophan was isolated. The isolated sample had a counting rate of 204 cpm in
the same counter. What was the weight of tryptophan in the original sample?
A. 6.28 mg
B. 20.8 mg
C. 50.5 mg
D. 58.4 mg
Help me! Why is C correct. Please explain! Thank you.
Transcribed Image Text:To a sample of a protein hydroysate, an analyst added 1.00 mg of tryptophan, which was labeled with 14C and exhibited a counting rate of 584 cpm above background. The labeled compound was thoroughly mixed with the sample and the mixture was passed through an ion-exchange column. The fraction effluent contaning only tryptophan was collected, and from it an 18.0 mg sample of pure tryptophan was isolated. The isolated sample had a counting rate of 204 cpm in the same counter. What was the weight of tryptophan in the original sample? A. 6.28 mg B. 20.8 mg C. 50.5 mg D. 58.4 mg Help me! Why is C correct. Please explain! Thank you.
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