A student researcher performed a chromatographic seperation of caffine and aspar

Question

A student researcher performed a chromatographic seperation of caffine and aspartame with menthol as the mobile phase using C-18 column. The retention times for menthol, caffine, and aspartame were 29.7 s, 98.0 s, 140.0 s, respectively.

A.)What is the selectivity factor (relative retention) for this column?  

B.) Calculate the hypothetical retention times for caffine and aspartame if their retention factors were reduced by 28.0 %. Assume the mobile phase retention time (tm= 29.7 s) remains constant.

Explanation / Answer

A) selectivity factor = kb/ka

retention factor kb = (tb-tM ) /tM

retention factor ka = (ta-tM) /tM

selectivity factor = (tb-tM) / (ta-tM )

here tb is retention time of aspartame

ta is retention time of caffine .

selectivity factor = (140-29.7 ) /( 98-29.7)

selectivity factor = 1.615

selctivity factor = 1.615

B) given retention factor is reduced by 28%

Let the initial retention factor be k

then final retention factor kf = k - 0.28 k

final retention factor kf = 0.72k

retention factor k = (tR -tM )/tM

initial retention facotr k= ( tR1 - tM )/ tM

final retention factor kf = ( tR2 - tM ) / tM

kf/k =( tR2-tM ) / ( tR1 - tM )   

( tR2-tM ) / ( tR1 - tM ) = 0.72

tR2 -tM = 0.72 tR1 - 0.72 tM

tR2 = 0.72 tR1 + 0.28 tM

so tR2 = 0.72 tR1 + 0.28 tM

for caffeine

tR2 = 0.72 x 98 + 0.28 x 29.7

tR2 = 78.876

final retention time of caffine is 78.876 s

for aspartame

tR2 = 0.72 x 140 + 0.28 x 29.7

tR2 = 109.116

final retention time of aspartame is 109.116 s

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