upon the bromination of an alkene to create stillbene dibromide 2.89g of the pro

Question

upon the bromination of an alkene to create stillbene dibromide 2.89g of the product was collected and has a melting point of 209c(once it started melting) and 211c (when the product finsihed melting.

? calculate the percent yield.

below is the experimental procedure and all the information that i have.

Place 2.0 g of (E)-stilbene, boiling chips, and 40 mL of ethanol in a 100 mL round-bottom flask. Heat the flask to dissolve the stilbene. Add 4.0 g of pyridinium tribromide (if solid material adheres to the interior walls of the flask, use few milliliters of ethanol to rinse it down). Continue to heat for additional 5 minutes after the addition of reagents is complete, and then remove the flask from the hot plate. The product dibromide should quickly begin to precipitate or crystallize. Let the reaction cool to room temperature, and then chill the mixture in an ice bath. Collect the product by vacuum filtration. Set aside a small amount of this

Explanation / Answer

Easy. All you have to do is calculate the moles of Stillbene and the moles of stillbene dibromide:

Molar Mass of Stillbene: 180.25 g/mol

Molar Mass of Stillbene Dibromide: 337.9 g/mol

Moles of Stillbene = 2 / 180.25 = 0.011 moles

Moles of Pyridinum Tribromide: 4 / 305 = 0.013 moles

It's an 1:1 reaction, so the limitant reactant would be the stillbene. Therefore:

moles of Stillbene = moles of Stillbene dibromide = 0.011 moles

mass of dibromide stillbene = 0.011 x 337.9 = 3.7158 g

% = (2.89 / 3.7158) x 100 = 77.78%

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