A chemist reacted 300 g of 1-butene with excess Br2 (in CCl4) at 25 8C. He isolated 418 g of
1,2-dibromobutane. What is the percent yield?

Question

A chemist reacted 300 g of 1-butene with excess Br2 (in CCl4) at 25 8C. He isolated 418 g of
1,2-dibromobutane. What is the percent yield?

matt101 · Answer

Br2 is added across the double bond in 1-butene to form a dihalide. The balanced reaction would be:

$$C_{4}H_{8}+Br_{2} \rightarrow C_{4}H_{8}Br_{2}$$
We have 300 g of 1-butene, which is 300 g / 56 g/mol = 5.36 mol. According to the reaction, this means we should also get 5.36 mol of 1,2-dibromobutane, which has a mass of 5.36 mol x 216 g/mol = 1158 g.

The formula for percent yield is:

$$percent \space yield = \frac{mass \space of \space actual \space yield}{mass \space of \space theoretical \space yield} \times 100$$
So let's just plug our numbers in to find the answer:

$$percent \space yield = \frac{418 \space g}{1158 \space g} \times 100 =36.1$$
The percent yield for this reaction is 36.1%.