substitution with cyanide 一問－ohe09nz81r的部落格

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substitution with cyanide 一問

發問:

In an experiment, 46.3g of 1-chlorobutane reacts with 30g of sodium cyanide to give 35.2g of pentanenitrile. Question: calculate the percentage yield of pentanenitrile 我想問點解唔可以直接 (35.2/ 46.3) x 100% 而是 no of moles of (C4H9CN/ no of moles of C4H9Cl) x 100%

最佳解答:

Why is the percentage yield of pentanenitrile = no of moles of (C4H9CN/ no of moles of C4H9Cl) x 100% ? In the reaction: CH3CH2CH2CH2Cl + NaCN → CH3CH2CH2CH2CN + NaCl Percentage yield of pentanenitrile (CH3CH2CH2CH2CN) = [(No. of moles of pentanenitrile formed)/(Max. no. of moles of pentanenitrile formed)] x 100% ...... (1) But Maximum amount of pentanenitrile formed = No. of moles 1-chlorobutane used Hence, percentage yield of pentanenitrile (CH3CH2CH2CH2CN) = [(No. of moles of pentanenitrile formed)/(No. of moles of 1-chlorobutane used) x 100% ...... (2) You had better use (1) instead of (2). This is because (Max. no. of moles of pentanenitrile formed) = (No. of moles of 1-chlorobutane used) is true only when the mole ratio of the reactant : product = 1 : 1 --------------- Calculate the percentage yield in the question. CH3CH2CH2CH2Cl + NaCN → CH3CH2CH2CH2CN + NaCl or C4H9Cl + NaCN → C5H9N + NaCl Molar mass of C4H9Cl = 12x4 + 1x9 + 35.45 = 92.45 No. of moles of C4H9Cl used = 46.3/92.45 = 0.501 mol Maximum no. of moles of C4H9Cl formed = 0.501 mol Molar mass of C5H9N = 12x5 + 1x9 + 14.01 = 83.01 Actual no. of moles of C5H9N formed = 35.2/83.01 = 0.424 mol Percentage yield of C5H9N = [(No. of moles of C5H9-N formed)/(Max. no. of moles of C5H9N formed)] x 100% = [0.424/0.501] x 100% = 84.6% --------------- Why the percentage yield of pentanenitrile is not (35.2 / 46.3) x 100% ? If you want to use the mass in calculation of the percentage yield, you can write that: Percentage yield of pentanenitrile = [(Actual mass of pentanenitrile formed)/(Maximum mass of pentanenitrile formed)] x 100% Actual mass of pentanenitrile formed is 35.2 g, but the maximum mass of pentanenitrile formed is NOT 46.3 g. Therefore, the yield is NOT (35.2 / 46.3) x 100%. Actually, the maximum mass of pentanenitrile formed is 0.501 x 83.01 = 41.6 g Therefore, percentage yield of pentanenitrile = [(Actual mass of pentanenitrile formed)/(Maximum mass of pentanenitrile formed)] x 100% = [(35.2/41.6)] x 100% = 84.6% =

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