Question

Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction? Remember: percent yield is your (experimental yield/theoretical yield)x100.

Answer 1

Answer: The percent yield of the reaction is 61.5 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For butanol:

Given mass of butanol = 15.0 g

Molar mass of butanol = 74 g/mol

Putting values in equation 1, we get:

$\text{Moles of butanol}=\frac{15.0g}{74g/mol}=0.203mol$

• For NaBr:

Given mass of NaBr = 22.4 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{22.4g}{103g/mol}=0.217mol$

• For sulfuric acid:

Given mass of sulfuric acid = 32.7 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{32.7g}{98g/mol}=0.334mol$

For the given chemical reaction:

$C_4H_9OH+NaBr+H_2SO_4\rightarrow C_4H_9Br+NaHSO_4+H_2O$

As, the reactants are present in 1 : 1 : 1 ratio. So, the reactant having minium number of moles will be considered as the limiting reagent.

Here, the limiting reagent is butanol because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of butanol produces 1 mole of bromobutane

So, 0.203 moles of butanol will produce = $\frac{1}{1}\times 0.203=0.203mol$ of bromobutane

• Now, calculating the mass of bromobutane from equation 1, we get:

Molar mass of bromobutane = 137 g/mol

Moles of bromobutane = 0.203 moles

Putting values in equation 1, we get:

$0.203mol=\frac{\text{Mass of bromobutane}}{137g/mol}\\\\\text{Mass of bromobutane}=(0.203mol\times 137g/mol)=27.81g$

• To calculate the percentage yield of bromobutane, we use the equation:

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

Experimental yield of bromobutane = 17.1 g

Theoretical yield of bromobutane = 27.81 g

Putting values in above equation, we get:

$\%\text{ yield of bromobutane}=\frac{17.1g}{27.81g}\times 100\\\\\% \text{yield of bromobutane}=61.5\%$

Hence, the percent yield of the reaction is 61.5 %.