Question

A researcher has prepared a sample of 1-Bromopropane from 10g of 1-propanol.
After purification he had made 12g of product. Which of the following is percentage
yield?

Answer 1

Answer:

The percentage yield of the reaction is 51.80 %.

Explanation:

$C_3H_7OH+HBr\rightarrow C_3H_7-Br+H_2O$

Moles of propanol = $\frac{10g}{60 g/mol}=0.1667 mol$

According to reaction, 1 mole of propanol gives 1 mole of bromo-propane. Then 0.1667 moles of propanol will give:

$\frac{1}{1}\times 1.667 mol = 0.1667 mol$

Theoretical yield of 0.1667 moles of bromo-propane:

0.1667 mol× 139 g/mol = 23.17 g

Experimental yield of bromo-propane = 12 g

Percentage yield =$\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%yield=\frac{12 g}{23.17 g}\times 100=51.80\%$

The percentage yield of the reaction is 51.80 %.

Answer 2

Answer:

The RMM of 1-Bromopropane (C₃H₇Br) is 3×12 + 7×1 + 1×79.9 = 122.9 so 12g is 12/122.9 = 0.09764 moles

The RMM of 1 -propanol (C₃H₇OH) is 3×12 + 7×1 + 1×16 + 1×1 = 60 so 10g is 10/60 = 0.1667 moles

0.09764 ÷ 0.1667 = 0.5858 = 58.58%

In other words, the percentage yield is 59%