Question

A reaction was performed in which 1.500 g of camphor was reduced by an excess of sodium borohydride to make 1.036 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Answer: The percent yield of the reaction is 68.16 %.

Step-by-step explanation:

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

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

• For camphor:

Given mass of camphor = 1.500 g

Molar mass of camphor = 152.23 g/mol

Putting values in equation 1, we get:

`\text{Moles of camphor}=(1.500g)/(152.23g/mol)=9.85* 10^(-3)mol`

The chemical equation for the reaction of camphor and sodium borohydride follows:

`\text{Camphor}+NaBH_4\rightarrow \text{Isoborneol}`

As, sodium borohydride is present in excess. It is an excess reagent. So, camphor is the limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of camphor produces 1 mole of isoborneol

So,
`9.85* 10^(-3)mol` of camphor will produce =
`(1)/(1)* 9.85* 10^(-3)mol=9.85* 10^(-3)mol` of isoborneol

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

Molar mass of isoborneol = 154.25 g/mol

Moles of isoborneol =
`9.85* 10^(-3)` moles

Putting values in equation 1, we get:

`9.85* 10^(-3)mol=\frac{\text{Mass of isoborneol}}{154.25g/mol}\\\\\text{Mass of isoborneol}=(9.85* 10^(-3)mol* 154.25g/mol)=1.52g`

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

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of isoborneol = 1.036 g

Theoretical yield of isoborneol = 1.52 g

Putting values in above equation, we get:

`\%\text{ yield of isoborneol}=(1.036g)/(1.52g)* 100\\\\\% \text{yield of isoborneol}=68.16\%`

Hence, the percent yield of the reaction is 68.16 %.