Question

Calculate the percent yield for the bromination you performed. You must use significant figures correctly. You must show all work. You must use masses in the calculation, not moles.

Reaction was the bromination of stilbenes masses: trans-stilbene: .102g / pyridinium tribromide: .200g / acetic acid:1.867 / flask and mixture: 8.599g / / mass of product: .120g

Answer 1

The reaction scheme is shown below (the scheme is taken from the internet).

Molar mass of trans-stilbene, C₁₄H₁₂ = (14*12.01 + 12*1.008) g/mol = 180.236 g/mol.

It is given that trans-stilbene is the limiting reactant. The yield of the product, meso stilbene dibromide is decided by the amount of the limiting reactant taken.

Mass of trans-stilbene taken = 0.102 g.

Mole(s) of trans-stilbene taken = (0.102 g)/(180.236 g/mol) = 5.65925*10^-4 mole.

As per the stoichiometric equation,

1 mole trans-stilbene = 1 mole meso-stilbene dibromide.

Mole(s) of meso-stilbene dibromide formed = mole(s) of trans-stilbene taken = 5.65925*10^-4 mole.

Molar mass of meso-stilbene dibromide, C₁₄H₁₂Br₂ = (14*12.01 + 12*1.008 + 2*79.904) g/mol = 340.044 g/mol.

Theoretical yield of meso-stilbene dibromide = (moles of meso-stilbene dibromide)*(molar mass of meso-stilbene dibromide) = (5.65925*10^-4 mole)*(340.044 g/mol) = 0.1914 g (ans)

Percent yield = (0.120 g)/(0.1914 g)*100 = 62.6959% ≈ 62.696% (ans).