Question

When 10 mL of 1-heptene was reacted with 5.1 grams of anhydrous hydrogen bromide, 8.7 grams of 2-bromoheptane was obtained after workup. Calculate the percent yield of 2-bromoheptane.

Please show all your work, thanks!

Answer 1

Solution :-

Using the volume and the density lets calculate the mass of the 1-heptene

Mass = volume * density

          = 10 ml * 0.67 g per ml = 6.7 g 1-heptene

Now lets calculate the moles of the both reactants

Moles = mass / molar mass

Moles of 1-heptene = 6.7 g / 98.186 g per mol = 0.0682 mol

Moles of HBr = 5.1 g / 80.91 g per mol = 0.063 mol HBr

Mole ratio of the both reactants with product is 1 : 1 and moles of the HBr are less therefore HBr is the limiting reactant

So moles of the product that can be formed are same as moles of HBr

So moles of the 2-bromoheptane = 0.063 moles

Now lets convert moles of the product to its mass

Mass = moles * molar mass

Mass of 2-bromoheptane = 0.063 mol * 179 g per mol

                                            = 11.3 g

Therefore theoretical yield = 11.3 g

Now lets calculate the percent yield

Percent yield = (actual yield / thereotical yield )* 100 %

                        = (8.7 g / 11.3 g)*100%

                        = 76.99 %

Therefore the percent yield = 76.99 %