Question

In performing the Grignard reaction under the conditions of Experiment 16, if you were to use 121 mg of benzophenone (182.21 g/mol), 18 mg of magnesium (24.30 g/mol), and 76 microliters of bromobenzene (157.02 g/mol, 1.50 g/mL), calculate the theoretical yield of triphenylmethanol (260.33 g/mol), in mg. Benzophenone is the limiting reactant. Enter your answer as digits only, no units, using the proper number of significant figures.

Answer 1

Answer: The theoretical yield of triphenylmethanol is 0.173 grams

Explanation:

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

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

Given mass of benzophenone = 121 mg = 0.121 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of benzophenone = 182.21 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzophenone}=\frac{0.121g}{182.21g/mol}=6.64\times 10^{-4}mol$

The chemical equation for the formation of triphenylmethanol from benzophenone follows:

$\text{Benzophenone}+Mg+\text{Bromobenzene}\rightarrow \text{Triphenylmethanol}+MgBr$

As, bromobenzene is the limiting reagent. So, it will limit the formation of products

By Stoichiometry of the reaction:

1 mole of bromobenzene produces 1 mole of triphenylmethanol

So, $6.64\times 10^{-4}mol$ of bromobenzene will produce = $\frac{1}{1}\times 6.64\times 10^{-4}=6.64\times 10^{-4}$ moles of triphenylmethanol

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

Molar mass of triphenylmethanol = 260.33 g/mol

Moles of triphenylmethanol = $6.64\times 10^{-4}$ moles

Putting values in equation 1, we get:

$6.64\times 10^{-4}mol=\frac{\text{Mass of triphenylmethanol}}{260.33g/mol}\\\\\text{Mass of triphenylmethanol}=(6.64\times 10^{-4}mol\times 260.33g/mol)=0.173g$

Hence, the theoretical yield of triphenylmethanol is 0.173 grams