1. The procedure in the manual calls for 1 (molar) equivalent of dibenzyl ketone

Question

1. The procedure in the manual calls for 1 (molar) equivalent of dibenzyl ketone relative to the amount of benzil that you made in step 1 (i.e. 1 mole of dibenzyl ketone for every 1 mole of benzil). If you synthesized 0.22 g of benzil in the first step, how many grams of dibenzyl ketone will you need in the second step? Also, calculate how many drops of the potassium hydroxide solution (0.1 g/mL potassium hydroxide in absolute ethanol; 70 drops/mL) would be needed to add 0.5 equivalents of ethanolic potassium hydroxide to the reaction mixture.

2. Calculate the required amount of 1.25 equivalents of anthranilic acid and 1.5 equivalents of isopentyl nitrite if the experiment starts with 100 mg of tetraphenylcyclopentadienone. How many drops of isopentyl nitrite (density, 0.872 g/mL at 25C, 75 drops/mL) should be dissolved in DME?

Explanation / Answer

1. the ratio of moles of benzil to dibenzylketone is 1 : 1

moles of benzil = 0.22 g/210.23 g/mol = 0.001 mols

So moles of dibenzylketone = 0.001 mols

mass of dibenzylketone required = 0.001 mols x 210.28 g/mol = 0.22 g

0.5 equivalent of ethanolic KOH = 0.001/2 = 0.0005 mols

KOH needed = 0.0005 mol x 56.1056 g/mol =0.028 g

Volume of KOH needed = 0.028 g/0.1 g/ml = 0.28 ml

1 drop = 0.05 ml

70 drops/ml ethanolic solution has = 70 x 0.05 x 0.1 = 0.35 g of KOH

So we would need = 0.028 x 70/0.35 = 5.6 drops of ethanolic KOH solution

2. We have,

moles of tetraphenylcyclopentadieneone = 0.1 g/384.48 g/mol = 0.00026 mols

anthranilic acid = 0.00026 mols x 1.25 eq. x 137.136 g/mol = 0.044 g

isopentylnitrite = 0.00026 mols x 1.5 eq x 117.15 g/mol = 0.046 g

Volume of isopentyl nitrite needed = 0.046/0.872 = 0.0524 ml = 1.05 drops to be dissolved in DME

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